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To solve the expression \( A^8 - 5A^7 + 7A^6 - 3A^5 + A^4 - 5A^3 + 8A^2 - 2A + I \), we need to work with the matrix \( A \) given by:

\[
A = \begin{bmatrix} 2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 2 \end{bmatrix}
\]

The calculation involves finding high powers of \( A \) up to \( A^8 \). To simplify this process, we might look for patterns in the powers of \( A \), such as identifying any cyclic or null properties (for example, if a certain power of \( A \) equals zero, or a power repeats previous results), or we might try diagonalization if applicable.

Here's the step-by-step outline of the solution process:

1. **Calculate Initial Powers of \( A \)**: Begin with \( A^2, A^3, \) and so forth, checking for any repetitive or simplifying patterns.

2. **Identify Patterns or Use Minimal Polynomial**: If we find a minimal polynomial of \( A \), this could reduce the computational workload by allowing us to express higher powers in terms of lower ones.

3. **Combine Terms**: Using the powers calculated, substitute back into the expression to find the resulting matrix.

Let’s proceed with calculating these powers explicitly and examining possible simplifications. I'll begin the calculations.The powers of \( A \) up to \( A^8 \) are as follows:

\[
\begin{align*}
A^2 &= \begin{bmatrix} 5 & 4 & 4 \\ 0 & 1 & 0 \\ 4 & 4 & 5 \end{bmatrix}, \\
A^3 &= \begin{bmatrix} 14 & 13 & 13 \\ 0 & 1 & 0 \\ 13 & 13 & 14 \end{bmatrix}, \\
A^4 &= \begin{bmatrix} 41 & 40 & 40 \\ 0 & 1 & 0 \\ 40 & 40 & 41 \end{bmatrix}, \\
A^5 &= \begin{bmatrix} 122 & 121 & 121 \\ 0 & 1 & 0 \\ 121 & 121 & 122 \end{bmatrix}, \\
A^6 &= \begin{bmatrix} 365 & 364 & 364 \\ 0 & 1 & 0 \\ 364 & 364 & 365 \end{bmatrix}, \\
A^7 &= \begin{bmatrix} 1094 & 1093 & 1093 \\ 0 & 1 & 0 \\ 1093 & 1093 & 1094 \end{bmatrix}, \\
A^8 &= \begin{bmatrix} 3281 & 3280 & 3280 \\ 0 & 1 & 0 \\ 3280 & 3280 & 3281 \end{bmatrix}.
\end{align*}
\]

We observe that each power of \( A \) maintains a specific pattern, which may be utilized in simplifying the expression \( A^8 - 5A^7 + 7A^6 - 3A^5 + A^4 - 5A^3 + 8A^2 - 2A + I \).

Now, let's substitute these matrices and compute the final matrix.The resulting matrix from the expression \( A^8 - 5A^7 + 7A^6 - 3A^5 + A^4 - 5A^3 + 8A^2 - 2A + I \) is:

\[
\begin{bmatrix} 8 & 5 & 5 \\ 0 & 3 & 0 \\ 5 & 5 & 8 \end{bmatrix}
\]

Would you like a detailed breakdown of each step? Or do you have any specific questions?

Here are some related questions for further exploration:
1. What is the minimal polynomial of matrix \( A \) and how does it help in simplifying expressions?
2. How does the Cayley-Hamilton theorem assist in evaluating high powers of matrices?
3. Can this result be verified using another method, such as diagonalization?
4. How do patterns in matrix powers aid in simplifying large polynomial expressions?
5. What properties of the matrix \( A \) lead to such predictable patterns in its powers?

**Tip:** When handling high powers of matrices, look for recurring patterns or minimal polynomials to reduce computational effort.

Find the matrix obtained by A^8 - 5A^7 + 7A^6 - 3A^5 + A^4 - 5A^3 + 8A^2 - 2A + I for matrix A = [[2, 1, 1], [0, 1, 0], [1, 1, 2]].

Learn how to compute the matrix resulting from A^8 - 5A^7 + 7A^6 - 3A^5 + A^4 - 5A^3 + 8A^2 - 2A + I for a 3x3 matrix A. This solution covers matrix powers, algebraic expressions, and key linear algebra theorems, such as the Cayley-Hamilton theorem. Suitable for undergraduate students studying linear algebra.

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