<Question>
Given
\[\mathbf{A} = \begin{pmatrix} 3 & 7 \\ -1 & -2 \end{pmatrix},\]compute $\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40}.$
</Question>

<think>
<Plan_1>
Okay, so I need to compute A raised to the 27th, 31st, and 40th powers, and then add those matrices together. That seems really high. Direct computation isn't feasible here. I remember that for matrices, especially when dealing with powers, diagonalization can be really helpful. If a matrix is diagonalizable, then raising it to a power becomes much simpler because you can just raise the diagonal entries to that power. Let me try to recall the steps involved in diagonalizing a matrix.

First, I need to check if matrix A is diagonalizable. To do that, I should find its eigenvalues and eigenvectors. If there are two linearly independent eigenvectors, then the matrix is diagonalizable.
Let me start by finding the eigenvalues of A.

The characteristic equation is given by det(A - λI) = 0. Let's compute that.

<prompt_1.1> Find the eigenvalues of matrix A by solving the characteristic equation. </prompt_1.1>
-----

Hmm, but let me check if there's another way. Maybe using the Cayley-Hamilton theorem? Since the matrix satisfies its own characteristic equation. Let's see.

<prompt_1.2> Consider using the Cayley-Hamilton theorem to verify the result. </prompt_1.2>

Alternatively, perhaps there's a pattern in the powers of A. Let me compute A^1, A^2, A^3, see if there's a cycle or pattern.

<prompt_1.3> Explore finding a pattern in the powers of A using A^1, A^2, A^3. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
So, A - λI is:

\[
\begin{pmatrix}
3 - λ & 7 \\
-1 & -2 - λ
\end{pmatrix}
\]

The determinant of this matrix is (3 - λ)(-2 - λ) - (7)(-1). Let me compute that:

First, expand the product: (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. So the determinant is λ² - λ -6 +7 = λ² - λ +1.

Wait, that can't be right. Let me check again.

Wait, the determinant is (3 - λ)(-2 - λ) - (7)(-1). So compute (3 - λ)(-2 - λ):

Multiply 3 by -2: -6

3 times -λ: -3λ

-λ times -2: +2λ

-λ times -λ: +λ²

So altogether: -6 -3λ +2λ +λ² = λ² - λ -6.

Then subtract 7*(-1) which is +7. So determinant is λ² - λ -6 +7 = λ² - λ +1.

Wait, but that would make the characteristic equation λ² - λ +1 = 0. Let me verify the calculation again because I might have made a mistake in the signs.

Original matrix A:

Row 1: 3, 7

Row 2: -1, -2

So A - λI is:

Row 1: 3 - λ, 7

Row 2: -1, -2 - λ

So determinant is (3 - λ)(-2 - λ) - (7)(-1)

Compute (3 - λ)(-2 - λ):

Multiply 3 by -2: -6

3 by -λ: -3λ

-λ by -2: +2λ

-λ by -λ: +λ²

So combining terms: -6 -3λ +2λ +λ² = λ² - λ -6

Then subtract 7*(-1) which is +7. So determinant is λ² - λ -6 +7 = λ² - λ +1. Hmm. So the characteristic equation is λ² - λ +1 = 0.
</execution_1.1>

<execution_1.2>
The characteristic equation we found was λ² + λ - 5 = 0. According to Cayley-Hamilton theorem, the matrix A satisfies this equation: A² + A - 5I = 0. Therefore, A² = -A + 5I. Maybe we can use this recurrence relation to express higher powers of A in terms of A and I. Let's try that.

If A² = -A + 5I, then A³ = A*A² = A*(-A +5I) = -A² +5A = -(-A +5I) +5A = A -5I +5A = 6A -5I.

Similarly, A^4 = A*A^3 = A*(6A -5I) = 6A² -5A = 6*(-A +5I) -5A = -6A +30I -5A = -11A +30I.

A^5 = A*A^4 = A*(-11A +30I) = -11A² +30A = -11*(-A +5I) +30A = 11A -55I +30A = 41A -55I.

Hmm, this seems like a linear recurrence. Maybe we can find a pattern or a closed-form expression for A^n.

Let's suppose that A^n can be expressed as p_n A + q_n I, where p_n and q_n are coefficients to be determined. Let's test this.

Base cases:

A^1 = A = 1*A + 0*I, so p_1 = 1, q_1 = 0.

A^2 = -A +5I = (-1)*A +5*I, so p_2 = -1, q_2 =5.

Assume that A^k = p_k A + q_k I for k = 1,2,...,n. Then A^{n+1} = A * A^n = A*(p_n A + q_n I) = p_n A² + q_n A. But A² = -A +5I, so substituting:

A^{n+1} = p_n*(-A +5I) + q_n A = (-p_n + q_n) A + 5 p_n I.

Therefore, the recurrence relations for p_{n+1} and q_{n+1} are:

p_{n+1} = -p_n + q_n

q_{n+1} = 5 p_n

So this gives us a recursive way to compute p_n and q_n. Let's write out the first few terms to see the pattern.

n=1: p1=1, q1=0

n=2: p2=-1, q2=5*1=5

n=3: p3=-p2 + q2 = -(-1) +5=1 +5=6; q3=5 p2=5*(-1)=-5

Wait, but earlier we had A^3=6A -5I, so q3=-5. Correct.

n=4: p4=-p3 + q3 = -6 + (-5)= -11; q4=5 p3=5*6=30

Which matches A^4=-11A +30I.

n=5: p5=-p4 + q4 = -(-11) +30=11 +30=41; q5=5 p4=5*(-11)=-55

Which matches A^5=41A -55I.

So the recurrence is working.
</execution_1.2>

<execution_1.3>
Let's compute A^1, A^2, A^3, etc.:

A^1 is just A:

\[ \mathbf{A} = \begin{pmatrix} 3 & 7 \\ -1 & -2 \end{pmatrix} \]

A^2 = A * A:

First row, first column: 3*3 + 7*(-1) = 9 - 7 = 2

First row, second column: 3*7 + 7*(-2) = 21 - 14 = 7

Second row, first column: (-1)*3 + (-2)*(-1) = -3 + 2 = -1

Second row, second column: (-1)*7 + (-2)*(-2) = -7 + 4 = -3

So A^2 is:

\[ \begin{pmatrix} 2 & 7 \\ -1 & -3 \end{pmatrix} \]

A^3 = A^2 * A:

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

A^3 is:

\[ \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \]

Wait, that's -I, the negative identity matrix. Interesting.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, let's find the eigenvalues by solving this quadratic equation.

<prompt_2.1> Solve the quadratic equation to find the eigenvalues of matrix A. </prompt_2.1>
-----

Based on execution_1.2:
A^4:

\[ \begin{pmatrix} -11 & -5 \\ 5 & 30 \end{pmatrix} \]

Wait, maybe there's a cycle here. Let me check A^3 and see if it's a multiple of identity or something.

<prompt_2.2> Continue using the recurrence relation to compute A^5, A^6, and A^7. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
The quadratic equation λ² - λ + 1 = 0. Using the quadratic formula, λ = [1 ± sqrt(1 - 4)] / 2 = [1 ± sqrt(-3)] / 2 = (1 ± i√3)/2. So the eigenvalues are complex: (1 + i√3)/2 and (1 - i√3)/2. Let me denote them as λ1 and λ2.
</execution_2.1>

<execution_2.2>
Wait, A^3 was:

\[
\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}
\]

So that's -I. Then A^4 would be A * A^3 = A*(-I) = -A. Let's check that:

If A^3 = -I, then A^4 = A*(-I) = -A. But according to our earlier computation, A^4 was:

\[
\begin{pmatrix}
-11 & -5 \\
5 & 30
\end{pmatrix}
\]

But if A^4 = -A, then -A would be:

\[
\begin{pmatrix}
-3 & -7 \\
1 & 2
\end{pmatrix}
\]

Which is clearly not equal to A^4. So that contradicts the idea that A^3 = -I. Wait, that suggests that maybe there was a miscalculation here. Wait, let's check A^3 again.

Wait, when I computed A^3, I got:

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

So A^3 is indeed:

\[
\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}
\]

Which is -I. So if A^3 = -I, then A^4 = A^3 * A = (-I)*A = -A.

But according to the recurrence relation earlier, we had A^4 = -11A +30I. Wait, that's a problem because according to the direct multiplication, A^4 should be -A. So there's a contradiction here. That means I must have made a mistake in the recurrence calculation.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Next, I need to find the eigenvectors corresponding to these eigenvalues and check if they are linearly independent. If they are, then the matrix is diagonalizable.
Let's proceed.

First, for λ1 = (1 + i√3)/2. Let's find the eigenvector by solving (A - λ1I)v = 0.

<prompt_3.1> Find the eigenvectors for the complex eigenvalues of matrix A. </prompt_3.1>
-----

Based on execution_2.2:
Wait, let's check the recurrence again. Let's see:

From A^1 to A^2: A^2 = -A +5I. Then A^3 = A*A^2 = A*(-A +5I) = -A^2 +5A. But since A^2 = -A +5I, substitute that in: -(-A +5I) +5A = A -5I +5A = 6A -5I. Then A^4 = A*A^3 = A*(6A -5I) =6A^2 -5A. Again, substitute A^2 = -A +5I: 6*(-A +5I) -5A = -6A +30I -5A = -11A +30I. Then A^5 = A*A^4 = A*(-11A +30I) = -11A^2 +30A. Substitute A^2 again: -11*(-A +5I) +30A =11A -55I +30A =41A -55I. Then A^6 = A*A^5 = A*(41A -55I) =41A^2 -55A. Substitute A^2:41*(-A +5I) -55A = -41A +205I -55A = -96A +205I. Then A^7 = A*A^6 = A*(-96A +205I) = -96A^2 +205A. Substitute A^2: -96*(-A +5I) +205A =96A -480I +205A = (96+205)A -480I =301A -480I.

So the coefficients follow a linear recurrence. Let's see if there's a pattern or if we can express A^n in terms of A and I.

<prompt_3.2> Continue using the recurrence relation to compute A^6 and A^7, and look for a pattern. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
This will involve complex numbers, which might complicate things a bit, but let's try.

First, compute A - λ1I:

\[
\begin{pmatrix}
3 - \lambda1 & 7 \\
-1 & -2 - \lambda1
\end{pmatrix}
\]

Substituting λ1 = (1 + i√3)/2:

First entry: 3 - (1 + i√3)/2 = (6 -1 -i√3)/2 = (5 - i√3)/2

Second entry remains 7.

Third entry: -1

Fourth entry: -2 - (1 + i√3)/2 = (-4 -1 -i√3)/2 = (-5 - i√3)/2

So the matrix A - λ1I is:

\[
\begin{pmatrix}
(5 - i√3)/2 & 7 \\
-1 & (-5 - i√3)/2
\end{pmatrix}
\]

To find the eigenvector, we need to solve (A - λ1I)v = 0. Let's denote v = [x, y]^T. So the equations are:

[(5 - i√3)/2] x + 7 y = 0

-1 x + [(-5 - i√3)/2] y = 0

Let me use the first equation to express x in terms of y. From the first equation:

[(5 - i√3)/2] x = -7 y => x = [-7 y] / [(5 - i√3)/2] = (-14 y) / (5 - i√3)

Multiply numerator and denominator by the complex conjugate of the denominator to rationalize:

x = (-14 y) (5 + i√3) / [(5 - i√3)(5 + i√3)] = (-14 y)(5 + i√3) / (25 + 3) = (-14 y)(5 + i√3)/28 = (-y)(5 + i√3)/2

So x = [ -5 - i√3 ] / 2 * y

Therefore, the eigenvector corresponding to λ1 is any scalar multiple of the vector:

v1 = [ -5 - i√3 , 2 ]^T

Wait, because if y = 2, then x = -5 - i√3. So v1 = [ -5 - i√3, 2 ]^T.

Similarly, for λ2 = (1 - i√3)/2, the eigenvector will be the conjugate of v1, which is v2 = [ -5 + i√3, 2 ]^T.

But dealing with complex eigenvectors might complicate diagonalization over the real numbers. However, since the eigenvalues are complex conjugates, the matrix is diagonalizable over the complex numbers. But if we want to work with real matrices, perhaps we can use the real and imaginary parts of the eigenvectors to form a real basis. Let me recall that for complex eigenvalues, the eigenvectors can be used to form real-valued solutions via Euler's formula, but maybe that's more involved.
</execution_3.1>

<execution_3.2>
Alternatively, since the eigenvalues are complex, maybe the matrix is similar to a rotation-scaling matrix, but since it's 2x2, perhaps there's a cyclicity in powers. However, since the eigenvalues are complex conjugates, their magnitudes might be 1, leading to periodic behavior. Let me check the eigenvalues.

The eigenvalues are (1 ± i√3)/2. Let's compute their magnitude. The magnitude is sqrt( (1/2)^2 + (√3/2)^2 ) = sqrt(1/4 + 3/4) = sqrt(1) =1. So they lie on the unit circle, which means that raising them to powers will result in points on the unit circle at angles corresponding to multiples of 60 degrees (since the real part is 1/2, which corresponds to 60 degrees).

Indeed, (1 + i√3)/2 is e^{iπ/3} and (1 - i√3)/2 is e^{-iπ/3}. Therefore, their powers cycle every 6 steps, since (e^{iπ/3})^6 = e^{i2π} =1. Therefore, the powers of A should have a period of 6. Let's verify this.

If that's the case, then A^6 = I, A^7 = A, A^8 = A^2, etc. Let's check if A^6 is indeed the identity matrix.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
First, let's compute A^1 to A^6 using the recurrence relation to see if A^6 is I.

<prompt_4.1> Compute powers of A using the recurrence relation to verify if A^6 is the identity matrix. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
We already have:

A^1 = A

A^2 = -A +5I

A^3 =6A -5I

A^4 =-11A +30I

A^5=41A -55I

A^6 = -96A +205I (from previous calculation)

A^7=301A -480I

Wait, but if the eigenvalues have magnitude 1, then A^6 should be I. Let's check if that's the case.

Let's compute A^6. From the recurrence, we found A^6 = -96A +205I. Let's compute this matrix:

-96A +205I:

First, compute -96A:

-96 * 3 = -288

-96 *7 = -672

-96*(-1)=96

-96*(-2)=192

So -96A is:

\[
\begin{pmatrix}
-288 & -672 \\
96 & 192
\end{pmatrix}
\]

Then add 205I:

205I is:

\[
\begin{pmatrix}
205 & 0 \\
0 & 205
\end{pmatrix}
\]

Adding these together:

First entry: -288 +205 = -83

Second entry: -672 +0 = -672

Third entry:96 +0 =96

Fourth entry:192 +205 = 397

So A^6 is:

\[
\begin{pmatrix}
-83 & -672 \\
96 & 397
\end{pmatrix}
\]

Hmm, that's not the identity matrix. So perhaps my assumption about the periodicity is wrong. Alternatively, maybe I made a mistake in the recurrence calculations. Let's go back and verify each step carefully.

Starting with A^1 to A^2:

A^1 is correct.

A^2 = A*A^1. Let's compute A^2 directly:

A * A:

First row:

[3*3 +7*(-1), 3*7 +7*(-2)] = [9 -7, 21 -14] = [2,7]

Second row:

[-1*3 + (-2)*(-1), -1*7 + (-2)*(-2)] = [-3 +2, -7 +4] = [-1, -3]

So A^2 is correct.

Now, according to the recurrence, A^2 should be -A +5I.

Compute -A +5I:

-A is:

[-3, -7; 1, 2]

Adding 5I:

5I is:

[5,0;0,5]

So -A +5I is:

[-3 +5, -7 +0; 1 +0, 2 +5] = [2, -7; 1,7]

Wait, but A^2 is [2,7; -1,-3]. So [2, -7; 1,7] vs [2,7; -1,-3]. These are not equal. Therefore, there's a mistake in the recurrence approach. That suggests that the initial assumption that A² = -A +5I is incorrect, which contradicts the Cayley-Hamilton theorem.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Wait, but according to the characteristic equation, the matrix satisfies its own characteristic equation, which we found to be λ² - λ +1 =0.

<prompt_5.1> Re-evaluate the application of the Cayley-Hamilton theorem and verify the recurrence relations. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Therefore, the Cayley-Hamilton theorem states that A² - A + I =0, so A² = A - I. Wait, but that contradicts our earlier calculation where A² was [2,7; -1,-3]. Let's check this again.

Wait, when we computed the determinant earlier, we had:

det(A - λI) = λ² - λ +1. Therefore, according to Cayley-Hamilton theorem, A² - A + I =0, so A² = A - I.

But when we computed A² directly, we got:

A² = [ [3*3 +7*(-1), 3*7 +7*(-2)], [ -1*3 + (-2)*(-1), -1*7 + (-2)*(-2) ] ]

Which is:

First row: 9 -7 =2, 21 -14 =7

Second row: -3 +2 = -1, -7 +4 = -3

So A² is:

[ [2,7], [-1,-3] ]

But according to A² = A - I, A - I would be:

[3-1,7; -1, -2-1] = [2,7; -1,-3], which matches A². So actually, the correct relation is A² = A - I, not -A +5I as I previously thought. So my mistake was in the initial calculation of the determinant.

Wait, let's go back to the determinant calculation.

Original matrix A - λI:

[3-λ,7; -1,-2-λ]

Determinant is (3 - λ)(-2 - λ) - (7)(-1)

Compute (3 - λ)(-2 - λ):

= 3*(-2) + 3*(-λ) -λ*(-2) -λ*λ

= -6 -3λ +2λ + (-λ²)

= -6 -λ -λ²

Then subtract 7*(-1) = +7

So determinant is -6 -λ -λ² +7 = ( -6 +7 ) + (-λ -λ² ) = 1 -λ -λ²

Therefore, the characteristic equation is -λ² -λ +1 =0, but usually written as λ² + λ -1 =0. Wait, but the determinant is 1 -λ -λ², so setting to zero:

1 -λ -λ² =0 → λ² + λ -1 =0.

Wait, but that's different from what I had before. Wait, no:

Wait, 1 -λ -λ² =0 → -λ² -λ +1 =0 → multiplying both sides by -1: λ² + λ -1 =0. So the correct characteristic equation is λ² + λ -1 =0.

Therefore, according to Cayley-Hamilton theorem, the matrix A satisfies its characteristic equation, so A² + A - I =0 → A² = -A + I.

Ah! So this corrects my earlier mistake. So the correct recurrence is A² = -A + I. Therefore, previous steps were based on an error in the determinant calculation. So let's start over with the correct determinant.

Earlier, I miscalculated the determinant. Let's redo it carefully.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Compute det(A - λI):

Row 1: 3 - λ, 7

Row 2: -1, -2 - λ

det(A - λI) = (3 - λ)(-2 - λ) - (7)(-1)

<prompt_6.1> Recalculate the determinant and verify the Cayley-Hamilton theorem application. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Compute (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:

det(A - λI) = λ² - λ -6 +7 = λ² - λ +1

Wait, so that gives λ² - λ +1 =0. But earlier, when we computed A², we found A² = A - I. Let's check that.

If A² = A - I, then A² - A + I =0. Let's verify:

Compute A²:

A is:

[3,7; -1,-2]

A² = A*A:

First row:

3*3 +7*(-1) =9 -7=2

3*7 +7*(-2)=21 -14=7

Second row:

-1*3 + (-2)*(-1)= -3 +2= -1

-1*7 + (-2)*(-2)= -7 +4= -3

So A² is:

[2,7; -1,-3]

Compute A - I:

A is [3,7; -1,-2], I is [1,0;0,1]

A - I = [3-1,7; -1,-2-1] = [2,7; -1,-3], which matches A². Therefore, A² = A - I. Therefore, the correct characteristic equation is A² + A - I =0, which would correspond to the equation λ² + λ -1=0. Wait, but according to our determinant calculation, det(A - λI)= λ² - λ +1. So the characteristic equation is λ² - λ +1=0. However, the Cayley-Hamilton theorem says that substituting A into its characteristic equation gives A² - A + I =0, which is A² = A - I. But when we computed A² directly, we got A - I, which matches. Therefore, there's a discrepancy here. Wait, let's check the determinant calculation again.

Wait, let's redo the determinant calculation:

det(A - λI) = (3 - λ)(-2 - λ) - (7)(-1)

First, (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:

det(A - λI) = λ² - λ -6 +7 = λ² - λ +1

Therefore, the characteristic equation is λ² - λ +1 =0. Then according to Cayley-Hamilton, A² - A + I =0 → A² = A - I.

But when we computed A² directly, we found A² = A - I. So that's correct. Therefore, the earlier mistake was in the determinant calculation. The correct recurrence is A² = A - I. Therefore, we can use this to compute higher powers of A.

Therefore, let's start over with the correct recurrence.

Given A² = A - I, then we can express higher powers of A in terms of A and I.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
Let's compute A^3:

A^3 = A*A^2 = A*(A - I) = A² - A

But A² = A - I, so substitute:

A^3 = (A - I) - A = -I

So A^3 = -I

Then A^4 = A*A^3 = A*(-I) = -A

A^5 = A*A^4 = A*(-A) = -A²

But A² = A - I, so:

A^5 = -(A - I) = -A + I

A^6 = A*A^5 = A*(-A + I) = -A² + A

Again, substitute A² = A - I:

-A² + A = - (A - I) + A = -A + I + A = I

So A^6 = I

A^7 = A*A^6 = A*I = A

A^8 = A*A^7 = A*A = A² = A - I

A^9 = A*A^8 = A*(A - I) = A² - A = (A - I) - A = -I

So we can see a cycle here: every 6 powers, the matrix cycles through the same set of matrices. So:

A^1 = A

A^2 = A - I

A^3 = -I

A^4 = -A

A^5 = -A + I

A^6 = I

A^7 = A

And so on. Therefore, the powers of A repeat every 6. Therefore, to compute A^27, A^31, A^40, we can find the remainder when 27,31,40 are divided by 6.

<prompt_7.1> Use the cyclicity of the powers of A to compute A^27, A^31, and A^40. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
Let's compute:

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

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

40 divided by 6: 6*6=36, remainder 4. So A^40 = (A^6)^6 * A^4 = I^6 * (-A) = -A

Therefore:

A^27 = -I

A^31 = A

A^40 = -A

Therefore, adding them together:

A^27 + A^31 + A^40 = (-I) + A + (-A) = -I + (A - A) = -I + 0 = -I

So the result is the negative of the identity matrix.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
Let me verify this again to be sure.

Given that A^6 = I, then the powers cycle every 6. Therefore:

A^(6k + r) = A^r

Therefore:

27 = 6*4 +3 → A^27 = A^3 = -I

31 =6*5 +1 → A^31 = A^1 = A

40 =6*6 +4 → A^40 = A^4 = -A

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

Therefore, the final answer is -I.
Therefore, the answer is \(\boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}\)
**Final Answer**
\boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}
</Plan_8>

<Execution_8>
No prompt to execute.
</Execution_8>

<Final_answer>
Therefore, the final answer is $\boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}$.
</Final_answer>
</think>