Question

Let \[ S = \left\{ AX : A = \begin{bmatrix} 2 & -4 \\ 1 & 1 \\ 1 & -1 \end{bmatrix}, \, X = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \right\}. \] If \[ \begin{bmatrix} -1 \\ \alpha \\ 1 \end{bmatrix} \in S, \text{ then the value of } \alpha \text{ is} \]

• A. -4
• B. -2
• C. 2
• D. 4

Hint:

To solve matrix equations, break them down into a system of linear equations and solve step by step.

Answer 1

The Correct Option is D

We are given the matrix equation: \[ AX = \begin{bmatrix} -1 \\ \alpha \\ 1 \end{bmatrix}. \] We need to solve for \( \alpha \). Using the matrix \( A \) and the given vector, we can set up the system of equations by performing the matrix multiplication: \[ \begin{bmatrix} 2 & -4 \\ 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} -1 \\ \alpha \\ 1 \end{bmatrix}. \] This results in the following system of equations: \[ 2x_1 - 4x_2 = -1, x_1 + x_2 = \alpha, x_1 - x_2 = 1. \] From the third equation, we have: \[ x_1 - x_2 = 1 $\Rightarrow$ x_1 = x_2 + 1. \] Substitute this expression for \( x_1 \) into the second equation: \[ x_2 + 1 + x_2 = \alpha $\Rightarrow$ 2x_2 + 1 = \alpha $\Rightarrow$ x_2 = \frac{\alpha - 1}{2}. \] Now, substitute \( x_2 = \frac{\alpha - 1}{2} \) into the first equation: \[ 2(x_2 + 1) - 4x_2 = -1 $\Rightarrow$ 2\left( \frac{\alpha - 1}{2} + 1 \right) - 4\left( \frac{\alpha - 1}{2} \right) = -1. \] Simplifying this equation: \[ 2 \times \frac{\alpha + 1}{2} - 4 \times \frac{\alpha - 1}{2} = -1 $\Rightarrow$ (\alpha + 1) - 2(\alpha - 1) = -1, \] \[ \alpha + 1 - 2\alpha + 2 = -1 $\Rightarrow$ -\alpha + 3 = -1 $\Rightarrow$ \alpha = 4. \] Final Answer: \text{(D) 4} \\