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To find the eigenvector corresponding to the eigenvalue \( \lambda = 4 \) for the matrix:

\[
\begin{bmatrix} 1 & 0 & 0 \\ 3 & -2 & 0 \\ 2 & 3 & 4 \end{bmatrix}
\]

we'll proceed as follows:

### Step 1: Set Up the Equation
To find the eigenvector, we need to solve:
\[
(A - 4I)\mathbf{x} = 0
\]
where \( A \) is the given matrix and \( I \) is the identity matrix of the same dimension.

The matrix \( 4I \) is:
\[
4I = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}
\]

Now, calculate \( A - 4I \):
\[
A - 4I = \begin{bmatrix} 1 - 4 & 0 & 0 \\ 3 & -2 - 4 & 0 \\ 2 & 3 & 4 - 4 \end{bmatrix} = \begin{bmatrix} -3 & 0 & 0 \\ 3 & -6 & 0 \\ 2 & 3 & 0 \end{bmatrix}
\]

### Step 2: Solve \( (A - 4I)\mathbf{x} = 0 \)
We now need to solve the following system:
\[
\begin{bmatrix} -3 & 0 & 0 \\ 3 & -6 & 0 \\ 2 & 3 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
\]

This system gives us three equations:
1. \( -3x_1 = 0 \Rightarrow x_1 = 0 \)
2. \( 3x_1 - 6x_2 = 0 \Rightarrow -6x_2 = 0 \Rightarrow x_2 = 0 \)
3. \( 2x_1 + 3x_2 = 0 \Rightarrow x_3 \) can be any value (let \( x_3 = -1 \)).

Thus, a corresponding eigenvector is:
\[
\mathbf{x} = \begin{bmatrix} 

Find the eigenvector of the matrix [1 0 0; 3 -2 0; 2 3 4] corresponding to the eigenvalue 4.

Learn how to find the eigenvector of the matrix [1 0 0; 3 -2 0; 2 3 4] corresponding to the eigenvalue 4. This problem covers key concepts in linear algebra, including eigenvalues and eigenvectors. Suitable for college-level students.

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