Question

If the given matrices \[ A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}, \quad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] satisfy \( A^{2} = kA - 2I\), \(\text{ the value of coefficient }\) k \(\text{ is}\) _________.

• A. 1
• B. 2
• C. 0
• D. 4

Hint:

To solve matrix equations, square the matrix, then equate the resulting expression with the given equation.

Answer 1

The Correct Option is A

Step 1: Square the matrix \( A \).
\[ A^{2} = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \times \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \] \[ = \begin{bmatrix} 3(3) + (-2)(4) & 3(-2) + (-2)(-2) \\ 4(3) + (-2)(4) & 4(-2) + (-2)(-2) \end{bmatrix} \] \[ = \begin{bmatrix} 9 - 8 & -6 + 4 \\ 12 - 8 & -8 + 4 \end{bmatrix} \] \[ = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} \] Step 2: Use the equation \( A^{2} = kA - 2I \).
From the given equation, substitute the known values: \[ \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = k \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 3k - 2 & -2k \\ 4k & -2k - 2 \end{bmatrix} \] Step 3: Equate the corresponding elements.
By comparing the two matrices, we get the following system of equations: - \( 3k - 2 = 1 \)
- \( -2k = -2 \)
- \( 4k = 4 \)
- \( -2k - 2 = -4 \)
Solving \( 3k - 2 = 1 \) gives: \[ 3k = 3 \quad \Rightarrow \quad k = 1 \] Conclusion:
Thus, the value of \( k \) is \( 1 \). The correct answer is option (A).