The matrix X in the equation AX = B, such that $A=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]$ is given by

$\left[\begin{array}{cc}1 & -4 \\ 0 & 1\end{array}\right]$

Given $A=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] \Rightarrow|A|=1 \neq 0 \Rightarrow A^{-1}$ exists

$A^{-1}=\frac{adj~ A}{|A|}=\left[\begin{array}{cc} 1 & -3 \\ 0 & 1 \end{array}\right]$

$AX=B \Rightarrow X=A^{-1} B$

$=\left[\begin{array}{cc} 1 & -3 \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & -4 \\ 0 & 1 \end{array}\right]$

Hence (2) is the correct answer.