A is a 2 ✕ 2 matrix with eigenvectors 
v1 = 

1	

−1
 and 
v2 = 

1	

1
 corresponding to eigenvalues 
𝜆1 = 
1
2
 and 𝜆2 = 2,
 respectively, and 
x =  

5	

−1
.
Find A9x.

To solve for \( A^9x \), we can leverage the fact that \( A \) is a diagonalizable matrix. Given the eigenvectors \( v_1 \) and \( v_2 \), and their corresponding eigenvalues \( \lambda_1 \) and \( \lambda_2 \), the matrix \( A \) can be represented as:

\[
A = PDP^{-1}
\]

where \( P \) is the matrix of eigenvectors and \( D \) is the diagonal matrix of eigenvalues.

Given:
- Eigenvectors: \( v_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \) and \( v_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \)
- Eigenvalues: \( \lambda_1 = \frac{1}{2} \) and \( \lambda_2 = 2 \)
- Vector \( x = \begin{pmatrix} 5 \\ -1 \end{pmatrix} \)

1. **Construct the matrix \( P \) and \( D \):**

\[
P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & 2 \end{pmatrix}
\]

2. **Find the inverse of \( P \):**

\[
P^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}
\]

3. **Compute \( A^9 \):**

Since \( A = PDP^{-1} \), then \( A^9 = PD^9P^{-1} \).

\[
D^9 = \begin{pmatrix} \left(\frac{1}{2}\right)^9 & 0 \\ 0 & 2^9 \end{pmatrix} = \begin{pmatrix} \frac{1}{512} & 0 \\ 0 & 512 \end{pmatrix}
\]

\[
A^9 = P \begin{pmatrix} \frac{1}{512} & 0 \\ 0 & 512 \end{pmatrix} P^{-1}
\]

4. **Calculate \( A^9x \):**

\[
A^9x = PD^9P^{-1}x
\]

First, find \( P^{-1}x \):

\[
P^{-1}x = \frac{1}{2} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 5 \\ -1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 5 - (-1) \\ 5 + (-1) \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 6 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}
\]

Then, multiply by \( D^9 \):

\[
D^9 \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{512} & 0 \\ 0 & 512 \end{pmatrix} \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{512} \\ 1024 \end{pmatrix}
\]

Finally, multiply by \( P \):

\[
A^9x = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} \frac{3}{512} \\ 1024 \end{pmatrix} = \begin{pmatrix} \frac{3}{512} + 1024 \\ -\frac{3}{512} + 1024 \end{pmatrix} = \begin{pmatrix} 1024.005859375 \\ 1023.994140625 \end{pmatrix}
\]

So,

\[
A^9x \approx \begin{pmatrix} 1024.01 \\ 1023.99 \end{pmatrix}
\]

Would you like more details on any part of this process?

### Related Questions:
1. How do you diagonalize a matrix?
2. What are the properties of eigenvalues and eigenvectors in matrix multiplication?
3. How do you compute matrix powers using eigenvalues and eigenvectors?
4. What are some practical applications of diagonalization in engineering?
5. How can one find the inverse of a 2x2 matrix quickly?

### Tip:
When dealing with matrix exponentiation, diagonalization is a powerful tool that simplifies the computation, especially for high powers.

A is a 2 ✕ 2 matrix with eigenvectors v1 = [1, −1] and v2 = [1, 1] corresponding to eigenvalues λ1 = 1/2 and λ2 = 2, respectively, and x = [5, −1]. Find A^9x.

Solve for A^9x using matrix diagonalization. This problem explores eigenvalues, eigenvectors, and matrix exponentiation to compute the power of a matrix. Ideal for linear algebra students at the college level.

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