Question

For the matrix \[ \begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix} \] , eigenvalue corresponding to the eigenvector \[ \begin{bmatrix} 2 \\ -3 \end{bmatrix} \] is ................. (in integer).

Hint:

To find the eigenvalue, multiply the matrix by the eigenvector and compare the result to the original eigenvector, yielding the scalar eigenvalue.

Answer 1

We are given the matrix:
\[ A = \begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix} \] and the eigenvector:
\[ v = \begin{bmatrix} 2 \\ -3 \end{bmatrix} \] We need to find the eigenvalue corresponding to this eigenvector.
The relationship between the matrix \(A\), eigenvector \(v\), and eigenvalue \(\lambda\) is given by:
\[ A v = \lambda v \] Substitute the values:
\[ \begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ -3 \end{bmatrix} = \lambda \begin{bmatrix} 2 \\ -3 \end{bmatrix} \] First, multiply the matrix \(A\) by the vector \(v\):
\[ \begin{bmatrix} 4 \times 2 + 2 \times (-3) \\ 3 \times 2 + 3 \times (-3) \end{bmatrix} = \begin{bmatrix} 8 - 6 \\ 6 - 9 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix} \] So we have:
\[ \begin{bmatrix} 2 \\ -3 \end{bmatrix} = \lambda \begin{bmatrix} 2 \\ -3 \end{bmatrix} \] This implies that:
\[ \lambda = 1 \] Thus, the eigenvalue is 1.