Question

Let \[ A = \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}. \] If \( XA = B \), then \( X \) is:

• A. \[ \begin{pmatrix} -3 & -1 \\ 1 & 1 \end{pmatrix} \]
• B. \[ \begin{pmatrix} -3 & 1 \\ 3 & -1 \end{pmatrix} \]
• C. \[ \begin{pmatrix} 3 & -1 \\ -3 & 1 \end{pmatrix} \]
• D. \[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
• E. \[ \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \]

Hint:

When solving matrix equations of the form \( XA = B \), you can isolate \( X \) by multiplying both sides by \( A^{-1} \) from the right. Be sure to compute the inverse correctly using the determinant and the adjoint matrix.

Answer 1

The Correct Option is C

We are given that:

\[ XA = B. \] To solve for \( X \), we multiply both sides by the inverse of \( A \) on the right:

\[ X = B A^{-1}. \] Step 1: Find the inverse of \( A \). The formula for the inverse of a 2x2 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is:

\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. \] For \( A = \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix} \), we have:

\[ \text{Determinant of } A = (0)(2) - (1)(-1) = 1. \] So, the inverse is:

\[ A^{-1} = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}. \] Step 2: Now, calculate \( X = B A^{-1} \):

\[ B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}, \quad A^{-1} = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}. \] Multiply \( B \) and \( A^{-1} \):

\[ X = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} (1)(2) + (1)(1) & (1)(-1) + (1)(0) \\ (-1)(2) + (-1)(1) & (-1)(-1) + (-1)(0) \end{pmatrix}. \] \[ X = \begin{pmatrix} 3 & -1 \\ -3 & 1 \end{pmatrix}. \] Thus, the value of \( X \) is \( \begin{pmatrix} 3 & -1 \\ -3 & 1 \end{pmatrix} \).

Therefore, the correct answer is option (C).