Question

It is given that \[ X \begin{bmatrix} 3 & 2 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}. \text{ Then matrix X is:} \]

• A. $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
• B. $\begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}$
• C. $\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$
• D. $\begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}$

Hint:

Use basic matrix multiplication rules to set up linear equations.

Answer 1

The Correct Option is C

Let $X = \begin{bmatrix} a & b
c & d \end{bmatrix}$. Then, \[ X \begin{bmatrix} 3 & 2 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}. \] So, \[ \begin{cases} 3a + 1b = 4\\ 2a - 1b = 1\\ 3c + 1d = 2\\ 2c - 1d = 3 \end{cases} \] Solve: \[ 3a + b = 4, 2a - b = 1. \text{Add:} 5a = 5 \implies a = 1. \text{So,} b = 1. \] Next: \[ 3c + d = 2, 2c - d = 3. \text{Add:} 5c = 5 \implies c = 1. \text{So,} d = -1. \] So, \[ X = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. \]