Question

If one of the eigenvectors of the matrix

\[ A = \begin{bmatrix} 1 & 1 \\ -4 & x \end{bmatrix} \]

is along the direction of

\[ \begin{bmatrix} 2\alpha \\ \alpha \end{bmatrix} \]

where \( \alpha \) is any non-zero real number, then the value of \( x \) is __________ (in integer).

Hint:

If a vector is an eigenvector of a matrix, then multiplying the matrix by that vector must result in a scalar multiple of the same vector. Use this property to derive unknowns like eigenvalues or entries of the matrix.

Answer 1

Correct Answer: 2

Step 1: Let the eigenvector be

\[ v = \begin{bmatrix} 2\alpha \\ \alpha \end{bmatrix} = \alpha \begin{bmatrix} 2 \\ 1 \end{bmatrix} \] Since an eigenvector is defined up to a scalar multiple, we may ignore \( \alpha \) and take \[ v = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \]

Step 2: Use the eigenvector property.

If \( v \) is an eigenvector of matrix \( A \), then \[ A v = \lambda v \]

Given \[ A = \begin{bmatrix} 1 & 1 \\ -4 & x \end{bmatrix} \]
Compute \( A v \):

\[ A \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1(2) + 1(1) \\ -4(2) + x(1) \end{bmatrix} = \begin{bmatrix} 3 \\ -8 + x \end{bmatrix} \]

Step 3: Equate with \( \lambda v \).

\[ \begin{bmatrix} 3 \\ -8 + x \end{bmatrix} = \lambda \begin{bmatrix} 2 \\ 1 \end{bmatrix} \]

This gives the system:

\[ 3 = 2\lambda \] \[ -8 + x = \lambda \]

Step 4: Solve for \( x \).

From the first equation:
\[ \lambda = \frac{3}{2} \]
Substitute into the second equation:

\[ -8 + x = \frac{3}{2} \Rightarrow x = \frac{19}{2} \]

Final Answer:
\[ x = \frac{19}{2} \]