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

Given $A = \begin{bmatrix} 1 & 3 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}$, we need to find matrix $X$ such that $AX = B$.

From $AX = B$:

$X = A^{-1}B$

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For a 2×2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is:

$A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

For matrix $A$ where $a = 1, b = 3, c = 0, d = 2$:

$ad - bc = (1)(2) - (3)(0)$

$= 2$

Therefore:

$A^{-1} = \frac{1}{2}\begin{bmatrix} 2 & -3 \\ 0 & 1 \end{bmatrix}$

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Now calculate $X = A^{-1}B$:

$X = \frac{1}{2}\begin{bmatrix} 2 & -3 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}$

Position (1,1): $2(1) + (-3)(0) = 2$

Position (1,2): $2(-1) + (-3)(2) = -2 - 6 = -8$

Position (2,1): $0(1) + 1(0) = 0$

Position (2,2): $0(-1) + 1(2) = 2$

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$X = \frac{1}{2}\begin{bmatrix} 2 & -8 \\ 0 & 2 \end{bmatrix}$