We are given the following information:

• $A$ is a $3 \times 3$ matrix with eigenvalues 2 and 1.
• $u = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ is an eigenvector corresponding to eigenvalue 2.
• $v = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}$ is an eigenvector corresponding to eigenvalue 1.
• $w = \begin{bmatrix} 7 \\ 2 \\ -3 \end{bmatrix}$.
• We are tasked with computing $A^5 w$.

Step 1: Diagonalization Insight

Since $A$ has distinct eigenvalues and corresponding eigenvectors, it can be diagonalized, meaning: $A = PDP^{-1}$ where $D$ is a diagonal matrix containing the eigenvalues, and $P$ is a matrix whose columns are the eigenvectors $u$, $v$, etc. But we don’t need to explicitly construct $A$. Instead, we can take advantage of the following property:

$A^k v = \lambda^k v$ for any eigenvector $v$ corresponding to the eigenvalue $\lambda$, where $A^k$ is a matrix raised to the power $k$.

Step 2: Decomposition of $w$

We need to express $w$ as a linear combination of the eigenvectors $u$ and $v$. Let’s assume: $w = c_1 u + c_2 v$ where $c_1$ and $c_2$ are scalars we need to solve for.

This means: $\begin{bmatrix} 7 \\ 2 \\ -3 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} + c_2 \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}$ We now solve for $c_1$ and $c_2$.

Step 3: Solve for $c_1$ and $c_2$

By comparing each component, we can set up a system of equations:

1. $c_1 + 2c_2 = 7$
2. $c_2 = 2$
3. $-c_1 = -3$

From equation (3), we immediately get $c_1 = 3$.

Now substitute $c_2 = 2$ and $c_1 = 3$ into equation (1): $3 + 2(2) = 7$ This is true, so the solution is $c_1 = 3$ and $c_2 = 2$.

Thus: $w = 3u + 2v$

Step 4: Compute $A^5 w$

Now that we know $w = 3u + 2v$, we can use the eigenvalue property to compute $A^5 w$.

Since $A u = 2u$ and $A v = v$, we have: $A^5 u = 2^5 u = 32u$ $A^5 v = 1^5 v = v$

Thus: $A^5 w = A^5 (3u + 2v) = 3A^5 u + 2A^5 v = 3(32u) + 2v = 96u + 2v$

Step 5: Substitute $u$ and $v$

Now substitute the values of $u$ and $v$: