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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 \( v \):
\[
A^5 w = 96 \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} + 2 \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}
= \begin{bmatrix} 96 \\ 0 \\ -96 \end{bmatrix} + \begin{bmatrix} 4 \\ 2 \\ 0 \end{bmatrix}
= \begin{bmatrix} 100 \\ 2 \\ -96 \end{bmatrix}
\]

Thus, the result is:
\[
A^5 w = \begin{bmatrix} 100 \\ 2 \\ -96 \end{bmatrix}
\]

Would you like further details or have any questions?

Here are 5 related questions you might find interesting:
1. How can you generalize this method to compute \( A^k w \) for any \( k \)?
2. How would you compute \( A^5 \) if \( A \) was not diagonalizable?
3. What are the conditions under which a matrix is diagonalizable?
4. Can a matrix with repeated eigenvalues be diagonalizable?
5. How can you find eigenvalues and eigenvectors of larger matrices?

**Tip:** Eigenvalues and eigenvectors provide a powerful tool to simplify matrix operations, especially for large powers of matrices. Always check for diagonalizability to ease computations.

Let A be a 3 x 3 matrix. Suppose that A has eigenvalues 2 and 1, and suppose that u and v are eigenvectors corresponding to 2 and 1, respectively. Compute A^5w, where w is a vector provided.

Learn how to compute A^5w for a 3x3 matrix with given eigenvalues and eigenvectors. This linear algebra problem involves diagonalization, eigenvalue properties, and matrix power computations. Suitable for university students studying linear algebra.

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