Explanation:
Okay, so I need to find the remainder when this big polynomial f(x) is divided by x² - 1. The problem says not to use long division, which makes sense because that would be really tedious with such a high-degree polynomial. Hmm, I remember that when you divide a polynomial by a quadratic like x² - 1, the remainder should be a linear polynomial, right? So the remainder will be of the form ax + b. My job is to find the coefficients a and b.

Now, how do I find a and b without long division? I think there's something about the Remainder Theorem or maybe using roots of the divisor. Since x² - 1 factors into (x - 1)(x + 1), maybe I can use the fact that if I plug in x = 1 and x = -1 into f(x), those should equal the remainder evaluated at those points. Wait, let me recall: if you divide f(x) by (x - c), the remainder is f(c). But here, the divisor is quadratic, so maybe I need to use both roots?

Yes, I think that's the way. If I have a remainder of ax + b, then when I plug in x = 1, f(1) should equal a(1) + b, and when I plug in x = -1, f(-1) should equal a(-1) + b. Then I can set up two equations and solve for a and b. That seems manageable. Let me try that.

First, let me compute f(1). Let's substitute x = 1 into the polynomial:

f(1) = (1)^10 + 5*(1)^9 - 8*(1)^8 + 7*(1)^7 - (1)^6 - 12*(1)^5 + 4*(1)^4 - 8*(1)^3 + 12*(1)^2 - 5*(1) - 5.

Calculating each term:

1^10 = 1

5*1^9 = 5*1 = 5

-8*1^8 = -8*1 = -8

7*1^7 = 7*1 = 7

-1^6 = -1

-12*1^5 = -12*1 = -12

4*1^4 = 4*1 = 4

-8*1^3 = -8*1 = -8

12*1^2 = 12*1 = 12

-5*1 = -5

-5 remains as is.

Now add all these up:

1 + 5 = 6

6 - 8 = -2

-2 + 7 = 5

5 - 1 = 4

4 - 12 = -8

-8 + 4 = -4

-4 - 8 = -12

-12 + 12 = 0

0 - 5 = -5

-5 - 5 = -10

So f(1) = -10. Therefore, according to the remainder, when x = 1, ax + b = a(1) + b = a + b = -10. So equation one: a + b = -10.

Now let's compute f(-1). Substitute x = -1 into f(x):

f(-1) = (-1)^10 + 5*(-1)^9 - 8*(-1)^8 + 7*(-1)^7 - (-1)^6 - 12*(-1)^5 + 4*(-1)^4 - 8*(-1)^3 + 12*(-1)^2 - 5*(-1) - 5.

Let me compute each term step by step:

(-1)^10 = 1 (since even exponent)

5*(-1)^9 = 5*(-1) = -5

-8*(-1)^8 = -8*(1) = -8

7*(-1)^7 = 7*(-1) = -7

-(-1)^6 = -1 (since (-1)^6 is 1, so negative of that is -1)

-12*(-1)^5 = -12*(-1) = 12

4*(-1)^4 = 4*(1) = 4

-8*(-1)^3 = -8*(-1) = 8

12*(-1)^2 = 12*(1) = 12

-5*(-1) = 5

-5 remains as is.

Now let's add them up:

Start with 1.

1 - 5 = -4

-4 - 8 = -12

-12 - 7 = -19

-19 - 1 = -20

-20 + 12 = -8

-8 + 4 = -4

-4 + 8 = 4

4 + 12 = 16

16 + 5 = 21

21 - 5 = 16

So f(-1) = 16. Therefore, according to the remainder, when x = -1, ax + b = a*(-1) + b = -a + b = 16. So equation two: -a + b = 16.

Now we have two equations:

1) a + b = -10

2) -a + b = 16

Let me solve these equations. Let's subtract equation 1 from equation 2? Wait, maybe add them? Let me see:

If I subtract equation 1 from equation 2:

(-a + b) - (a + b) = 16 - (-10)

Left side: -a + b - a - b = -2a

Right side: 16 + 10 = 26

So -2a = 26 => a = -13

Then plug back into equation 1: -13 + b = -10 => b = -10 + 13 = 3

So the remainder is ax + b = -13x + 3. Let me check if this makes sense.

Wait, let me verify my calculations for f(1) and f(-1) again because it's easy to make a mistake with all those signs.

Starting with f(1):

Original polynomial:

x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

At x=1, all terms become 1, except the coefficients:

So:

1 +5 -8 +7 -1 -12 +4 -8 +12 -5 -5

Let me compute step by step:

Start with 1 +5 = 6

6 -8 = -2

-2 +7 = 5

5 -1 = 4

4 -12 = -8

-8 +4 = -4

-4 -8 = -12

-12 +12 = 0

0 -5 = -5

-5 -5 = -10. Okay, that's correct. So f(1) = -10.

Now f(-1):

Original polynomial:

x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

At x=-1:

(-1)^10 = 1

5*(-1)^9 = 5*(-1) = -5

-8*(-1)^8 = -8*(1) = -8

7*(-1)^7 = 7*(-1) = -7

-(-1)^6 = -1

-12*(-1)^5 = -12*(-1) = 12

4*(-1)^4 = 4*(1) = 4

-8*(-1)^3 = -8*(-1) = 8

12*(-1)^2 = 12*(1) = 12

-5*(-1) = 5

-5 remains.

So adding them up:

1 -5 -8 -7 -1 +12 +4 +8 +12 +5 -5

Let me compute step by step:

Start with 1.

1 -5 = -4

-4 -8 = -12

-12 -7 = -19

-19 -1 = -20

-20 +12 = -8

-8 +4 = -4

-4 +8 = 4

4 +12 = 16

16 +5 = 21

21 -5 = 16. Okay, that's correct. So f(-1) = 16.

So equations:

a + b = -10

- a + b = 16

Subtracting first equation from second:

(-a + b) - (a + b) = 16 - (-10)

Which is -2a = 26 => a = -13. Then b = -10 - a = -10 - (-13) = 3. So remainder is -13x + 3. Let me check if that makes sense.

Alternatively, maybe I can use another method to verify. For example, since x² - 1 divides into f(x) with remainder ax + b, then f(x) = (x² - 1)Q(x) + ax + b. Therefore, if I can express f(x) in terms of x² - 1, but maybe that's similar to long division. Alternatively, maybe I can use substitution.

Alternatively, since x² ≡ 1 mod (x² - 1), so any higher power of x can be reduced by replacing x² with 1. For example, x³ = x*x² ≡ x*1 = x, x^4 = x²*x² ≡ 1*1 = 1, x^5 = x^4*x ≡ 1*x = x, and so on. So in general, even exponents can be reduced to 1, odd exponents to x. Therefore, maybe I can rewrite f(x) by replacing x² with 1, x³ with x, x^4 with 1, etc., and then simplify.

Let me try that approach. Let me write f(x) as:

x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

Now, replace x² with 1, so:

For each term:

x^10: since x² = 1, x^10 = (x²)^5 = 1^5 = 1

Similarly, x^9 = x^(2*4 + 1) = (x²)^4 * x = 1^4 * x = x

x^8 = (x²)^4 = 1^4 = 1

x^7 = x^(2*3 + 1) = (x²)^3 * x = 1^3 * x = x

x^6 = (x²)^3 = 1^3 = 1

x^5 = x^(2*2 + 1) = (x²)^2 * x = 1^2 * x = x

x^4 = (x²)^2 = 1^2 = 1

x^3 = x^(2*1 + 1) = (x²)^1 * x = 1 * x = x

x^2 = 1

So substituting all these into f(x):

1 + 5x -8*1 +7x -1 -12x +4*1 -8x +12*1 -5x -5

Now let me compute each term:

1 (from x^10)

+5x (from 5x^9)

-8*1 = -8 (from -8x^8)

+7x (from 7x^7)

-1 (from -x^6)

-12x (from -12x^5)

+4*1 = +4 (from 4x^4)

-8x (from -8x^3)

+12*1 = +12 (from 12x^2)

-5x (from -5x)

-5 (constant term)

Now combine like terms:

Constants: 1 -8 -1 +4 +12 -5

Let me compute that:

Start with 1 -8 = -7

-7 -1 = -8

-8 +4 = -4

-4 +12 = 8

8 -5 = 3

So constants sum to 3.

Now the x terms: 5x +7x -12x -8x -5x

Compute coefficients:

5 +7 = 12; 12 -12 = 0; 0 -8 = -8; -8 -5 = -13. So total x term is -13x.

Therefore, the entire expression simplifies to -13x + 3, which matches the earlier result. So that confirms that the remainder is indeed -13x + 3. Therefore, the answer should be -13x + 3. Wait, but let me check if the problem wants it in a specific form? The question says "find the remainder", so as a polynomial, which is linear, so -13x + 3 is correct. But maybe they want it written as 3 -13x? Either way, both are correct. But let me check if I made any mistake in substitution.

Wait, let me check again the substitution steps. For example, x^10: since x²=1, x^10 = (x²)^5 = 1^5 = 1. Correct. Similarly, x^9 = x^(2*4 + 1) = (x²)^4 * x = 1^4 * x = x. Correct. Similarly for others. Then substituting all terms:

Original f(x):

x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

After substitution:

1 +5x -8 +7x -1 -12x +4 -8x +12 -5x -5

Yes, that's correct. Then constants: 1 -8 -1 +4 +12 -5. Let me check again:

1 -8 is -7, -7 -1 is -8, -8 +4 is -4, -4 +12 is 8, 8 -5 is 3. Correct. Then x terms: 5x +7x is 12x, minus 12x is 0, minus 8x is -8x, minus 5x is -13x. Correct. So yes, remainder is -13x + 3. Therefore, the answer is -13x + 3.

But let me check with another method just to be thorough. For example, suppose I use the fact that if f(x) divided by x² -1 gives remainder ax + b, then f(x) - (ax + b) is divisible by x² -1. Therefore, f(x) - (ax + b) has roots at x = 1 and x = -1. Which is exactly what I used earlier with f(1) = a + b and f(-1) = -a + b. So that's consistent. Therefore, solving gives a = -13, b = 3. So that's correct.

Alternatively, maybe I can use Taylor series expansion or something? But that might be more complicated. Alternatively, think of f(x) = (x² -1)Q(x) + ax + b. Then, if I take derivative? Not sure. Maybe not necessary.

Alternatively, think of f(x) evaluated at x = 1 and x = -1, which we did. So seems solid.

Therefore, I think the remainder is -13x + 3. Let me check once more with another approach. Suppose I write f(x) as:

f(x) = x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

Let me group terms in pairs or something. But maybe not helpful. Alternatively, note that x² = 1, so x^even = 1, x^odd = x. So as before, which we already did. So that gives the same result. So I think confident that the remainder is -13x + 3.

But just to make sure, let me plug in x = 1 into the remainder: -13(1) + 3 = -10, which matches f(1) = -10. Similarly, x = -1: -13(-1) + 3 = 13 + 3 = 16, which matches f(-1) = 16. Perfect. So that's correct.

Therefore, the remainder is -13x + 3. So the answer is \boxed{-13x + 3}?

Wait, but sometimes problems expect the answer in a certain format. Let me check if the question specifies. The problem says "find the remainder when f(x) is divided by x² -1". The remainder is a linear polynomial, so -13x + 3 is correct. However, sometimes people prefer writing it with the x term first, so maybe -13x + 3 or 3 -13x. But both are equivalent. However, in standard form, it's usually written with descending degree, so -13x + 3. But maybe the problem expects it in a specific way. Let me check if there's any convention. In most cases, the answer would be written as ax + b, so -13x + 3. So I think that's acceptable.

Alternatively, maybe they want it as 3 -13x? But both are correct. However, since the question didn't specify, either is fine. But let me check if my calculation of a and b is correct. From the equations:

a + b = -10

- a + b = 16

Subtracting first equation from second: (-a + b) - (a + b) = 16 - (-10) => -2a = 26 => a = -13. Then b = -10 - a = -10 - (-13) = 3. So yes, correct. So the remainder is -13x + 3. So I think that's the answer.

**Final Answer**
The remainder is \boxed{-13x + 3}.

To find the remainder when the polynomial
$$
f(x) = x^{10} + 5x^9 - 8x^8 + 7x^7 - x^6 - 12x^5 + 4x^4 - 8x^3 + 12x^2 - 5x - 5
$$
is divided by $ x^2 - 1 $, we can use the **Remainder Theorem** for polynomials divided by a quadratic divisor.

---

### Step 1: Understand the Structure of the Remainder

When dividing a polynomial $ f(x) $ by a quadratic polynomial $ x^2 - 1 $, the **remainder** will be a **linear polynomial** of the form:
$$
R(x) = ax + b
$$
This is because the degree of the remainder must be less than the degree of the divisor, which is 2.

---

### Step 2: Use the Roots of the Divisor

The roots of $ x^2 - 1 = 0 $ are $ x = 1 $ and $ x = -1 $. By the **Remainder Theorem**, we know:
$$
f(1) = R(1) = a(1) + b = a + b
$$
$$
f(-1) = R(-1) = a(-1) + b = -a + b
$$

So we can set up the following system of equations:
$$
\begin{cases}
a + b = f(1) \\
-a + b = f(-1)
\end{cases}
$$

---

### Step 3: Evaluate $ f(1) $ and $ f(-1) $

#### Compute $ f(1) $:
$$
f(1) = 1^{10} + 5 \cdot 1^9 - 8 \cdot 1^8 + 7 \cdot 1^7 - 1^6 - 12 \cdot 1^5 + 4 \cdot 1^4 - 8 \cdot 1^3 + 12 \cdot 1^2 - 5 \cdot 1 - 5
$$
$$
= 1 + 5 - 8 + 7 - 1 - 12 + 4 - 8 + 12 - 5 - 5 = -10
$$

#### Compute $ f(-1) $:
$$
f(-1) = (-1)^{10} + 5 \cdot (-1)^9 - 8 \cdot (-1)^8 + 7 \cdot (-1)^7 - (-1)^6 - 12 \cdot (-1)^5 + 4 \cdot (-1)^4 - 8 \cdot (-1)^3 + 12 \cdot (-1)^2 - 5 \cdot (-1) - 5
$$
$$
= 1 - 5 - 8 - 7 - 1 + 12 + 4 + 8 + 12 + 5 - 5 = 16
$$

---

### Step 4: Solve the System of Equations

We now solve:
$$
\begin{cases}
a + b = -10 \\
-a + b = 16
\end{cases}
$$

Subtract the first equation from the second:
$$
(-a + b) - (a + b) = 16 - (-10) \Rightarrow -2a = 26 \Rightarrow a = -13
$$

Substitute $ a = -13 $ into the first equation:
$$
-13 + b = -10 \Rightarrow b = 3
$$

---

### Step 5: Write the Remainder

The remainder is:
$$
R(x) = ax + b = -13x + 3
$$

---

### Final Answer

$$
\boxed{-13x + 3}
$$

[Assistant]
Explanation:
Okay, so I need to find the remainder when this big polynomial f(x) is divided by x² - 1. The problem says not to use long division, which makes sense because that would be really tedious with such a high-degree polynomial. Hmm, I remember that when you divide a polynomial by a quadratic like x² - 1, the remainder should be a linear polynomial, right? So the remainder will be of the form ax + b. My job is to find the coefficients a and b.

Now, how do I find a and b without long division? I think there's something about the Remainder Theorem or maybe using roots of the divisor. Since x² - 1 factors into (x - 1)(x + 1), maybe I can use the fact that if I plug in x = 1 and x = -1 into f(x), those should equal the remainder evaluated at those points. Wait, let me recall: if you divide f(x) by (x - c), the remainder is f(c). But here, the divisor is quadratic, so maybe I need to use both roots?

Yes, I think that's the way. If I have a remainder of ax + b, then when I plug in x = 1, f(1) should equal a(1) + b, and when I plug in x = -1, f(-1) should equal a(-1) + b. Then I can set up two equations and solve for a and b. That seems manageable. Let me try that.

First, let me compute f(1). Let's substitute x = 1 into the polynomial:

f(1) = (1)^10 + 5*(1)^9 - 8*(1)^8 + 7*(1)^7 - (1)^6 - 12*(1)^5 + 4*(1)^4 - 8*(1)^3 + 12*(1)^2 - 5*(1) - 5.

Calculating each term:

1^10 = 1

5*1^9 = 5*1 = 5

-8*1^8 = -8*1 = -8

7*1^7 = 7*1 = 7

-1^6 = -1

-12*1^5 = -12*1 = -12

4*1^4 = 4*1 = 4

-8*1^3 = -8*1 = -8

12*1^2 = 12*1 = 12

-5*1 = -5

-5 remains as is.

Now add all these up:

1 + 5 = 6

6 - 8 = -2

-2 + 7 = 5

5 - 1 = 4

4 - 12 = -8

-8 + 4 = -4

-4 - 8 = -12

-12 + 12 = 0

0 - 5 = -5

-5 - 5 = -10

So f(1) = -10. Therefore, according to the remainder, when x = 1, ax + b = a(1) + b = a + b = -10. So equation one: a + b = -10.

Now let's compute f(-1). Substitute x = -1 into f(x):

f(-1) = (-1)^10 + 5*(-1)^9 - 8*(-1)^8 + 7*(-1)^7 - (-1)^6 - 12*(-1)^5 + 4*(-1)^4 - 8*(-1)^3 + 12*(-1)^2 - 5*(-1) - 5.

Let me compute each term step by step:

(-1)^10 = 1 (since even exponent)

5*(-1)^9 = 5*(-1) = -5

-8*(-1)^8 = -8*(1) = -8

7*(-1)^7 = 7*(-1) = -7

-(-1)^6 = -1 (since (-1)^6 is 1, so negative of that is -1)

-12*(-1)^5 = -12*(-1) = 12

4*(-1)^4 = 4*(1) = 4

-8*(-1)^3 = -8*(-1) = 8

12*(-1)^2 = 12*(1) = 12

-5*(-1) = 5

-5 remains as is.

Now let's add them up:

Start with 1.

1 - 5 = -4

-4 - 8 = -12

-12 - 7 = -19

-19 - 1 = -20

-20 + 12 = -8

-8 + 4 = -4

-4 + 8 = 4

4 + 12 = 16

16 + 5 = 21

21 - 5 = 16

So f(-1) = 16. Therefore, according to the remainder, when x = -1, ax + b = a*(-1) + b = -a + b = 16. So equation two: -a + b = 16.

Now we have two equations:

1) a + b = -10

2) -a + b = 16

Let me solve these equations. Let's subtract equation 1 from equation 2? Wait, maybe add them? Let me see:

If I subtract equation 1 from equation 2:

(-a + b) - (a + b) = 16 - (-10)

Left side: -a + b - a - b = -2a

Right side: 16 + 10 = 26

So -2a = 26 => a = -13

Then plug back into equation 1: -13 + b = -10 => b = -10 + 13 = 3

So the remainder is ax + b = -13x + 3. Let me check if this makes sense.

Wait, let me verify my calculations for f(1) and f(-1) again because it's easy to make a mistake with all those signs.

Starting with f(1):

Original polynomial:

x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

At x=1, all terms become 1, except the coefficients:

So:

1 +5 -8 +7 -1 -12 +4 -8 +12 -5 -5

Let me compute step by step:

Start with 1 +5 = 6

6 -8 = -2

-2 +7 = 5

5 -1 = 4

4 -12 = -8

-8 +4 = -4

-4 -8 = -12

-12 +12 = 0

0 -5 = -5

-5 -5 = -10. Okay, that's correct. So f(1) = -10.

Now f(-1):

Original polynomial:

x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

At x=-1:

(-1)^10 = 1

5*(-1)^9 = 5*(-1) = -5

-8*(-1)^8 = -8*(1) = -8

7*(-1)^7 = 7*(-1) = -7

-(-1)^6 = -1

-12*(-1)^5 = -12*(-1) = 12

4*(-1)^4 = 4*(1) = 4

-8*(-1)^3 = -8*(-1) = 8

12*(-1)^2 = 12*(1) = 12

-5*(-1) = 5

-5 remains.

So adding them up:

1 -5 -8 -7 -1 +12 +4 +8 +12 +5 -5

Let me compute step by step:

Start with 1.

1 -5 = -4

-4 -8 = -12

-12 -7 = -19

-19 -1 = -20

-20 +12 = -8

-8 +4 = -4

-4 +8 = 4

4 +12 = 16

16 +5 = 21

21 -5 = 16. Okay, that's correct. So f(-1) = 16.

So equations:

a + b = -10

- a + b = 16

Subtracting first equation from second:

(-a + b) - (a + b) = 16 - (-10)

Which is -2a = 26 => a = -13. Then b = -10 - a = -10 - (-13) = 3. So remainder is -13x + 3. Let me check if that makes sense.

Alternatively, maybe I can use another method to verify. For example, since x² - 1 divides into f(x) with remainder ax + b, then f(x) = (x² - 1)Q(x) + ax + b. Therefore, if I can express f(x) in terms of x² - 1, but maybe that's similar to long division. Alternatively, maybe I can use substitution.

Alternatively, since x² ≡ 1 mod (x² - 1), so any higher power of x can be reduced by replacing x² with 1. For example, x³ = x*x² ≡ x*1 = x, x^4 = x²*x² ≡ 1*1 = 1, x^5 = x^4*x ≡ 1*x = x, and so on. So in general, even exponents can be reduced to 1, odd exponents to x. Therefore, maybe I can rewrite f(x) by replacing x² with 1, x³ with x, x^4 with 1, etc., and then simplify.

Let me try that approach. Let me write f(x) as:

x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

Now, replace x² with 1, so:

For each term:

x^10: since x² = 1, x^10 = (x²)^5 = 1^5 = 1

Similarly, x^9 = x^(2*4 + 1) = (x²)^4 * x = 1^4 * x = x

x^8 = (x²)^4 = 1^4 = 1

x^7 = x^(2*3 + 1) = (x²)^3 * x = 1^3 * x = x

x^6 = (x²)^3 = 1^3 = 1

x^5 = x^(2*2 + 1) = (x²)^2 * x = 1^2 * x = x

x^4 = (x²)^2 = 1^2 = 1

x^3 = x^(2*1 + 1) = (x²)^1 * x = 1 * x = x

x^2 = 1

So substituting all these into f(x):

1 + 5x -8*1 +7x -1 -12x +4*1 -8x +12*1 -5x -5

Now let me compute each term:

1 (from x^10)

+5x (from 5x^9)

-8*1 = -8 (from -8x^8)

+7x (from 7x^7)

-1 (from -x^6)

-12x (from -12x^5)

+4*1 = +4 (from 4x^4)

-8x (from -8x^3)

+12*1 = +12 (from 12x^2)

-5x (from -5x)

-5 (constant term)

Now combine like terms:

Constants: 1 -8 -1 +4 +12 -5

Let me compute that:

Start with 1 -8 = -7

-7 -1 = -8

-8 +4 = -4

-4 +12 = 8

8 -5 = 3

So constants sum to 3.

Now the x terms: 5x +7x -12x -8x -5x

Compute coefficients:

5 +7 = 12; 12 -12 = 0; 0 -8 = -8; -8 -5 = -13. So total x term is -13x.

Therefore, the entire expression simplifies to -13x + 3, which matches the earlier result. So that confirms that the remainder is indeed -13x + 3. Therefore, the answer should be -13x + 3. Wait, but let me check if the problem wants it in a specific form? The question says "find the remainder", so as a polynomial, which is linear, so -13x + 3 is correct. But maybe they want it written as 3 -13x? Either way, both are correct. But let me check if I made any mistake in substitution.

Wait, let me check again the substitution steps. For example, x^10: since x²=1, x^10 = (x²)^5 = 1^5 = 1. Correct. Similarly, x^9 = x^(2*4 + 1) = (x²)^4 * x = 1^4 * x = x. Correct. Similarly for others. Then substituting all terms:

Original f(x):

x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

After substitution:

1 +5x -8 +7x -1 -12x +4 -8x +12 -5x -5

Yes, that's correct. Then constants: 1 -8 -1 +4 +12 -5. Let me check again:

1 -8 is -7, -7 -1 is -8, -8 +4 is -4, -4 +12 is 8, 8 -5 is 3. Correct. Then x terms: 5x +7x is 12x, minus 12x is 0, minus 8x is -8x, minus 5x is -13x. Correct. So yes, remainder is -13x + 3. Therefore, the answer is -13x + 3.

But let me check with another method just to be thorough. For example, suppose I use the fact that if f(x) divided by x² -1 gives remainder ax + b, then f(x) - (ax + b) is divisible by x² -1. Therefore, f(x) - (ax + b) has roots at x = 1 and x = -1. Which is exactly what I used earlier with f(1) = a + b and f(-1) = -a + b. So that's consistent. Therefore, solving gives a = -13, b = 3. So that's correct.

Alternatively, maybe I can use Taylor series expansion or something? But that might be more complicated. Alternatively, think of f(x) = (x² -1)Q(x) + ax + b. Then, if I take derivative? Not sure. Maybe not necessary.

Alternatively, think of f(x) evaluated at x = 1 and x = -1, which we did. So seems solid.

Therefore, I think the remainder is -13x + 3. Let me check once more with another approach. Suppose I write f(x) as:

f(x) = x^10 +5x^9 -8x^8 +7x^7 -x^6 -12x^5 +4x^4 -8x^3 +12x^2 -5x -5

Let me group terms in pairs or something. But maybe not helpful. Alternatively, note that x² = 1, so x^even = 1, x^odd = x. So as before, which we already did. So that gives the same result. So I think confident that the remainder is -13x + 3.

But just to make sure, let me plug in x = 1 into the remainder: -13(1) + 3 = -10, which matches f(1) = -10. Similarly, x = -1: -13(-1) + 3 = 13 + 3 = 16, which matches f(-1) = 16. Perfect. So that's correct.

Therefore, the remainder is -13x + 3. So the answer is \boxed{-13x + 3}?

Wait, but sometimes problems expect the answer in a certain format. Let me check if the question specifies. The problem says "find the remainder when f(x) is divided by x² -1". The remainder is a linear polynomial, so -13x + 3 is correct. However, sometimes people prefer writing it with the x term first, so maybe -13x + 3 or 3 -13x. But both are equivalent. However, in standard form, it's usually written with descending degree, so -13x + 3. But maybe the problem expects it in a specific way. Let me check if there's any convention. In most cases, the answer would be written as ax + b, so -13x + 3. So I think that's acceptable.

Alternatively, maybe they want it as 3 -13x? But both are correct. However, since the question didn't specify, either is fine. But let me check if my calculation of a and b is correct. From the equations:

a + b = -10

- a + b = 16

Subtracting first equation from second: (-a + b) - (a + b) = 16 - (-10) => -2a = 26 => a = -13. Then b = -10 - a = -10 - (-13) = 3. So yes, correct. So the remainder is -13x + 3. So I think that's the answer.

**Final Answer**
The remainder is \boxed{-13x + 3}.

To find the remainder when the polynomial
$$
f(x) = x^{10} + 5x^9 - 8x^8 + 7x^7 - x^6 - 12x^5 + 4x^4 - 8x^3 + 12x^2 - 5x - 5
$$
is divided by $ x^2 - 1 $, we can use the **Remainder Theorem** for polynomials divided by a quadratic divisor.

---

### Step 1: Understand the Structure of the Remainder

When dividing a polynomial $ f(x) $ by a quadratic polynomial $ x^2 - 1 $, the **remainder** will be a **linear polynomial** of the form:
$$
R(x) = ax + b
$$
This is because the degree of the remainder must be less than the degree of the divisor, which is 2.

---

### Step 2: Use the Roots of the Divisor

The roots of $ x^2 - 1 = 0 $ are $ x = 1 $ and $ x = -1 $. By the **Remainder Theorem**, we know:
$$
f(1) = R(1) = a(1) + b = a + b
$$
$$
f(-1) = R(-1) = a(-1) + b = -a + b
$$

So we can set up the following system of equations:
$$
\begin{cases}
a + b = f(1) \\
-a + b = f(-1)
\end{cases}
$$

---

### Step 3: Evaluate $ f(1) $ and $ f(-1) $

#### Compute $ f(1) $:
$$
f(1) = 1^{10} + 5 \cdot 1^9 - 8 \cdot 1^8 + 7 \cdot 1^7 - 1^6 - 12 \cdot 1^5 + 4 \cdot 1^4 - 8 \cdot 1^3 + 12 \cdot 1^2 - 5 \cdot 1 - 5
$$
$$
= 1 + 5 - 8 + 7 - 1 - 12 + 4 - 8 + 12 - 5 - 5 = -10
$$

#### Compute $ f(-1) $:
$$
f(-1) = (-1)^{10} + 5 \cdot (-1)^9 - 8 \cdot (-1)^8 + 7 \cdot (-1)^7 - (-1)^6 - 12 \cdot (-1)^5 + 4 \cdot (-1)^4 - 8 \cdot (-1)^3 + 12 \cdot (-1)^2 - 5 \cdot (-1) - 5
$$
$$
= 1 - 5 - 8 - 7 - 1 + 12 + 4 + 8 + 12 + 5 - 5 = 16
$$

---

### Step 4: Solve the System of Equations

We now solve:
$$
\begin{cases}
a + b = -10 \\
-a + b = 16
\end{cases}
$$

Subtract the first equation from the second:
$$
(-a + b) - (a + b) = 16 - (-10) \Rightarrow -2a = 26 \Rightarrow a = -13
$$

Substitute $ a = -13 $ into the first equation:
$$
-13 + b = -10 \Rightarrow b = 3
$$

---

### Step 5: Write the Remainder

The remainder is:
$$
R(x) = ax + b = -13x + 3
$$

---

### Final Answer

$$
\boxed{-13x + 3}
$$

Solution:
We have
$$f(x) = q(x)(x^2-1) + r(x),$$where $q(x)$ is the quotient and $r(x)$ is the remainder. Since $x^2-1$ is quadratic, the remainder is at most linear; let us write $r(x) = ax+b$.

Observe that $x=-1$ and $x=1$ are both zeroes of $x^2-1$. Thus $f(1)=r(1)$ and $f(-1)=r(-1)$.

We can use the given formula for $f(x)$ to compute $f(1)=-10$ and $f(-1)=16$. Thus we have the system of equations
$$\begin{cases}
-10 = a\cdot (1) + b,\\
\phantom{-}16 = a\cdot (-1) + b.
\end{cases}$$Adding these equations yields $6=2b$ and hence $b=3$. Substituting into either equation then yields $a=-13$.

Therefore, $r(x) = ax+b = \boxed{-13x+3}$.


Explanation:
Okay, so I need to compute A^27 + A^31 + A^40 where A is the matrix [[3, 7], [-1, -2]]. Hmm, matrix exponentiation can be tricky, especially with such high exponents. I remember that for diagonal matrices, you can just raise the diagonal entries to the power, but this matrix isn't diagonal. Maybe I can diagonalize it? If I can find eigenvalues and eigenvectors, then maybe I can express A as PDP^{-1} and then A^n would be PD^nP^{-1}. That sounds like a plan. Let me try that.

First, let me find the eigenvalues of A. The eigenvalues are solutions to the characteristic equation det(A - λI) = 0. So let me compute that determinant.

The matrix A - λI is [[3 - λ, 7], [-1, -2 - λ]]. The determinant is (3 - λ)(-2 - λ) - (7)(-1). Let me compute that:

(3 - λ)(-2 - λ) = 3*(-2) + 3*(-λ) - λ*(-2) - λ*λ = -6 - 3λ + 2λ + λ² = λ² - λ - 6.

Then subtract (7)(-1) which is -7*(-1) = +7? Wait, no. Wait, determinant is (top left * bottom right) - (top right * bottom left). So it's (3 - λ)(-2 - λ) - (7)(-1). So that's (λ² - λ - 6) + 7. Wait, let me recalculate:

First, expand (3 - λ)(-2 - λ):

Multiply 3 by -2: -6

Multiply 3 by -λ: -3λ

Multiply -λ by -2: +2λ

Multiply -λ by -λ: +λ²

So altogether: -6 -3λ + 2λ + λ² = λ² - λ -6. Then subtract (7)(-1) which is -7*(-1) = +7? Wait, no. Wait, determinant is (top left * bottom right) - (top right * bottom left). So the first term is (3 - λ)(-2 - λ) and the second term is 7*(-1). So determinant is (3 - λ)(-2 - λ) - (7)(-1) = (λ² - λ -6) +7 = λ² - λ +1. Wait, is that right?

Wait, let me check again:

(3 - λ)(-2 - λ) = 3*(-2) + 3*(-λ) + (-λ)*(-2) + (-λ)*(-λ) = -6 -3λ + 2λ + λ² = λ² - λ -6. Then subtract (7)(-1) which is -7*(-1) = +7? Wait, no. Wait, determinant is (top left * bottom right) - (top right * bottom left). So the second term is 7*(-1). So determinant is (λ² - λ -6) - (7*(-1))? Wait, no. Wait, the formula is ad - bc. So for matrix [[a, b], [c, d]], determinant is ad - bc. So here, a = 3 - λ, d = -2 - λ, b = 7, c = -1. So determinant is (3 - λ)(-2 - λ) - (7)(-1). So that is (λ² - λ -6) - (-7) = λ² - λ -6 +7 = λ² - λ +1. Yes, that's correct. So the characteristic equation is λ² - λ +1 = 0. Let me solve that.

Using quadratic formula: λ = [1 ± sqrt(1 - 4*1*1)] / 2 = [1 ± sqrt(-3)] / 2. So the eigenvalues are complex: (1 ± i√3)/2. Hmm, complex eigenvalues. That complicates things a bit. But maybe I can still diagonalize it using complex numbers? Or perhaps there's another approach?

Alternatively, maybe I can find a recurrence relation? Since the matrix is 2x2, maybe its powers can be expressed in terms of lower exponents? Let me think. For a 2x2 matrix, if we can find a relation between A^n and previous terms, maybe we can find a pattern or a formula.

Alternatively, since the eigenvalues are complex, maybe we can express A in terms of its eigenvalues and use some properties? Let me recall that if a matrix has eigenvalues λ and μ, then A^n can be expressed in terms of λ^n and μ^n if it's diagonalizable. But since the eigenvalues are complex, maybe we can write A^n as a combination of powers of those eigenvalues?

But since the eigenvalues are complex, perhaps we can use Euler's formula or something? Let me think. Let me first check if the matrix is diagonalizable. Since the eigenvalues are distinct (they are complex conjugates), the matrix is diagonalizable over the complex numbers. So, even though the eigenvalues are complex, we can write A = PDP^{-1}, where D is diagonal with entries (1 + i√3)/2 and (1 - i√3)/2. Then A^n = PD^nP^{-1}. But computing this might be complicated, but maybe manageable?

Alternatively, maybe we can find a closed-form expression for A^n using the eigenvalues. Let me recall that for a 2x2 matrix, if we have eigenvalues λ and μ, then A^n can be expressed as a linear combination of A^{n-1}, A^{n-2}, etc. Wait, maybe using the Cayley-Hamilton theorem? The Cayley-Hamilton theorem says that a matrix satisfies its own characteristic equation. The characteristic equation here is λ² - λ +1 = 0. Therefore, A² - A + I = 0. Therefore, A² = A - I. That seems useful! So A² = A - I. Then maybe we can find a recurrence relation for A^n?

Yes! If A² = A - I, then we can use this to express higher powers of A in terms of lower powers. Let me try to find a recurrence relation. Let me denote A^n = a_n A + b_n I. Then, using the relation A² = A - I, we can find a recurrence for a_n and b_n.

Let me check. Suppose A^n = a_n A + b_n I. Then A^{n+1} = A * A^n = A*(a_n A + b_n I) = a_n A² + b_n A. But since A² = A - I, substitute that in: a_n (A - I) + b_n A = a_n A - a_n I + b_n A = (a_n + b_n) A - a_n I. Therefore, A^{n+1} = (a_n + b_n) A - a_n I. Therefore, comparing with the form A^{n+1} = a_{n+1} A + b_{n+1} I, we have:

a_{n+1} = a_n + b_n

b_{n+1} = -a_n

So we have a recurrence relation:

a_{n+1} = a_n + b_n

b_{n+1} = -a_n

This seems manageable. Now, we need initial conditions. Let me find a_1 and b_1. Since A^1 = A = 1*A + 0*I, so a_1 = 1, b_1 = 0.

Similarly, A^0 = I = 0*A + 1*I, so a_0 = 0, b_0 = 1. Wait, but maybe we need to check for n=1. Let me confirm:

For n=1: A^1 = a_1 A + b_1 I => 1*A + 0*I, so yes, a_1=1, b_1=0.

For n=0: A^0 = I = 0*A + 1*I, so a_0=0, b_0=1.

So with that, we can compute a_n and b_n for any n. Let me try to compute some terms.

But since we need A^27, A^31, A^40, maybe we can find a general formula for a_n and b_n?

Alternatively, since the recurrence relations are linear, maybe we can solve them. Let me write the recurrence relations:

From above:

a_{n+1} = a_n + b_n

b_{n+1} = -a_n

Let me substitute b_n from the second equation into the first. From the second equation, b_n = -a_{n-1}. Therefore, substituting into the first equation:

a_{n+1} = a_n + b_n = a_n - a_{n-1}

Therefore, we have a second-order recurrence relation:

a_{n+1} = a_n - a_{n-1}

With initial conditions:

For n=0: a_0 = 0

For n=1: a_1 = 1

Wait, let me check. If n=1, then a_{1} = 1, and for n=0, a_0 = 0. Then for n=2, a_2 = a_1 - a_0 = 1 - 0 = 1. Then a_3 = a_2 - a_1 = 1 - 1 = 0. a_4 = a_3 - a_2 = 0 - 1 = -1. a_5 = a_4 - a_3 = -1 - 0 = -1. a_6 = a_5 - a_4 = -1 - (-1) = 0. a_7 = a_6 - a_5 = 0 - (-1) = 1. a_8 = a_7 - a_6 = 1 - 0 = 1. Wait, so the sequence is 0, 1, 1, 0, -1, -1, 0, 1, 1,... It seems periodic with period 6? Let me check:

a_0 = 0

a_1 = 1

a_2 = 1

a_3 = 0

a_4 = -1

a_5 = -1

a_6 = 0

a_7 = 1

a_8 = 1

Yes, so the sequence repeats every 6 terms. Therefore, the sequence a_n is periodic with period 6. Therefore, we can find a_n by computing n mod 6.

Similarly, since b_n = -a_{n-1}, then b_n is also periodic with the same period.

Therefore, if we can find a_n for any n, we can compute A^n = a_n A + b_n I.

So, given that, let me find a_n for n=27, 31, 40.

First, let me note that the period is 6. Therefore, for any n, a_n = a_{n mod 6}, but need to adjust for the offset. Let me check:

For example, n=0: 0 mod 6 = 0, a_0=0

n=1: 1 mod 6 =1, a_1=1

n=2: 2 mod 6=2, a_2=1

n=3: 3 mod 6=3, a_3=0

n=4: 4 mod 6=4, a_4=-1

n=5: 5 mod 6=5, a_5=-1

n=6: 6 mod 6=0, a_6=0, which matches a_0.

Therefore, the formula is a_n = a_{n mod 6}, with the understanding that if n mod 6 is 0, then it's a_0=0, etc.

Therefore, to compute a_n for n=27, 31, 40, let's compute n mod 6:

For n=27: 27 divided by 6 is 4*6=24, remainder 3. So 27 mod 6 = 3. Therefore, a_27 = a_3 = 0.

For n=31: 31 divided by 6 is 5*6=30, remainder 1. So 31 mod 6 =1. Therefore, a_31 = a_1 =1.

For n=40: 40 divided by 6 is 6*6=36, remainder 4. So 40 mod 6=4. Therefore, a_40 = a_4 = -1.

Similarly, compute b_n for each n. Since b_n = -a_{n-1}.

So for n=27: b_27 = -a_{26}. Let's compute 26 mod 6. 26 divided by 6 is 4*6=24, remainder 2. So 26 mod 6=2. Therefore, a_26 = a_2=1. Therefore, b_27 = -1.

Similarly, for n=31: b_31 = -a_{30}. 30 mod 6=0, so a_30 = a_0=0. Therefore, b_31 = -0 = 0.

For n=40: b_40 = -a_{39}. 39 mod 6: 6*6=36, remainder 3. So 39 mod 6=3. Therefore, a_39 = a_3=0. Therefore, b_40 = -0 = 0.

Therefore, we have:

A^27 = a_27 A + b_27 I = 0*A + (-1)I = -I

Wait, wait. Wait, A^n = a_n A + b_n I. So for n=27, a_27=0, b_27=-1. Therefore, A^27 = 0*A + (-1)I = -I.

Similarly, A^31 = a_31 A + b_31 I = 1*A + 0*I = A.

A^40 = a_40 A + b_40 I = (-1)A + 0*I = -A.

Therefore, the sum A^27 + A^31 + A^40 = (-I) + A + (-A) = -I + A - A = -I. Wait, that's interesting. So the sum simplifies to -I?

Wait, let me check again:

A^27 = -I

A^31 = A

A^40 = -A

Therefore, adding them: (-I) + A + (-A) = -I + (A - A) = -I + 0 = -I. So the result is -I?

But let me verify this with some steps. Maybe I made a mistake in computing a_n and b_n?

Let me check for n=2. Earlier, we saw that a_2=1. Let me compute A^2 using the original matrix and check if it's equal to a_2 A + b_2 I.

Original matrix A is [[3,7],[-1,-2]]

Compute A^2:

A^2 = A*A = [[3*3 + 7*(-1), 3*7 + 7*(-2)], [-1*3 + (-2)*(-1), -1*7 + (-2)*(-2)]]

Compute first row first column: 9 -7 = 2

First row second column: 21 -14 = 7

Second row first column: -3 + 2 = -1

Second row second column: -7 + 4 = -3

So A^2 = [[2, 7], [-1, -3]]

Now, according to our formula, A^2 = a_2 A + b_2 I. a_2=1, b_2=?

From earlier, b_n = -a_{n-1}. For n=2, b_2 = -a_1 = -1. Therefore, A^2 = 1*A + (-1)*I = A - I.

Compute A - I: [[3 -1, 7 - 0], [-1 - 0, -2 -1]] = [[2, 7], [-1, -3]], which matches the direct computation. So that's correct.

Similarly, check A^3. According to the recurrence, A^3 = a_3 A + b_3 I. a_3=0, b_3=-a_2=-1. Therefore, A^3 = 0*A + (-1)*I = -I. Let me compute A^3 directly.

A^3 = A^2 * A = [[2,7],[-1,-3]] * [[3,7],[-1,-2]]

First row first column: 2*3 + 7*(-1) = 6 -7 = -1

First row second column: 2*7 + 7*(-2) = 14 -14 = 0

Second row first column: -1*3 + (-3)*(-1) = -3 + 3 = 0

Second row second column: -1*7 + (-3)*(-2) = -7 + 6 = -1

Therefore, A^3 = [[-1, 0], [0, -1]] = -I. Which matches the formula. So that's correct. Therefore, our formula seems to work.

Similarly, check A^4. According to the recurrence, a_4=-1, b_4=-a_3= -0=0. Therefore, A^4 = -1*A + 0*I = -A. Let me compute A^4 directly. A^4 = A^3 * A = (-I)*A = -A. Which is correct. So that's correct.

Therefore, our formula seems valid. Therefore, proceeding with that, the conclusion that A^27 = -I, A^31 = A, A^40 = -A, so their sum is -I + A - A = -I. Therefore, the answer is -I, which is [[-1, 0], [0, -1]].

But let me check for another exponent to be sure. Let's check A^6. According to the sequence, a_6 = 0, b_6 = -a_5 = -(-1) = 1. Therefore, A^6 = 0*A + 1*I = I. Let me compute A^6 directly. Since A^3 = -I, then A^6 = (A^3)^2 = (-I)^2 = I. Correct. So that's correct. Therefore, the formula works.

Therefore, going back, the conclusion that A^27 + A^31 + A^40 = -I is correct? Wait, but let me check the exponents again.

Wait, n=27: 27 mod 6 is 3, so a_27 = a_3 = 0, b_27 = -a_{26} = -a_{26}. 26 mod 6 is 2, so a_26 = a_2 = 1. Therefore, b_27 = -1. Therefore, A^27 = 0*A + (-1)I = -I. Correct.

n=31: 31 mod 6 is 1, so a_31 = a_1 = 1, b_31 = -a_{30} = -a_0 = -0 = 0. Therefore, A^31 = 1*A + 0*I = A. Correct.

n=40: 40 mod 6 is 4, so a_40 = a_4 = -1, b_40 = -a_{39} = -a_3 = -0 = 0. Therefore, A^40 = -1*A + 0*I = -A. Correct.

Therefore, adding them: -I + A - A = -I. Therefore, the answer is -I. Therefore, the matrix [[-1, 0], [0, -1]].

But let me check if there's another way to see this. Since A^2 = A - I, then A^3 = A^2 - A = (A - I) - A = -I. Then A^4 = A^3 * A = -I * A = -A. A^5 = A^4 * A = -A^2 = -(A - I) = -A + I. A^6 = A^5 * A = (-A + I) * A = -A^2 + A = -(A - I) + A = -A + I + A = I. Then A^6 = I. Therefore, the powers cycle every 6. Therefore, A^6 = I, so A^{6k} = I. Therefore, exponents can be reduced modulo 6.

Therefore, for example:

A^27: 27 divided by 6 is 4*6=24, remainder 3. So A^27 = A^{6*4 + 3} = (A^6)^4 * A^3 = I^4 * A^3 = A^3 = -I. Correct.

Similarly, A^31: 31 divided by 6 is 5*6=30, remainder 1. So A^31 = A^{6*5 +1} = (A^6)^5 * A^1 = I^5 * A = A. Correct.

A^40: 40 divided by 6 is 6*6=36, remainder 4. So A^40 = A^{6*6 + 4} = (A^6)^6 * A^4 = I^6 * A^4 = A^4 = -A. Correct.

Therefore, the same conclusion. Therefore, the sum is -I + A - A = -I. Therefore, the answer is -I. Therefore, the matrix [[-1, 0], [0, -1]].

But let me check once more with another exponent. For example, A^3 = -I, A^4 = -A, A^5 = -A + I, A^6 = I, A^7 = A, etc. So the cycle is indeed every 6. Therefore, the exponents 27, 31, 40 correspond to positions in the cycle:

27 mod 6 = 3: A^3 = -I

31 mod 6 = 1: A^1 = A

40 mod 6 = 4: A^4 = -A

Therefore, sum is -I + A - A = -I. So yes, that's correct.

Therefore, the answer is -I, which is the matrix with -1 on the diagonal and 0 elsewhere. Therefore, boxed, it would be \boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}.

But let me check once more if there's a possibility that I made a mistake in the sign somewhere. For example, when computing b_n. Let me check for n=27, b_27 = -a_{26}. Since a_{26} is a_{26 mod 6} = a_2 = 1. Therefore, b_27 = -1. Therefore, A^27 = 0*A + (-1)I = -I. Correct. Similarly, for n=40, b_40 = -a_{39} = -a_{3} = -0 = 0. Correct. So A^40 = -A. Correct. So the sum is indeed -I + A - A = -I. Therefore, I think this is correct.

Alternatively, maybe the problem expects a different form? Let me check if -I is indeed the correct answer. Let me compute A^27 + A^31 + A^40 using another approach. For example, since A^6 = I, then A^27 = A^{6*4 + 3} = (A^6)^4 * A^3 = I^4 * A^3 = A^3 = -I. Similarly, A^31 = A^{6*5 + 1} = A^1 = A. A^40 = A^{6*6 + 4} = A^4 = -A. Therefore, sum is -I + A - A = -I. So same result. Therefore, I think this is correct.

Therefore, the answer is -I, which is [[-1, 0], [0, -1]]. Therefore, boxed as above.

**Final Answer**
\boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}

To compute the matrix expression
$$
\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40}
$$
where
$$
\mathbf{A} = \begin{pmatrix} 3 & 7 \\ -1 & -2 \end{pmatrix},
$$
we begin by analyzing the structure of the matrix and its powers.

---

### Step 1: Use the Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem states that a matrix satisfies its own characteristic equation. For matrix $\mathbf{A}$, we compute the characteristic polynomial:

$$
\det(\mathbf{A} - \lambda \mathbf{I}) = \det\begin{pmatrix} 3 - \lambda & 7 \\ -1 & -2 - \lambda \end{pmatrix}
= (3 - \lambda)(-2 - \lambda) - (7)(-1)
= \lambda^2 - \lambda + 1.
$$

Thus, the characteristic equation is:
$$
\lambda^2 - \lambda + 1 = 0.
$$

By the Cayley-Hamilton Theorem, we have:
$$
\mathbf{A}^2 - \mathbf{A} + \mathbf{I} = \mathbf{0} \quad \Rightarrow \quad \mathbf{A}^2 = \mathbf{A} - \mathbf{I}.
$$

This recurrence allows us to express higher powers of $\mathbf{A}$ in terms of $\mathbf{A}$ and $\mathbf{I}$.

---

### Step 2: Find a Recurrence for Powers of $\mathbf{A}$

Let us define:
$$
\mathbf{A}^n = a_n \mathbf{A} + b_n \mathbf{I}.
$$

Using the recurrence $\mathbf{A}^2 = \mathbf{A} - \mathbf{I}$, we derive the following recurrence relations for the coefficients $a_n$ and $b_n$:

$$
a_{n+1} = a_n + b_n, \quad b_{n+1} = -a_n.
$$

With initial conditions:
$$
a_0 = 0, \quad b_0 = 1, \quad a_1 = 1, \quad b_1 = 0.
$$

This recurrence leads to a periodic sequence for $a_n$ and $b_n$ with period 6. The sequence of $a_n$ is:

$$
a_0 = 0, \quad a_1 = 1, \quad a_2 = 1, \quad a_3 = 0, \quad a_4 = -1, \quad a_5 = -1, \quad a_6 = 0, \quad \text{and so on.}
$$

Similarly, $b_n = -a_{n-1}$, so the sequence for $b_n$ is also periodic with the same period.

---

### Step 3: Compute the Required Powers

We now compute the values of $a_n$ and $b_n$ for the exponents 27, 31, and 40.

- $27 \mod 6 = 3 \Rightarrow a_{27} = a_3 = 0, \quad b_{27} = -a_{26} = -a_2 = -1$
- $31 \mod 6 = 1 \Rightarrow a_{31} = a_1 = 1, \quad b_{31} = -a_{30} = -a_0 = 0$
- $40 \mod 6 = 4 \Rightarrow a_{40} = a_4 = -1, \quad b_{40} = -a_{39} = -a_3 = 0$

Thus, we have:
- $\mathbf{A}^{27} = a_{27} \mathbf{A} + b_{27} \mathbf{I} = 0 \cdot \mathbf{A} - 1 \cdot \mathbf{I} = -\mathbf{I}$
- $\mathbf{A}^{31} = a_{31} \mathbf{A} + b_{31} \mathbf{I} = 1 \cdot \mathbf{A} + 0 \cdot \mathbf{I} = \mathbf{A}$
- $\mathbf{A}^{40} = a_{40} \mathbf{A} + b_{40} \mathbf{I} = -1 \cdot \mathbf{A} + 0 \cdot \mathbf{I} = -\mathbf{A}$

---

### Step 4: Add the Results

Now compute the sum:
$$
\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40} = -\mathbf{I} + \mathbf{A} - \mathbf{A} = -\mathbf{I}.
$$

---

### Final Answer

$$
\boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}
$$

[Assistant]
Explanation:
Okay, so I need to compute A^27 + A^31 + A^40 where A is the matrix [[3, 7], [-1, -2]]. Hmm, matrix exponentiation can be tricky, especially with such high exponents. I remember that for diagonal matrices, you can just raise the diagonal entries to the power, but this matrix isn't diagonal. Maybe I can diagonalize it? If I can find eigenvalues and eigenvectors, then maybe I can express A as PDP^{-1} and then A^n would be PD^nP^{-1}. That sounds like a plan. Let me try that.

First, let me find the eigenvalues of A. The eigenvalues are solutions to the characteristic equation det(A - λI) = 0. So let me compute that determinant.

The matrix A - λI is [[3 - λ, 7], [-1, -2 - λ]]. The determinant is (3 - λ)(-2 - λ) - (7)(-1). Let me compute that:

(3 - λ)(-2 - λ) = 3*(-2) + 3*(-λ) - λ*(-2) - λ*λ = -6 - 3λ + 2λ + λ² = λ² - λ - 6.

Then subtract (7)(-1) which is -7*(-1) = +7? Wait, no. Wait, determinant is (top left * bottom right) - (top right * bottom left). So it's (3 - λ)(-2 - λ) - (7)(-1). So that's (λ² - λ - 6) + 7. Wait, let me recalculate:

First, expand (3 - λ)(-2 - λ):

Multiply 3 by -2: -6

Multiply 3 by -λ: -3λ

Multiply -λ by -2: +2λ

Multiply -λ by -λ: +λ²

So altogether: -6 -3λ + 2λ + λ² = λ² - λ -6. Then subtract (7)(-1) which is -7*(-1) = +7? Wait, no. Wait, determinant is (top left * bottom right) - (top right * bottom left). So the first term is (3 - λ)(-2 - λ) and the second term is 7*(-1). So determinant is (3 - λ)(-2 - λ) - (7)(-1) = (λ² - λ -6) +7 = λ² - λ +1. Wait, is that right?

Wait, let me check again:

(3 - λ)(-2 - λ) = 3*(-2) + 3*(-λ) + (-λ)*(-2) + (-λ)*(-λ) = -6 -3λ + 2λ + λ² = λ² - λ -6. Then subtract (7)(-1) which is -7*(-1) = +7? Wait, no. Wait, determinant is (top left * bottom right) - (top right * bottom left). So the second term is 7*(-1). So determinant is (λ² - λ -6) - (7*(-1))? Wait, no. Wait, the formula is ad - bc. So for matrix [[a, b], [c, d]], determinant is ad - bc. So here, a = 3 - λ, d = -2 - λ, b = 7, c = -1. So determinant is (3 - λ)(-2 - λ) - (7)(-1). So that is (λ² - λ -6) - (-7) = λ² - λ -6 +7 = λ² - λ +1. Yes, that's correct. So the characteristic equation is λ² - λ +1 = 0. Let me solve that.

Using quadratic formula: λ = [1 ± sqrt(1 - 4*1*1)] / 2 = [1 ± sqrt(-3)] / 2. So the eigenvalues are complex: (1 ± i√3)/2. Hmm, complex eigenvalues. That complicates things a bit. But maybe I can still diagonalize it using complex numbers? Or perhaps there's another approach?

Alternatively, maybe I can find a recurrence relation? Since the matrix is 2x2, maybe its powers can be expressed in terms of lower exponents? Let me think. For a 2x2 matrix, if we can find a relation between A^n and previous terms, maybe we can find a pattern or a formula.

Alternatively, since the eigenvalues are complex, maybe we can express A in terms of its eigenvalues and use some properties? Let me recall that if a matrix has eigenvalues λ and μ, then A^n can be expressed in terms of λ^n and μ^n if it's diagonalizable. But since the eigenvalues are complex, maybe we can write A^n as a combination of powers of those eigenvalues?

But since the eigenvalues are complex, perhaps we can use Euler's formula or something? Let me think. Let me first check if the matrix is diagonalizable. Since the eigenvalues are distinct (they are complex conjugates), the matrix is diagonalizable over the complex numbers. So, even though the eigenvalues are complex, we can write A = PDP^{-1}, where D is diagonal with entries (1 + i√3)/2 and (1 - i√3)/2. Then A^n = PD^nP^{-1}. But computing this might be complicated, but maybe manageable?

Alternatively, maybe we can find a closed-form expression for A^n using the eigenvalues. Let me recall that for a 2x2 matrix, if we have eigenvalues λ and μ, then A^n can be expressed as a linear combination of A^{n-1}, A^{n-2}, etc. Wait, maybe using the Cayley-Hamilton theorem? The Cayley-Hamilton theorem says that a matrix satisfies its own characteristic equation. The characteristic equation here is λ² - λ +1 = 0. Therefore, A² - A + I = 0. Therefore, A² = A - I. That seems useful! So A² = A - I. Then maybe we can find a recurrence relation for A^n?

Yes! If A² = A - I, then we can use this to express higher powers of A in terms of lower powers. Let me try to find a recurrence relation. Let me denote A^n = a_n A + b_n I. Then, using the relation A² = A - I, we can find a recurrence for a_n and b_n.

Let me check. Suppose A^n = a_n A + b_n I. Then A^{n+1} = A * A^n = A*(a_n A + b_n I) = a_n A² + b_n A. But since A² = A - I, substitute that in: a_n (A - I) + b_n A = a_n A - a_n I + b_n A = (a_n + b_n) A - a_n I. Therefore, A^{n+1} = (a_n + b_n) A - a_n I. Therefore, comparing with the form A^{n+1} = a_{n+1} A + b_{n+1} I, we have:

a_{n+1} = a_n + b_n

b_{n+1} = -a_n

So we have a recurrence relation:

a_{n+1} = a_n + b_n

b_{n+1} = -a_n

This seems manageable. Now, we need initial conditions. Let me find a_1 and b_1. Since A^1 = A = 1*A + 0*I, so a_1 = 1, b_1 = 0.

Similarly, A^0 = I = 0*A + 1*I, so a_0 = 0, b_0 = 1. Wait, but maybe we need to check for n=1. Let me confirm:

For n=1: A^1 = a_1 A + b_1 I => 1*A + 0*I, so yes, a_1=1, b_1=0.

For n=0: A^0 = I = 0*A + 1*I, so a_0=0, b_0=1.

So with that, we can compute a_n and b_n for any n. Let me try to compute some terms.

But since we need A^27, A^31, A^40, maybe we can find a general formula for a_n and b_n?

Alternatively, since the recurrence relations are linear, maybe we can solve them. Let me write the recurrence relations:

From above:

a_{n+1} = a_n + b_n

b_{n+1} = -a_n

Let me substitute b_n from the second equation into the first. From the second equation, b_n = -a_{n-1}. Therefore, substituting into the first equation:

a_{n+1} = a_n + b_n = a_n - a_{n-1}

Therefore, we have a second-order recurrence relation:

a_{n+1} = a_n - a_{n-1}

With initial conditions:

For n=0: a_0 = 0

For n=1: a_1 = 1

Wait, let me check. If n=1, then a_{1} = 1, and for n=0, a_0 = 0. Then for n=2, a_2 = a_1 - a_0 = 1 - 0 = 1. Then a_3 = a_2 - a_1 = 1 - 1 = 0. a_4 = a_3 - a_2 = 0 - 1 = -1. a_5 = a_4 - a_3 = -1 - 0 = -1. a_6 = a_5 - a_4 = -1 - (-1) = 0. a_7 = a_6 - a_5 = 0 - (-1) = 1. a_8 = a_7 - a_6 = 1 - 0 = 1. Wait, so the sequence is 0, 1, 1, 0, -1, -1, 0, 1, 1,... It seems periodic with period 6? Let me check:

a_0 = 0

a_1 = 1

a_2 = 1

a_3 = 0

a_4 = -1

a_5 = -1

a_6 = 0

a_7 = 1

a_8 = 1

Yes, so the sequence repeats every 6 terms. Therefore, the sequence a_n is periodic with period 6. Therefore, we can find a_n by computing n mod 6.

Similarly, since b_n = -a_{n-1}, then b_n is also periodic with the same period.

Therefore, if we can find a_n for any n, we can compute A^n = a_n A + b_n I.

So, given that, let me find a_n for n=27, 31, 40.

First, let me note that the period is 6. Therefore, for any n, a_n = a_{n mod 6}, but need to adjust for the offset. Let me check:

For example, n=0: 0 mod 6 = 0, a_0=0

n=1: 1 mod 6 =1, a_1=1

n=2: 2 mod 6=2, a_2=1

n=3: 3 mod 6=3, a_3=0

n=4: 4 mod 6=4, a_4=-1

n=5: 5 mod 6=5, a_5=-1

n=6: 6 mod 6=0, a_6=0, which matches a_0.

Therefore, the formula is a_n = a_{n mod 6}, with the understanding that if n mod 6 is 0, then it's a_0=0, etc.

Therefore, to compute a_n for n=27, 31, 40, let's compute n mod 6:

For n=27: 27 divided by 6 is 4*6=24, remainder 3. So 27 mod 6 = 3. Therefore, a_27 = a_3 = 0.

For n=31: 31 divided by 6 is 5*6=30, remainder 1. So 31 mod 6 =1. Therefore, a_31 = a_1 =1.

For n=40: 40 divided by 6 is 6*6=36, remainder 4. So 40 mod 6=4. Therefore, a_40 = a_4 = -1.

Similarly, compute b_n for each n. Since b_n = -a_{n-1}.

So for n=27: b_27 = -a_{26}. Let's compute 26 mod 6. 26 divided by 6 is 4*6=24, remainder 2. So 26 mod 6=2. Therefore, a_26 = a_2=1. Therefore, b_27 = -1.

Similarly, for n=31: b_31 = -a_{30}. 30 mod 6=0, so a_30 = a_0=0. Therefore, b_31 = -0 = 0.

For n=40: b_40 = -a_{39}. 39 mod 6: 6*6=36, remainder 3. So 39 mod 6=3. Therefore, a_39 = a_3=0. Therefore, b_40 = -0 = 0.

Therefore, we have:

A^27 = a_27 A + b_27 I = 0*A + (-1)I = -I

Wait, wait. Wait, A^n = a_n A + b_n I. So for n=27, a_27=0, b_27=-1. Therefore, A^27 = 0*A + (-1)I = -I.

Similarly, A^31 = a_31 A + b_31 I = 1*A + 0*I = A.

A^40 = a_40 A + b_40 I = (-1)A + 0*I = -A.

Therefore, the sum A^27 + A^31 + A^40 = (-I) + A + (-A) = -I + A - A = -I. Wait, that's interesting. So the sum simplifies to -I?

Wait, let me check again:

A^27 = -I

A^31 = A

A^40 = -A

Therefore, adding them: (-I) + A + (-A) = -I + (A - A) = -I + 0 = -I. So the result is -I?

But let me verify this with some steps. Maybe I made a mistake in computing a_n and b_n?

Let me check for n=2. Earlier, we saw that a_2=1. Let me compute A^2 using the original matrix and check if it's equal to a_2 A + b_2 I.

Original matrix A is [[3,7],[-1,-2]]

Compute A^2:

A^2 = A*A = [[3*3 + 7*(-1), 3*7 + 7*(-2)], [-1*3 + (-2)*(-1), -1*7 + (-2)*(-2)]]

Compute first row first column: 9 -7 = 2

First row second column: 21 -14 = 7

Second row first column: -3 + 2 = -1

Second row second column: -7 + 4 = -3

So A^2 = [[2, 7], [-1, -3]]

Now, according to our formula, A^2 = a_2 A + b_2 I. a_2=1, b_2=?

From earlier, b_n = -a_{n-1}. For n=2, b_2 = -a_1 = -1. Therefore, A^2 = 1*A + (-1)*I = A - I.

Compute A - I: [[3 -1, 7 - 0], [-1 - 0, -2 -1]] = [[2, 7], [-1, -3]], which matches the direct computation. So that's correct.

Similarly, check A^3. According to the recurrence, A^3 = a_3 A + b_3 I. a_3=0, b_3=-a_2=-1. Therefore, A^3 = 0*A + (-1)*I = -I. Let me compute A^3 directly.

A^3 = A^2 * A = [[2,7],[-1,-3]] * [[3,7],[-1,-2]]

First row first column: 2*3 + 7*(-1) = 6 -7 = -1

First row second column: 2*7 + 7*(-2) = 14 -14 = 0

Second row first column: -1*3 + (-3)*(-1) = -3 + 3 = 0

Second row second column: -1*7 + (-3)*(-2) = -7 + 6 = -1

Therefore, A^3 = [[-1, 0], [0, -1]] = -I. Which matches the formula. So that's correct. Therefore, our formula seems to work.

Similarly, check A^4. According to the recurrence, a_4=-1, b_4=-a_3= -0=0. Therefore, A^4 = -1*A + 0*I = -A. Let me compute A^4 directly. A^4 = A^3 * A = (-I)*A = -A. Which is correct. So that's correct.

Therefore, our formula seems valid. Therefore, proceeding with that, the conclusion that A^27 = -I, A^31 = A, A^40 = -A, so their sum is -I + A - A = -I. Therefore, the answer is -I, which is [[-1, 0], [0, -1]].

But let me check for another exponent to be sure. Let's check A^6. According to the sequence, a_6 = 0, b_6 = -a_5 = -(-1) = 1. Therefore, A^6 = 0*A + 1*I = I. Let me compute A^6 directly. Since A^3 = -I, then A^6 = (A^3)^2 = (-I)^2 = I. Correct. So that's correct. Therefore, the formula works.

Therefore, going back, the conclusion that A^27 + A^31 + A^40 = -I is correct? Wait, but let me check the exponents again.

Wait, n=27: 27 mod 6 is 3, so a_27 = a_3 = 0, b_27 = -a_{26} = -a_{26}. 26 mod 6 is 2, so a_26 = a_2 = 1. Therefore, b_27 = -1. Therefore, A^27 = 0*A + (-1)I = -I. Correct.

n=31: 31 mod 6 is 1, so a_31 = a_1 = 1, b_31 = -a_{30} = -a_0 = -0 = 0. Therefore, A^31 = 1*A + 0*I = A. Correct.

n=40: 40 mod 6 is 4, so a_40 = a_4 = -1, b_40 = -a_{39} = -a_3 = -0 = 0. Therefore, A^40 = -1*A + 0*I = -A. Correct.

Therefore, adding them: -I + A - A = -I. Therefore, the answer is -I. Therefore, the matrix [[-1, 0], [0, -1]].

But let me check if there's another way to see this. Since A^2 = A - I, then A^3 = A^2 - A = (A - I) - A = -I. Then A^4 = A^3 * A = -I * A = -A. A^5 = A^4 * A = -A^2 = -(A - I) = -A + I. A^6 = A^5 * A = (-A + I) * A = -A^2 + A = -(A - I) + A = -A + I + A = I. Then A^6 = I. Therefore, the powers cycle every 6. Therefore, A^6 = I, so A^{6k} = I. Therefore, exponents can be reduced modulo 6.

Therefore, for example:

A^27: 27 divided by 6 is 4*6=24, remainder 3. So A^27 = A^{6*4 + 3} = (A^6)^4 * A^3 = I^4 * A^3 = A^3 = -I. Correct.

Similarly, A^31: 31 divided by 6 is 5*6=30, remainder 1. So A^31 = A^{6*5 +1} = (A^6)^5 * A^1 = I^5 * A = A. Correct.

A^40: 40 divided by 6 is 6*6=36, remainder 4. So A^40 = A^{6*6 + 4} = (A^6)^6 * A^4 = I^6 * A^4 = A^4 = -A. Correct.

Therefore, the same conclusion. Therefore, the sum is -I + A - A = -I. Therefore, the answer is -I. Therefore, the matrix [[-1, 0], [0, -1]].

But let me check once more with another exponent. For example, A^3 = -I, A^4 = -A, A^5 = -A + I, A^6 = I, A^7 = A, etc. So the cycle is indeed every 6. Therefore, the exponents 27, 31, 40 correspond to positions in the cycle:

27 mod 6 = 3: A^3 = -I

31 mod 6 = 1: A^1 = A

40 mod 6 = 4: A^4 = -A

Therefore, sum is -I + A - A = -I. So yes, that's correct.

Therefore, the answer is -I, which is the matrix with -1 on the diagonal and 0 elsewhere. Therefore, boxed, it would be \boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}.

But let me check once more if there's a possibility that I made a mistake in the sign somewhere. For example, when computing b_n. Let me check for n=27, b_27 = -a_{26}. Since a_{26} is a_{26 mod 6} = a_2 = 1. Therefore, b_27 = -1. Therefore, A^27 = 0*A + (-1)I = -I. Correct. Similarly, for n=40, b_40 = -a_{39} = -a_{3} = -0 = 0. Correct. So A^40 = -A. Correct. So the sum is indeed -I + A - A = -I. Therefore, I think this is correct.

Alternatively, maybe the problem expects a different form? Let me check if -I is indeed the correct answer. Let me compute A^27 + A^31 + A^40 using another approach. For example, since A^6 = I, then A^27 = A^{6*4 + 3} = (A^6)^4 * A^3 = I^4 * A^3 = A^3 = -I. Similarly, A^31 = A^{6*5 + 1} = A^1 = A. A^40 = A^{6*6 + 4} = A^4 = -A. Therefore, sum is -I + A - A = -I. So same result. Therefore, I think this is correct.

Therefore, the answer is -I, which is [[-1, 0], [0, -1]]. Therefore, boxed as above.

**Final Answer**
\boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}

To compute the matrix expression
$$
\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40}
$$
where
$$
\mathbf{A} = \begin{pmatrix} 3 & 7 \\ -1 & -2 \end{pmatrix},
$$
we begin by analyzing the structure of the matrix and its powers.

---

### Step 1: Use the Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem states that a matrix satisfies its own characteristic equation. For matrix $\mathbf{A}$, we compute the characteristic polynomial:

$$
\det(\mathbf{A} - \lambda \mathbf{I}) = \det\begin{pmatrix} 3 - \lambda & 7 \\ -1 & -2 - \lambda \end{pmatrix}
= (3 - \lambda)(-2 - \lambda) - (7)(-1)
= \lambda^2 - \lambda + 1.
$$

Thus, the characteristic equation is:
$$
\lambda^2 - \lambda + 1 = 0.
$$

By the Cayley-Hamilton Theorem, we have:
$$
\mathbf{A}^2 - \mathbf{A} + \mathbf{I} = \mathbf{0} \quad \Rightarrow \quad \mathbf{A}^2 = \mathbf{A} - \mathbf{I}.
$$

This recurrence allows us to express higher powers of $\mathbf{A}$ in terms of $\mathbf{A}$ and $\mathbf{I}$.

---

### Step 2: Find a Recurrence for Powers of $\mathbf{A}$

Let us define:
$$
\mathbf{A}^n = a_n \mathbf{A} + b_n \mathbf{I}.
$$

Using the recurrence $\mathbf{A}^2 = \mathbf{A} - \mathbf{I}$, we derive the following recurrence relations for the coefficients $a_n$ and $b_n$:

$$
a_{n+1} = a_n + b_n, \quad b_{n+1} = -a_n.
$$

With initial conditions:
$$
a_0 = 0, \quad b_0 = 1, \quad a_1 = 1, \quad b_1 = 0.
$$

This recurrence leads to a periodic sequence for $a_n$ and $b_n$ with period 6. The sequence of $a_n$ is:

$$
a_0 = 0, \quad a_1 = 1, \quad a_2 = 1, \quad a_3 = 0, \quad a_4 = -1, \quad a_5 = -1, \quad a_6 = 0, \quad \text{and so on.}
$$

Similarly, $b_n = -a_{n-1}$, so the sequence for $b_n$ is also periodic with the same period.

---

### Step 3: Compute the Required Powers

We now compute the values of $a_n$ and $b_n$ for the exponents 27, 31, and 40.

- $27 \mod 6 = 3 \Rightarrow a_{27} = a_3 = 0, \quad b_{27} = -a_{26} = -a_2 = -1$
- $31 \mod 6 = 1 \Rightarrow a_{31} = a_1 = 1, \quad b_{31} = -a_{30} = -a_0 = 0$
- $40 \mod 6 = 4 \Rightarrow a_{40} = a_4 = -1, \quad b_{40} = -a_{39} = -a_3 = 0$

Thus, we have:
- $\mathbf{A}^{27} = a_{27} \mathbf{A} + b_{27} \mathbf{I} = 0 \cdot \mathbf{A} - 1 \cdot \mathbf{I} = -\mathbf{I}$
- $\mathbf{A}^{31} = a_{31} \mathbf{A} + b_{31} \mathbf{I} = 1 \cdot \mathbf{A} + 0 \cdot \mathbf{I} = \mathbf{A}$
- $\mathbf{A}^{40} = a_{40} \mathbf{A} + b_{40} \mathbf{I} = -1 \cdot \mathbf{A} + 0 \cdot \mathbf{I} = -\mathbf{A}$

---

### Step 4: Add the Results

Now compute the sum:
$$
\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40} = -\mathbf{I} + \mathbf{A} - \mathbf{A} = -\mathbf{I}.
$$

---

### Final Answer

$$
\boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}
$$

Solution:
Note that
\[\mathbf{A}^2 = \begin{pmatrix} 2 & 7 \\ -1 & -3 \end{pmatrix}\]and
\[\mathbf{A}^3 = \mathbf{A} \mathbf{A}^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -\mathbf{I}.\]Then
\begin{align*}
\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40} &= (\mathbf{A}^3)^9 + (\mathbf{A}^3)^{10} \mathbf{A} + (\mathbf{A}^3)^{13} \mathbf{A} \\
&= (-\mathbf{I})^9 + (-\mathbf{I})^{10} \mathbf{A} + (-\mathbf{I})^{13} \mathbf{A} \\
&= -\mathbf{I} + \mathbf{A} - \mathbf{A} \\
&= -\mathbf{I} = \boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}.
\end{align*}


Explanation:
Okay, so I need to find an odd integer t between 0 and 23 (so t is from 1 to 22, but since it's odd, t is 1, 3, 5,...,21) such that t + 2 is the inverse of t modulo 23. Hmm, let me recall what an inverse modulo means.

The inverse of a number t modulo 23 is another number x such that t * x ≡ 1 mod 23. So in this case, the problem says that t + 2 is the inverse of t. That means t * (t + 2) ≡ 1 mod 23. So I need to solve the congruence equation t(t + 2) ≡ 1 mod 23.

Let me write that out: t² + 2t ≡ 1 mod 23. If I rearrange that, it becomes t² + 2t - 1 ≡ 0 mod 23. So this is a quadratic congruence equation. Maybe I can solve it by completing the square or using the quadratic formula? Let me try completing the square first.

Starting with t² + 2t - 1 ≡ 0 mod 23. Let me add 1 to both sides to get t² + 2t ≡ 1 mod 23. Then, completing the square: take half of the coefficient of t, which is 1, square it, so 1² = 1. So add 1 to both sides: t² + 2t + 1 ≡ 1 + 1 mod 23. That gives (t + 1)² ≡ 2 mod 23.

So now, the equation is (t + 1)² ≡ 2 mod 23. Therefore, t + 1 is a square root of 2 modulo 23. So I need to find integers x such that x² ≡ 2 mod 23, then t = x - 1. Then check which of those t's are odd and between 0 and 23.

But how do I find square roots modulo 23? Let me recall that 23 is a prime number, so maybe I can use some methods for finding square roots modulo primes. One method is to check all numbers from 1 to 22 and see which ones squared give 2 mod 23. Since 23 is not too big, this might be feasible.

Alternatively, maybe there's a formula or something. Let me try the brute force approach first. Let me compute squares modulo 23:

Let me list squares mod 23:

0² = 0

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25 ≡ 25 - 23 = 2

Oh! Wait, 5² is 25, which is 2 mod 23. So 5 is a square root of 2 mod 23. Therefore, x ≡ 5 or x ≡ -5 mod 23. Since -5 mod 23 is 23 - 5 = 18. So the square roots of 2 mod 23 are 5 and 18.

Therefore, t + 1 ≡ 5 mod 23 or t + 1 ≡ 18 mod 23. Therefore, solving for t:

Case 1: t + 1 ≡ 5 mod 23 ⇒ t ≡ 5 - 1 = 4 mod 23. So t = 4 or 4 + 23k. But since t is between 0 and 23, t = 4.

Case 2: t + 1 ≡ 18 mod 23 ⇒ t ≡ 18 - 1 = 17 mod 23. So t = 17 or 17 + 23k. Again, between 0 and 23, t = 17.

So the possible solutions are t = 4 and t = 17. But the problem says that t must be odd. Let me check: 4 is even, 17 is odd. Therefore, the unique odd integer t is 17.

Wait, but let me verify this. Let me check if t = 17 satisfies the original condition.

First, check if t + 2 is the inverse of t modulo 23. So t + 2 = 17 + 2 = 19. So we need to check if 17 * 19 ≡ 1 mod 23. Let me compute 17 * 19.

17 * 19: Let me compute 17*20 = 340, subtract 17: 340 - 17 = 323. Now divide 323 by 23 and find the remainder.

Let me compute 23*14 = 322. Therefore, 323 - 322 = 1. Therefore, 323 ≡ 1 mod 23. So yes, 17 * 19 ≡ 1 mod 23. Therefore, 19 is indeed the inverse of 17 mod 23.

So t = 17 is a solution. Also, t = 4: check if t + 2 = 6 is the inverse of 4 mod 23. Let me check 4 * 6 = 24 ≡ 1 mod 23? 24 - 23 = 1, yes. So 4 * 6 ≡ 1 mod 23. Therefore, t = 4 is also a solution. But since the problem says "unique odd integer t", and 4 is even, so 17 is the answer.

But let me make sure there are no other solutions. Since we had the equation (t + 1)^2 ≡ 2 mod 23, which had two solutions, 5 and 18 mod 23, leading to t = 4 and t = 17. Since 23 is prime, the equation x² ≡ a mod p has either 0 or 2 solutions. Since 2 is a quadratic residue mod 23 (as we found 5² ≡ 2), there are exactly two solutions. Therefore, only t = 4 and t = 17. Since the problem says "unique odd integer", so 17 is the answer.

But just to be thorough, let me check if there are any other square roots of 2 mod 23. For example, maybe I missed some numbers? Let me check squares again:

Compute squares mod 23:

0² = 0

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25 ≡ 2

6² = 36 ≡ 36 - 23 = 13

7² = 49 ≡ 49 - 2*23 = 49 - 46 = 3

8² = 64 ≡ 64 - 2*23 = 64 - 46 = 18

9² = 81 ≡ 81 - 3*23 = 81 - 69 = 12

10² = 100 ≡ 100 - 4*23 = 100 - 92 = 8

11² = 121 ≡ 121 - 5*23 = 121 - 115 = 6

12² = 144 ≡ 144 - 6*23 = 144 - 138 = 6

Wait, 12² is 144. 23*6 is 138, so 144 - 138 = 6. So 12² ≡ 6 mod 23.

13²: 13² = 169. 23*7 = 161, so 169 - 161 = 8. So 13² ≡ 8 mod 23.

14²: 14² = 196. 23*8 = 184, 196 - 184 = 12. So 14² ≡ 12 mod 23.

15²: 225. 23*9 = 207, 225 - 207 = 18. So 15² ≡ 18 mod 23.

16²: 256. 23*11 = 253, 256 - 253 = 3. So 16² ≡ 3 mod 23.

17²: 289. 23*12 = 276, 289 - 276 = 13. So 17² ≡ 13 mod 23.

18²: 324. 23*14 = 322, 324 - 322 = 2. So 18² ≡ 2 mod 23.

19²: 361. 23*15 = 345, 361 - 345 = 16. So 19² ≡ 16 mod 23.

20²: 400. 23*17 = 391, 400 - 391 = 9. So 20² ≡ 9 mod 23.

21²: 441. 23*19 = 437, 441 - 437 = 4. So 21² ≡ 4 mod 23.

22²: 484. 23*21 = 483, 484 - 483 = 1. So 22² ≡ 1 mod 23.

So indeed, only 5² and 18² ≡ 2 mod 23. Therefore, only two solutions. Therefore, t = 4 and t = 17. Since the problem asks for the unique odd integer t, which is 17.

But just to make sure, let me check if there's another way to approach this problem. Maybe using the inverse formula?

The inverse of t modulo 23 is a number x such that t * x ≡ 1 mod 23. The problem says that x = t + 2. Therefore, t*(t + 2) ≡ 1 mod 23. Which is the same equation as before. So solving t² + 2t - 1 ≡ 0 mod 23.

Alternatively, maybe using the quadratic formula? In modular arithmetic, the quadratic formula works if the modulus is prime and the discriminant is a quadratic residue.

So for the equation t² + 2t - 1 ≡ 0 mod 23, the discriminant D is b² - 4ac = (2)² - 4*1*(-1) = 4 + 4 = 8. So D = 8.

So solutions are t ≡ [-b ± √D]/2a mod 23. Here, a = 1, b = 2, so t ≡ [-2 ± √8]/2 mod 23.

So need to compute √8 mod 23. Let me check if 8 is a quadratic residue mod 23. Let me check squares again. From earlier, squares mod 23 are 0,1,4,9,16,2,13,18,12,8,6, etc. Wait, looking back:

Earlier, when I computed squares, 10² ≡ 8 mod 23. So 10² ≡ 8 mod 23. Therefore, √8 mod 23 is 10 or -10 mod 23. -10 mod 23 is 13. So √8 ≡ 10 or 13 mod 23.

Therefore, solutions are t ≡ [-2 ± 10]/2 mod 23 and t ≡ [-2 ± 13]/2 mod 23. Let me compute each case:

First case: [-2 + 10]/2 = 8/2 = 4 mod 23.

Second case: [-2 - 10]/2 = -12/2 = -6 mod 23. -6 mod 23 is 17.

Third case: [-2 + 13]/2 = 11/2 mod 23. Wait, but 11/2 is equivalent to 11 * 2^{-1} mod 23. What is the inverse of 2 mod 23? Since 2*12 = 24 ≡ 1 mod 23, so 2^{-1} ≡ 12 mod 23. Therefore, 11 * 12 = 132. 132 divided by 23: 23*5 = 115, 132 - 115 = 17. So 11/2 ≡ 17 mod 23.

Fourth case: [-2 -13]/2 = -15/2 mod 23. Similarly, -15/2 = (-15)*12 mod 23. Let me compute -15*12: -180. Now, -180 mod 23. Let me divide 180 by 23. 23*7 = 161, 23*8 = 184. So 180 - 161 = 19. Therefore, -180 ≡ -19 mod 23. Which is 23 - 19 = 4 mod 23.

So the solutions are t ≡ 4, 17 mod 23. Which matches the previous result. Therefore, again, t = 4 and t = 17.

So again, since we need an odd integer, t = 17 is the answer.

Alternatively, maybe I can check all odd numbers between 1 and 22 and see which one satisfies the condition. Let me try that as another verification.

List of odd t: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21.

For each t, compute t + 2, then check if t*(t + 2) ≡ 1 mod 23.

Let me do this step by step:

t = 1: t + 2 = 3. 1*3 = 3 ≡ 3 mod 23 ≠ 1. Not good.

t = 3: t + 2 = 5. 3*5 = 15 ≡ 15 ≠ 1.

t = 5: t + 2 = 7. 5*7 = 35. 35 - 23 = 12 ≡ 12 ≠ 1.

t = 7: t + 2 = 9. 7*9 = 63. 63 - 2*23 = 63 - 46 = 17 ≡ 17 ≠ 1.

t = 9: t + 2 = 11. 9*11 = 99. 99 - 4*23 = 99 - 92 = 7 ≡ 7 ≠ 1.

t = 11: t + 2 = 13. 11*13 = 143. 143 - 6*23 = 143 - 138 = 5 ≡ 5 ≠ 1.

t = 13: t + 2 = 15. 13*15 = 195. 195 - 8*23 = 195 - 184 = 11 ≡ 11 ≠ 1.

t = 15: t + 2 = 17. 15*17 = 255. 255 - 11*23 = 255 - 253 = 2 ≡ 2 ≠ 1.

t = 17: t + 2 = 19. 17*19 = 323. 323 - 14*23 = 323 - 322 = 1 ≡ 1. Bingo! So this works.

t = 19: t + 2 = 21. 19*21 = 399. 399 - 17*23 = 399 - 391 = 8 ≡ 8 ≠ 1.

t = 21: t + 2 = 23. 21*23 = 483. 483 - 21*23 = 0 ≡ 0 ≠ 1.

So only t = 17 works. Therefore, the answer is 17.

Therefore, after multiple methods: solving the quadratic congruence, checking square roots, quadratic formula, and brute force checking all odd numbers, I arrive at the same conclusion. So I can be confident that the unique odd integer t is 17.

**Final Answer**
The unique odd integer $ t $ is \boxed{17}.

To solve the problem, we are given that $ t $ is an **odd integer** such that $ 0 < t < 23 $, and that $ t + 2 $ is the **inverse** of $ t $ modulo 23. That is, we are looking for an integer $ t $ satisfying:

$$
t \cdot (t + 2) \equiv 1 \pmod{23}
$$

---

### Step 1: Translate the condition into a congruence

We start by expanding the left-hand side:

$$
t(t + 2) = t^2 + 2t
$$

So the condition becomes:

$$
t^2 + 2t \equiv 1 \pmod{23}
$$

Rewriting:

$$
t^2 + 2t - 1 \equiv 0 \pmod{23}
$$

This is a quadratic congruence in $ t $.

---

### Step 2: Solve the quadratic congruence

We can complete the square to simplify the equation:

$$
t^2 + 2t - 1 \equiv 0 \pmod{23}
$$

Add 1 to both sides:

$$
t^2 + 2t \equiv 1 \pmod{23}
$$

Now complete the square:

$$
t^2 + 2t + 1 \equiv 2 \pmod{23}
\Rightarrow (t + 1)^2 \equiv 2 \pmod{23}
$$

So we are looking for values of $ t $ such that $ t + 1 $ is a square root of 2 modulo 23.

---

### Step 3: Find square roots of 2 modulo 23

We test small integers modulo 23 to find values $ x $ such that $ x^2 \equiv 2 \pmod{23} $:

- $ 5^2 = 25 \equiv 2 \pmod{23} $
- $ (-5) \equiv 18 \pmod{23} $, and $ 18^2 = 324 \equiv 2 \pmod{23} $

So the square roots of 2 modulo 23 are $ 5 $ and $ 18 $.

Thus:

$$
t + 1 \equiv 5 \pmod{23} \Rightarrow t \equiv 4 \pmod{23}
$$
$$
t + 1 \equiv 18 \pmod{23} \Rightarrow t \equiv 17 \pmod{23}
$$

So the possible values of $ t $ are $ 4 $ and $ 17 $.

---

### Step 4: Choose the odd value

We are told that $ t $ must be **odd** and satisfy $ 0 < t < 23 $. Among the two possible values:

- $ t = 4 $ is even
- $ t = 17 $ is odd

So the only valid solution is:

$$
\boxed{17}
$$

---

### Step 5: Verify the solution

Let’s verify that $ t = 17 $ satisfies the original condition:

- $ t + 2 = 19 $
- $ 17 \cdot 19 = 323 $
- $ 323 \mod 23 = 1 $ (since $ 23 \cdot 14 = 322 $, and $ 323 - 322 = 1 $)

So indeed, $ 17 \cdot 19 \equiv 1 \pmod{23} $, confirming that $ t = 17 $ is the inverse of $ t + 2 $ modulo 23.

---

### Final Answer

$$
\boxed{17}
$$

[Assistant]
Explanation:
Okay, so I need to find an odd integer t between 0 and 23 (so t is from 1 to 22, but since it's odd, t is 1, 3, 5,...,21) such that t + 2 is the inverse of t modulo 23. Hmm, let me recall what an inverse modulo means.

The inverse of a number t modulo 23 is another number x such that t * x ≡ 1 mod 23. So in this case, the problem says that t + 2 is the inverse of t. That means t * (t + 2) ≡ 1 mod 23. So I need to solve the congruence equation t(t + 2) ≡ 1 mod 23.

Let me write that out: t² + 2t ≡ 1 mod 23. If I rearrange that, it becomes t² + 2t - 1 ≡ 0 mod 23. So this is a quadratic congruence equation. Maybe I can solve it by completing the square or using the quadratic formula? Let me try completing the square first.

Starting with t² + 2t - 1 ≡ 0 mod 23. Let me add 1 to both sides to get t² + 2t ≡ 1 mod 23. Then, completing the square: take half of the coefficient of t, which is 1, square it, so 1² = 1. So add 1 to both sides: t² + 2t + 1 ≡ 1 + 1 mod 23. That gives (t + 1)² ≡ 2 mod 23.

So now, the equation is (t + 1)² ≡ 2 mod 23. Therefore, t + 1 is a square root of 2 modulo 23. So I need to find integers x such that x² ≡ 2 mod 23, then t = x - 1. Then check which of those t's are odd and between 0 and 23.

But how do I find square roots modulo 23? Let me recall that 23 is a prime number, so maybe I can use some methods for finding square roots modulo primes. One method is to check all numbers from 1 to 22 and see which ones squared give 2 mod 23. Since 23 is not too big, this might be feasible.

Alternatively, maybe there's a formula or something. Let me try the brute force approach first. Let me compute squares modulo 23:

Let me list squares mod 23:

0² = 0

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25 ≡ 25 - 23 = 2

Oh! Wait, 5² is 25, which is 2 mod 23. So 5 is a square root of 2 mod 23. Therefore, x ≡ 5 or x ≡ -5 mod 23. Since -5 mod 23 is 23 - 5 = 18. So the square roots of 2 mod 23 are 5 and 18.

Therefore, t + 1 ≡ 5 mod 23 or t + 1 ≡ 18 mod 23. Therefore, solving for t:

Case 1: t + 1 ≡ 5 mod 23 ⇒ t ≡ 5 - 1 = 4 mod 23. So t = 4 or 4 + 23k. But since t is between 0 and 23, t = 4.

Case 2: t + 1 ≡ 18 mod 23 ⇒ t ≡ 18 - 1 = 17 mod 23. So t = 17 or 17 + 23k. Again, between 0 and 23, t = 17.

So the possible solutions are t = 4 and t = 17. But the problem says that t must be odd. Let me check: 4 is even, 17 is odd. Therefore, the unique odd integer t is 17.

Wait, but let me verify this. Let me check if t = 17 satisfies the original condition.

First, check if t + 2 is the inverse of t modulo 23. So t + 2 = 17 + 2 = 19. So we need to check if 17 * 19 ≡ 1 mod 23. Let me compute 17 * 19.

17 * 19: Let me compute 17*20 = 340, subtract 17: 340 - 17 = 323. Now divide 323 by 23 and find the remainder.

Let me compute 23*14 = 322. Therefore, 323 - 322 = 1. Therefore, 323 ≡ 1 mod 23. So yes, 17 * 19 ≡ 1 mod 23. Therefore, 19 is indeed the inverse of 17 mod 23.

So t = 17 is a solution. Also, t = 4: check if t + 2 = 6 is the inverse of 4 mod 23. Let me check 4 * 6 = 24 ≡ 1 mod 23? 24 - 23 = 1, yes. So 4 * 6 ≡ 1 mod 23. Therefore, t = 4 is also a solution. But since the problem says "unique odd integer t", and 4 is even, so 17 is the answer.

But let me make sure there are no other solutions. Since we had the equation (t + 1)^2 ≡ 2 mod 23, which had two solutions, 5 and 18 mod 23, leading to t = 4 and t = 17. Since 23 is prime, the equation x² ≡ a mod p has either 0 or 2 solutions. Since 2 is a quadratic residue mod 23 (as we found 5² ≡ 2), there are exactly two solutions. Therefore, only t = 4 and t = 17. Since the problem says "unique odd integer", so 17 is the answer.

But just to be thorough, let me check if there are any other square roots of 2 mod 23. For example, maybe I missed some numbers? Let me check squares again:

Compute squares mod 23:

0² = 0

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25 ≡ 2

6² = 36 ≡ 36 - 23 = 13

7² = 49 ≡ 49 - 2*23 = 49 - 46 = 3

8² = 64 ≡ 64 - 2*23 = 64 - 46 = 18

9² = 81 ≡ 81 - 3*23 = 81 - 69 = 12

10² = 100 ≡ 100 - 4*23 = 100 - 92 = 8

11² = 121 ≡ 121 - 5*23 = 121 - 115 = 6

12² = 144 ≡ 144 - 6*23 = 144 - 138 = 6

Wait, 12² is 144. 23*6 is 138, so 144 - 138 = 6. So 12² ≡ 6 mod 23.

13²: 13² = 169. 23*7 = 161, so 169 - 161 = 8. So 13² ≡ 8 mod 23.

14²: 14² = 196. 23*8 = 184, 196 - 184 = 12. So 14² ≡ 12 mod 23.

15²: 225. 23*9 = 207, 225 - 207 = 18. So 15² ≡ 18 mod 23.

16²: 256. 23*11 = 253, 256 - 253 = 3. So 16² ≡ 3 mod 23.

17²: 289. 23*12 = 276, 289 - 276 = 13. So 17² ≡ 13 mod 23.

18²: 324. 23*14 = 322, 324 - 322 = 2. So 18² ≡ 2 mod 23.

19²: 361. 23*15 = 345, 361 - 345 = 16. So 19² ≡ 16 mod 23.

20²: 400. 23*17 = 391, 400 - 391 = 9. So 20² ≡ 9 mod 23.

21²: 441. 23*19 = 437, 441 - 437 = 4. So 21² ≡ 4 mod 23.

22²: 484. 23*21 = 483, 484 - 483 = 1. So 22² ≡ 1 mod 23.

So indeed, only 5² and 18² ≡ 2 mod 23. Therefore, only two solutions. Therefore, t = 4 and t = 17. Since the problem asks for the unique odd integer t, which is 17.

But just to make sure, let me check if there's another way to approach this problem. Maybe using the inverse formula?

The inverse of t modulo 23 is a number x such that t * x ≡ 1 mod 23. The problem says that x = t + 2. Therefore, t*(t + 2) ≡ 1 mod 23. Which is the same equation as before. So solving t² + 2t - 1 ≡ 0 mod 23.

Alternatively, maybe using the quadratic formula? In modular arithmetic, the quadratic formula works if the modulus is prime and the discriminant is a quadratic residue.

So for the equation t² + 2t - 1 ≡ 0 mod 23, the discriminant D is b² - 4ac = (2)² - 4*1*(-1) = 4 + 4 = 8. So D = 8.

So solutions are t ≡ [-b ± √D]/2a mod 23. Here, a = 1, b = 2, so t ≡ [-2 ± √8]/2 mod 23.

So need to compute √8 mod 23. Let me check if 8 is a quadratic residue mod 23. Let me check squares again. From earlier, squares mod 23 are 0,1,4,9,16,2,13,18,12,8,6, etc. Wait, looking back:

Earlier, when I computed squares, 10² ≡ 8 mod 23. So 10² ≡ 8 mod 23. Therefore, √8 mod 23 is 10 or -10 mod 23. -10 mod 23 is 13. So √8 ≡ 10 or 13 mod 23.

Therefore, solutions are t ≡ [-2 ± 10]/2 mod 23 and t ≡ [-2 ± 13]/2 mod 23. Let me compute each case:

First case: [-2 + 10]/2 = 8/2 = 4 mod 23.

Second case: [-2 - 10]/2 = -12/2 = -6 mod 23. -6 mod 23 is 17.

Third case: [-2 + 13]/2 = 11/2 mod 23. Wait, but 11/2 is equivalent to 11 * 2^{-1} mod 23. What is the inverse of 2 mod 23? Since 2*12 = 24 ≡ 1 mod 23, so 2^{-1} ≡ 12 mod 23. Therefore, 11 * 12 = 132. 132 divided by 23: 23*5 = 115, 132 - 115 = 17. So 11/2 ≡ 17 mod 23.

Fourth case: [-2 -13]/2 = -15/2 mod 23. Similarly, -15/2 = (-15)*12 mod 23. Let me compute -15*12: -180. Now, -180 mod 23. Let me divide 180 by 23. 23*7 = 161, 23*8 = 184. So 180 - 161 = 19. Therefore, -180 ≡ -19 mod 23. Which is 23 - 19 = 4 mod 23.

So the solutions are t ≡ 4, 17 mod 23. Which matches the previous result. Therefore, again, t = 4 and t = 17.

So again, since we need an odd integer, t = 17 is the answer.

Alternatively, maybe I can check all odd numbers between 1 and 22 and see which one satisfies the condition. Let me try that as another verification.

List of odd t: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21.

For each t, compute t + 2, then check if t*(t + 2) ≡ 1 mod 23.

Let me do this step by step:

t = 1: t + 2 = 3. 1*3 = 3 ≡ 3 mod 23 ≠ 1. Not good.

t = 3: t + 2 = 5. 3*5 = 15 ≡ 15 ≠ 1.

t = 5: t + 2 = 7. 5*7 = 35. 35 - 23 = 12 ≡ 12 ≠ 1.

t = 7: t + 2 = 9. 7*9 = 63. 63 - 2*23 = 63 - 46 = 17 ≡ 17 ≠ 1.

t = 9: t + 2 = 11. 9*11 = 99. 99 - 4*23 = 99 - 92 = 7 ≡ 7 ≠ 1.

t = 11: t + 2 = 13. 11*13 = 143. 143 - 6*23 = 143 - 138 = 5 ≡ 5 ≠ 1.

t = 13: t + 2 = 15. 13*15 = 195. 195 - 8*23 = 195 - 184 = 11 ≡ 11 ≠ 1.

t = 15: t + 2 = 17. 15*17 = 255. 255 - 11*23 = 255 - 253 = 2 ≡ 2 ≠ 1.

t = 17: t + 2 = 19. 17*19 = 323. 323 - 14*23 = 323 - 322 = 1 ≡ 1. Bingo! So this works.

t = 19: t + 2 = 21. 19*21 = 399. 399 - 17*23 = 399 - 391 = 8 ≡ 8 ≠ 1.

t = 21: t + 2 = 23. 21*23 = 483. 483 - 21*23 = 0 ≡ 0 ≠ 1.

So only t = 17 works. Therefore, the answer is 17.

Therefore, after multiple methods: solving the quadratic congruence, checking square roots, quadratic formula, and brute force checking all odd numbers, I arrive at the same conclusion. So I can be confident that the unique odd integer t is 17.

**Final Answer**
The unique odd integer $ t $ is \boxed{17}.

To solve the problem, we are given that $ t $ is an **odd integer** such that $ 0 < t < 23 $, and that $ t + 2 $ is the **inverse** of $ t $ modulo 23. That is, we are looking for an integer $ t $ satisfying:

$$
t \cdot (t + 2) \equiv 1 \pmod{23}
$$

---

### Step 1: Translate the condition into a congruence

We start by expanding the left-hand side:

$$
t(t + 2) = t^2 + 2t
$$

So the condition becomes:

$$
t^2 + 2t \equiv 1 \pmod{23}
$$

Rewriting:

$$
t^2 + 2t - 1 \equiv 0 \pmod{23}
$$

This is a quadratic congruence in $ t $.

---

### Step 2: Solve the quadratic congruence

We can complete the square to simplify the equation:

$$
t^2 + 2t - 1 \equiv 0 \pmod{23}
$$

Add 1 to both sides:

$$
t^2 + 2t \equiv 1 \pmod{23}
$$

Now complete the square:

$$
t^2 + 2t + 1 \equiv 2 \pmod{23}
\Rightarrow (t + 1)^2 \equiv 2 \pmod{23}
$$

So we are looking for values of $ t $ such that $ t + 1 $ is a square root of 2 modulo 23.

---

### Step 3: Find square roots of 2 modulo 23

We test small integers modulo 23 to find values $ x $ such that $ x^2 \equiv 2 \pmod{23} $:

- $ 5^2 = 25 \equiv 2 \pmod{23} $
- $ (-5) \equiv 18 \pmod{23} $, and $ 18^2 = 324 \equiv 2 \pmod{23} $

So the square roots of 2 modulo 23 are $ 5 $ and $ 18 $.

Thus:

$$
t + 1 \equiv 5 \pmod{23} \Rightarrow t \equiv 4 \pmod{23}
$$
$$
t + 1 \equiv 18 \pmod{23} \Rightarrow t \equiv 17 \pmod{23}
$$

So the possible values of $ t $ are $ 4 $ and $ 17 $.

---

### Step 4: Choose the odd value

We are told that $ t $ must be **odd** and satisfy $ 0 < t < 23 $. Among the two possible values:

- $ t = 4 $ is even
- $ t = 17 $ is odd

So the only valid solution is:

$$
\boxed{17}
$$

---

### Step 5: Verify the solution

Let’s verify that $ t = 17 $ satisfies the original condition:

- $ t + 2 = 19 $
- $ 17 \cdot 19 = 323 $
- $ 323 \mod 23 = 1 $ (since $ 23 \cdot 14 = 322 $, and $ 323 - 322 = 1 $)

So indeed, $ 17 \cdot 19 \equiv 1 \pmod{23} $, confirming that $ t = 17 $ is the inverse of $ t + 2 $ modulo 23.

---

### Final Answer

$$
\boxed{17}
$$

Solution:
We could find the answer by trial and error -- testing each candidate $t$ to see if $t\cdot (t+2)\equiv 1\pmod{23}$. However, here is another way:

We can easily see that $4\cdot 6=24\equiv 1\pmod{23}$, so $4$ fulfills the main requirement that its inverse is $2$ more than it. Unfortunately, $4$ isn't odd. But we also have \begin{align*}
(-4)\cdot (-6) &= 4\cdot 6 \\
&\equiv 1\pmod{23},
\end{align*} so $-4$ and $-6$ are each other's inverses $\pmod{23}$. Since $-4\equiv 19\pmod{23}$ and $-6\equiv 17\pmod{23}$, the answer $t=\boxed{17}$ satisfies the requirements of the problem.

(We can even check that $17\cdot 19 = 323 = 14\cdot 23 + 1$.)


Explanation:
Okay, so I need to find the x-intercept (a), the y-intercept (b), and the slope (m) of the line given by the equation x/4 + y/12 = 1. Then add them all together for the final answer. Let me start by recalling what each of these terms means.

First, the x-intercept is the point where the line crosses the x-axis. At that point, the y-coordinate is 0. Similarly, the y-intercept is where the line crosses the y-axis, so the x-coordinate there is 0. The slope, m, is the steepness of the line, calculated as rise over run, or change in y over change in x.

The given equation is x/4 + y/12 = 1. Hmm, this looks like the standard form of a line's intercept form. I remember that the intercept form of a line is x/a + y/b = 1, where a is the x-intercept and b is the y-intercept. So, comparing that to the given equation, it seems like a is 4 and b is 12? Wait, let me check that again.

If the equation is x/4 + y/12 = 1, then yes, that would mean the x-intercept is 4 because when y=0, x/4 = 1, so x=4. Similarly, when x=0, y/12 = 1, so y=12. Therefore, a is 4 and b is 12. That seems straightforward. So, a = 4 and b = 12. But let me confirm by solving for intercepts manually just to be sure.

For the x-intercept, set y = 0:

x/4 + 0/12 = 1 => x/4 = 1 => x = 4. Yep, that's correct. So a = 4.

For the y-intercept, set x = 0:

0/4 + y/12 = 1 => y/12 = 1 => y = 12. So b = 12. Okay, that checks out.

Now, the slope. How do I find the slope from this equation? Well, the standard form of a line is usually written as y = mx + c, where m is the slope and c is the y-intercept. So maybe I should convert the given equation into that slope-intercept form.

Starting with x/4 + y/12 = 1. Let me solve for y.

First, subtract x/4 from both sides:

y/12 = 1 - x/4

Then multiply both sides by 12 to solve for y:

y = 12*(1 - x/4) => y = 12*1 - 12*(x/4) => y = 12 - 3x

So, y = -3x + 12. Therefore, comparing to y = mx + c, the slope m is -3, and the y-intercept is 12. Wait, but we already found the y-intercept (b) as 12, which matches. So that's consistent.

Therefore, m = -3. So, putting it all together, a = 4, b = 12, m = -3. Therefore, a + b + m = 4 + 12 + (-3) = 4 + 12 is 16, minus 3 is 13. So the answer should be 13?

Wait, let me check again. Maybe I made a mistake in calculating the slope? Let me verify.

Original equation: x/4 + y/12 = 1. Let me rearrange it again step by step.

Multiply both sides by 12 to eliminate denominators:

12*(x/4) + 12*(y/12) = 12*1 => 3x + y = 12. So, 3x + y = 12. Then solving for y: y = -3x + 12. Yep, same result. So slope is -3. So that's correct.

Alternatively, another way to find the slope is using the intercept form. The intercept form is x/a + y/b = 1, and if we want to convert that to slope-intercept form, we can do:

y/b = 1 - x/a => y = (-b/a)x + b. Therefore, the slope m is -b/a. In this case, a is 4, b is 12, so m = -12/4 = -3. Yep, same answer. So that's correct.

Therefore, a = 4, b = 12, m = -3. Adding them together: 4 + 12 + (-3) = 13. So the answer is 13. Wait, but let me check once more to make sure I didn't mix up any numbers.

Original equation: x/4 + y/12 = 1. So intercepts are 4 and 12. Slope is -3. So 4 + 12 is 16, minus 3 is 13. Seems right.

Alternatively, maybe I can check with another method. Let's find two points on the line and compute the slope.

We already know the intercepts: (4, 0) and (0, 12). So the slope between these two points is (12 - 0)/(0 - 4) = 12 / (-4) = -3. Yep, same result. So that's correct.

Therefore, all steps check out. So the value of a + b + m is 4 + 12 + (-3) = 13. Therefore, the answer is 13.

**Final Answer**
The value of $a + b + m$ is \boxed{13}.

To solve the problem, we are given the equation of a line:

$$
\frac{x}{4} + \frac{y}{12} = 1
$$

We are asked to find the **x-intercept** $ a $, the **y-intercept** $ b $, and the **slope** $ m $, and then compute the sum $ a + b + m $.

---

### Step 1: Identify the x-intercept $ a $

The **x-intercept** is the point where the line crosses the x-axis, i.e., where $ y = 0 $. Substituting $ y = 0 $ into the equation:

$$
\frac{x}{4} + \frac{0}{12} = 1 \Rightarrow \frac{x}{4} = 1 \Rightarrow x = 4
$$

So, the x-intercept is:

$$
a = 4
$$

---

### Step 2: Identify the y-intercept $ b $

The **y-intercept** is the point where the line crosses the y-axis, i.e., where $ x = 0 $. Substituting $ x = 0 $ into the equation:

$$
\frac{0}{4} + \frac{y}{12} = 1 \Rightarrow \frac{y}{12} = 1 \Rightarrow y = 12
$$

So, the y-intercept is:

$$
b = 12
$$

---

### Step 3: Find the slope $ m $

To find the slope, we convert the given equation into **slope-intercept form** $ y = mx + c $, where $ m $ is the slope and $ c $ is the y-intercept.

Start with the original equation:

$$
\frac{x}{4} + \frac{y}{12} = 1
$$

Multiply both sides by 12 to eliminate denominators:

$$
12 \cdot \left( \frac{x}{4} + \frac{y}{12} \right) = 12 \cdot 1 \Rightarrow 3x + y = 12
$$

Now solve for $ y $:

$$
y = -3x + 12
$$

From this, we see that the slope $ m = -3 $.

---

### Step 4: Compute $ a + b + m $

Now that we have:

- $ a = 4 $
- $ b = 12 $
- $ m = -3 $

We compute:

$$
a + b + m = 4 + 12 + (-3) = 13
$$

---

### Final Answer

$$
\boxed{13}
$$

[Assistant]
Explanation:
Okay, so I need to find the x-intercept (a), the y-intercept (b), and the slope (m) of the line given by the equation x/4 + y/12 = 1. Then add them all together for the final answer. Let me start by recalling what each of these terms means.

First, the x-intercept is the point where the line crosses the x-axis. At that point, the y-coordinate is 0. Similarly, the y-intercept is where the line crosses the y-axis, so the x-coordinate there is 0. The slope, m, is the steepness of the line, calculated as rise over run, or change in y over change in x.

The given equation is x/4 + y/12 = 1. Hmm, this looks like the standard form of a line's intercept form. I remember that the intercept form of a line is x/a + y/b = 1, where a is the x-intercept and b is the y-intercept. So, comparing that to the given equation, it seems like a is 4 and b is 12? Wait, let me check that again.

If the equation is x/4 + y/12 = 1, then yes, that would mean the x-intercept is 4 because when y=0, x/4 = 1, so x=4. Similarly, when x=0, y/12 = 1, so y=12. Therefore, a is 4 and b is 12. That seems straightforward. So, a = 4 and b = 12. But let me confirm by solving for intercepts manually just to be sure.

For the x-intercept, set y = 0:

x/4 + 0/12 = 1 => x/4 = 1 => x = 4. Yep, that's correct. So a = 4.

For the y-intercept, set x = 0:

0/4 + y/12 = 1 => y/12 = 1 => y = 12. So b = 12. Okay, that checks out.

Now, the slope. How do I find the slope from this equation? Well, the standard form of a line is usually written as y = mx + c, where m is the slope and c is the y-intercept. So maybe I should convert the given equation into that slope-intercept form.

Starting with x/4 + y/12 = 1. Let me solve for y.

First, subtract x/4 from both sides:

y/12 = 1 - x/4

Then multiply both sides by 12 to solve for y:

y = 12*(1 - x/4) => y = 12*1 - 12*(x/4) => y = 12 - 3x

So, y = -3x + 12. Therefore, comparing to y = mx + c, the slope m is -3, and the y-intercept is 12. Wait, but we already found the y-intercept (b) as 12, which matches. So that's consistent.

Therefore, m = -3. So, putting it all together, a = 4, b = 12, m = -3. Therefore, a + b + m = 4 + 12 + (-3) = 4 + 12 is 16, minus 3 is 13. So the answer should be 13?

Wait, let me check again. Maybe I made a mistake in calculating the slope? Let me verify.

Original equation: x/4 + y/12 = 1. Let me rearrange it again step by step.

Multiply both sides by 12 to eliminate denominators:

12*(x/4) + 12*(y/12) = 12*1 => 3x + y = 12. So, 3x + y = 12. Then solving for y: y = -3x + 12. Yep, same result. So slope is -3. So that's correct.

Alternatively, another way to find the slope is using the intercept form. The intercept form is x/a + y/b = 1, and if we want to convert that to slope-intercept form, we can do:

y/b = 1 - x/a => y = (-b/a)x + b. Therefore, the slope m is -b/a. In this case, a is 4, b is 12, so m = -12/4 = -3. Yep, same answer. So that's correct.

Therefore, a = 4, b = 12, m = -3. Adding them together: 4 + 12 + (-3) = 13. So the answer is 13. Wait, but let me check once more to make sure I didn't mix up any numbers.

Original equation: x/4 + y/12 = 1. So intercepts are 4 and 12. Slope is -3. So 4 + 12 is 16, minus 3 is 13. Seems right.

Alternatively, maybe I can check with another method. Let's find two points on the line and compute the slope.

We already know the intercepts: (4, 0) and (0, 12). So the slope between these two points is (12 - 0)/(0 - 4) = 12 / (-4) = -3. Yep, same result. So that's correct.

Therefore, all steps check out. So the value of a + b + m is 4 + 12 + (-3) = 13. Therefore, the answer is 13.

**Final Answer**
The value of $a + b + m$ is \boxed{13}.

To solve the problem, we are given the equation of a line:

$$
\frac{x}{4} + \frac{y}{12} = 1
$$

We are asked to find the **x-intercept** $ a $, the **y-intercept** $ b $, and the **slope** $ m $, and then compute the sum $ a + b + m $.

---

### Step 1: Identify the x-intercept $ a $

The **x-intercept** is the point where the line crosses the x-axis, i.e., where $ y = 0 $. Substituting $ y = 0 $ into the equation:

$$
\frac{x}{4} + \frac{0}{12} = 1 \Rightarrow \frac{x}{4} = 1 \Rightarrow x = 4
$$

So, the x-intercept is:

$$
a = 4
$$

---

### Step 2: Identify the y-intercept $ b $

The **y-intercept** is the point where the line crosses the y-axis, i.e., where $ x = 0 $. Substituting $ x = 0 $ into the equation:

$$
\frac{0}{4} + \frac{y}{12} = 1 \Rightarrow \frac{y}{12} = 1 \Rightarrow y = 12
$$

So, the y-intercept is:

$$
b = 12
$$

---

### Step 3: Find the slope $ m $

To find the slope, we convert the given equation into **slope-intercept form** $ y = mx + c $, where $ m $ is the slope and $ c $ is the y-intercept.

Start with the original equation:

$$
\frac{x}{4} + \frac{y}{12} = 1
$$

Multiply both sides by 12 to eliminate denominators:

$$
12 \cdot \left( \frac{x}{4} + \frac{y}{12} \right) = 12 \cdot 1 \Rightarrow 3x + y = 12
$$

Now solve for $ y $:

$$
y = -3x + 12
$$

From this, we see that the slope $ m = -3 $.

---

### Step 4: Compute $ a + b + m $

Now that we have:

- $ a = 4 $
- $ b = 12 $
- $ m = -3 $

We compute:

$$
a + b + m = 4 + 12 + (-3) = 13
$$

---

### Final Answer

$$
\boxed{13}
$$

Solution: We can find the $x$-intercept by setting $y=0$ in the equation.  This gives us $\frac{x}{4} = 1$, so $x =4$, which means $a=4$.  Similarly, letting $x=0$ gives $\frac{y}{12} = 1$, so $y=12$, which means $b=12$.  There are several ways we can find the slope.  First, we could put the equation in slope-intercept form by subtracting $\frac{x}{4}$ from both sides, and then multiplying by 12.  This gives $y = -3x +12$, which tells us that the slope is $-3$ (and confirms our solution for the $y$-intercept).  We also could have noted that since we have already shown that $(4,0)$ and $(0,12)$ are on the line, the slope of the line is $\frac{12 -0}{0-4} = -3$.  Therefore, the desired sum is $4+12 -3 = \boxed{13}$.