Question

If \( A = \begin{bmatrix} 4 & -1 \\ 12 & x \end{bmatrix} \) and \( A^2 = A \), then the value of \( x \) is:

• A. -8
• B. -3
• C. 0
• D. 3
• E. 8

Hint:

For matrix equations like \( A^2 = A \), solve by multiplying the matrices and equating corresponding elements.

Answer 1

The Correct Option is B

Given \( A^2 = A \), we have: \[ A^2 = \begin{bmatrix} 4 & -1 \\ 12 & x \end{bmatrix}^2 = \begin{bmatrix} 4 & -1 \\ 12 & x \end{bmatrix} \] Calculating \( A^2 \): \[ A^2 = \begin{bmatrix} 4 & -1 \\ 12 & x \end{bmatrix} \times \begin{bmatrix} 4 & -1 \\ 12 & x \end{bmatrix} \] \[ A^2 = \begin{bmatrix} 16 - 12 & -4 + x \\ 48 + 12x & -12 + x^2 \end{bmatrix} \] Equating this to \( A \): \[ \begin{bmatrix} 16 - 12 & -4 + x \\ 48 + 12x & -12 + x^2 \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ 12 & x \end{bmatrix} \] 
We solve the system of equations: 
1. \( 16 - 12 = 4 \), so this is satisfied. 
2. \( -4 + x = -1 \Rightarrow x = 3 \). 
3. \( 48 + 12x = 12 \Rightarrow 12x = -36 \Rightarrow x = -3 \). 
4. \( -12 + x^2 = x \Rightarrow x^2 - x - 12 = 0 \Rightarrow (x - 3)(x + 4) = 0 \Rightarrow x = -3 \). 
Thus, \( x = -3 \).