If \( \mathrm{A}\begin{bmatrix}1 & -2\\1 & 4\end{bmatrix}=\begin{bmatrix}6 & 0\\0 & 6\end{bmatrix} \), then matrices \( A \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix} \)
Let M be the matrix multiplied on the right. Then \( \mathrm{A}\mathrm{M}=6\mathrm{I} \), so \( \mathrm{A}=6\mathrm{M}^{-1} \).
\( |\mathrm{M}|=4+2=6 \)
\( \displaystyle \mathrm{adj}\,\mathrm{M}=\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix} \)
\( \displaystyle \mathrm{A}=6\cdot\frac{1}{6}\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}=\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix} \)
Shortcut: Instead of finding the inverse, multiply each option by the given matrix. Option (3) gives
\( \displaystyle \begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}\begin{bmatrix}1 & -2\\1 & 4\end{bmatrix}=\begin{bmatrix}4+2 & -8+8\\-1+1 & 2+4\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}6 & 0\\0 & 6\end{bmatrix} \)