The matrix \(X\) in the equation \(AX = B\), such that \(A = \begin{bmatrix}1&3\\0&2\end{bmatrix}\) and \(B = \begin{bmatrix}1&-1\\0&2\end{bmatrix}\) is given by
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{2}\begin{bmatrix}2&-8\\0&2\end{bmatrix}\)
Since \(|A| = 1 \times 2 - 3 \times 0 = 2 \ne 0\), A is invertible and \(X = A^{-1}B\). For a 2 × 2 matrix, the inverse is \(\frac{1}{|A|}\,\mathrm{adj}\,A\):
\( \displaystyle A^{-1} = \frac{1}{2}\begin{bmatrix}2&-3\\0&1\end{bmatrix} \)
So:
\( \displaystyle X = \frac{1}{2}\begin{bmatrix}2&-3\\0&1\end{bmatrix}\begin{bmatrix}1&-1\\0&2\end{bmatrix} \)
\( \displaystyle = \frac{1}{2}\begin{bmatrix}2 + 0&-2 - 6\\0 + 0&0 + 2\end{bmatrix} = \frac{1}{2}\begin{bmatrix}2&-8\\0&2\end{bmatrix} \)
Shortcut: Multiply A by Option (1) and check that the product is B:
\( \displaystyle \begin{bmatrix}1&3\\0&2\end{bmatrix}\begin{bmatrix}1&-4\\0&1\end{bmatrix} = \begin{bmatrix}1&-4 + 3\\0&2\end{bmatrix} = \begin{bmatrix}1&-1\\0&2\end{bmatrix} = B \)