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Let's solve Problem 2 step-by-step:

### Problem 2: Find the characteristic equation of the matrix \( A \) and hence compute \( A^{-1} \). Also find the matrix represented by \( A^8 - 5A^7 + 7A^6 - 3A^5 + 8A^2 - 2A + I \).

Given matrix:
\[
A = \begin{pmatrix}
2 & 1 & 1 \\
0 & 1 & 0 \\
1 & 1 & 2
\end{pmatrix}
\]

#### Step 1: Find the Characteristic Equation of \( A \)

The characteristic equation of a matrix \( A \) is found by solving:
\[
\det(A - \lambda I) = 0
\]
where \( \lambda \) are the eigenvalues of \( A \), and \( I \) is the identity matrix.

\[
A - \lambda I = \begin{pmatrix}
2 - \lambda & 1 & 1 \\
0 & 1 - \lambda & 0 \\
1 & 1 & 2 - \lambda
\end{pmatrix}
\]

Calculate the determinant:
\[
\det(A - \lambda I) = \begin{vmatrix}
2 - \lambda & 1 & 1 \\
0 & 1 - \lambda & 0 \\
1 & 1 & 2 - \lambda
\end{vmatrix}
\]

Expanding the determinant along the second row:
\[
\det(A - \lambda I) = (2 - \lambda) \left[(1 - \lambda)(2 - \lambda) - 1\right] - 1 \left[0 \cdot (2 - \lambda) - 1 \cdot 1\right] + 1 \left[0 \cdot 1 - 1 \cdot (2 - \lambda)\right]
\]

Simplify the above expression:
\[
= (2 - \lambda) \left[(2 - \lambda)(1 - \lambda) - 1\right] - (2 - \lambda)
= (2 - \lambda) \left[2 - 3\lambda + \lambda^2 - 1\right] - (2 - \lambda)
= (2 - \lambda) \left[\lambda^2 - 3\lambda + 1\right] - (2 - \lambda)
= (2 - \lambda)(\lambda^2 - 3\lambda + 1) - (2 - \lambda)
= (2 - \lambda)(\lambda^2 - 3\lambda + 1 - 1)
= (2 - \lambda)(\lambda^2 - 3\lambda)
= (2 - \lambda)\lambda(\lambda - 3)
\]

Thus, the characteristic equation is:
\[
\lambda (\lambda - 3)(\lambda - 2) = 0
\]

#### Step 2: Compute \( A^{-1} \)

Using the characteristic polynomial, the eigenvalues are \(\lambda = 0, 3, 2\).

To find \( A^{-1} \), we use the fact that:
\[
A^{-1} = \frac{1}{\det(A)} \text{adj}(A)
\]

The determinant \(\det(A) = 6 - 1 = 5\).

Now, compute the adjugate of \( A \):
\[
\text{adj}(A) = \begin{pmatrix}
2 & -1 & 1 \\
0 & 5 & 0 \\
-1 & -1 & 2
\end{pmatrix}
\]

Therefore,
\[
A^{-1} = \frac{1}{5} \begin{pmatrix}
2 & -1 & 1 \\
0 & 5 & 0 \\
-1 & -1 & 2
\end{pmatrix}
\]

#### Step 3: Find the Matrix Represented by \( A^8 - 5A^7 + 7A^6 - 3A^5 + 8A^2 - 2A + I \)

Using the Cayley-Hamilton theorem, we know that \( A \) satisfies its own characteristic polynomial:
\[
A^3 - 5A^2 + 8A - 2I = 0
\]

So,
\[
A^8 = 5A^7 - 7A^6 + 3A^5 - 8A^2 + 2A - I
\]

Therefore,
\[
A^8 - 5A^7 + 7A^6 - 3A^5 + 8A^2 - 2A + I = 0
\]

Thus, the resulting matrix is the zero matrix:
\[
A^8 - 5A^7 + 7A^6 - 3A^5 + 8A^2 - 2A + I = \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\]

---

Would you like the detailed solution for another problem, or do you have any questions on this proof?

Here are some options for the next problem to solve:

1. Reduce a given matrix to diagonal form.
2. Reduce a quadratic form to canonical form.
3. Verify a given matrix identity.
4. Solve a system of linear equations by row operations.

**Tip:** When working with characteristic polynomials, leveraging the Cayley-Hamilton theorem can significantly simplify matrix computations.

Learn how to find the characteristic equation of a matrix and perform matrix operations. This problem involves matrix algebra, eigenvalues, characteristic polynomial, and the Cayley-Hamilton theorem. Suitable for advanced college-level students.

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