The matrix X such that it satisfies the equation: \( \mathrm{X}\begin{bmatrix}3 & 2\\1 & -1\end{bmatrix}=\begin{bmatrix}4 & 1\\2 & 3\end{bmatrix} \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} \)
Let \( \mathrm{B}=\begin{bmatrix}3 & 2\\1 & -1\end{bmatrix} \) and \( \mathrm{C}=\begin{bmatrix}4 & 1\\2 & 3\end{bmatrix} \), so the equation is \( \mathrm{XB}=\mathrm{C} \). Since X is on the left of B, multiplying by \( \mathrm{B}^{-1} \) on the right gives \( \mathrm{X}=\mathrm{C}\mathrm{B}^{-1} \).
\( \displaystyle |\mathrm{B}|=-3-2=-5 \)
\( \displaystyle \mathrm{B}^{-1}=\frac{1}{-5}\begin{bmatrix}-1 & -2\\-1 & 3\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}\frac{1}{5} & \frac{2}{5}\\\frac{1}{5} & -\frac{3}{5}\end{bmatrix} \)
\( \displaystyle \mathrm{X}=\begin{bmatrix}4 & 1\\2 & 3\end{bmatrix}\begin{bmatrix}\frac{1}{5} & \frac{2}{5}\\\frac{1}{5} & -\frac{3}{5}\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}\frac{4+1}{5} & \frac{8-3}{5}\\\frac{2+3}{5} & \frac{4-9}{5}\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} \)
Shortcut: Multiply each option by B instead of finding the inverse. For Option (4):
\( \displaystyle \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}\begin{bmatrix}3 & 2\\1 & -1\end{bmatrix}=\begin{bmatrix}3+1 & 2-1\\3-1 & 2+1\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}4 & 1\\2 & 3\end{bmatrix} \)
Tip: In \( \mathrm{XB}=\mathrm{C} \) the answer is \( \mathrm{CB}^{-1} \), not \( \mathrm{B}^{-1}\mathrm{C} \); matrix multiplication is not commutative, so the side matters.