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) → $\frac{1}{2}\begin{bmatrix}2&-8\\0&2\end{bmatrix}$
The matrix equation is:
$AX = B$
where $A = \begin{bmatrix} 1 & 3 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}$.
Since $X = A^{-1} B$, first find the inverse of $A$:
$A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$
$= \frac{1}{(1)(2) - (3)(0)} \begin{bmatrix} 2 & -3 \\ 0 & 1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 2 & -3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{3}{2} \\ 0 & \frac{1}{2} \end{bmatrix}$
Now, compute $X$:
$X = A^{-1} B = \begin{bmatrix} 1 & -\frac{3}{2} \\ 0 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}$
$= \begin{bmatrix} 1 \times 1 + \left(-\frac{3}{2}\right) \times 0 & 1 \times (-1) + \left(-\frac{3}{2}\right) \times 2 \\ 0 \times 1 + \frac{1}{2} \times 0 & 0 \times (-1) + \frac{1}{2} \times 2 \end{bmatrix}$
$= \begin{bmatrix} 1 & -1 - 3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -4 \\ 0 & 1 \end{bmatrix}$