If X=[ array ll 3 & -4 1 & -1 array ], B=[ array cc 5 & 2 -2 & 1...

\mathrm{X}=\left[$\begin{array}{cc}3 & -4 \\ 1 & -1\end{array}$\right], \mathrm{B}=\left[$\begin{array}{cc}5 & 2 \\ -2 & 1\end{array}$\right] and \mathrm{A}=\left[$\begin{array}{ll}\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{s}\end{array}$\right] Now, \mathrm{AX}=\mathrm{B}\therefore \quad\left[$\begin{array}{ll} \mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{ s} \end{array}$\right]\left[$\begin{array}{cc} 3 & -4 \\ 1 & -1 \end{array}$\right]=\left[$\begin{array}{cc} 5 & 2 \\ -2 & 1 \end{array}$\right]\Rightarrow\left[$\begin{array}{cc}3 p+q & -4 p-q \\ 3 r+s & -4 r-s\end{array}$\right]=\left[$\begin{array}{cc}5 & 2 \\ -2 & 1\end{array}$\right]$\begin{aligned} & 3 p+q=5 ...(i)\\ & -4 p-q=2 ...(ii)\\ & 3 r+s=-2 ...(iii)\\ & -4 r-s=1...(iv) \end{aligned}$$$From equations (i) and (ii), we get$$\begin{array}{ll} & -p=7 \\ \therefore \quad & p=-7 \\ \Rightarrow \quad q=5-3(-7) \\ & q=26 \end{array}$$From equations (iii) and (iv), -r=-1$$\begin{aligned}& \therefore \quad \mathrm{r}=1 \\ \text{ and } \Rightarrow \quad \mathrm{ s}=-2-3=-5 \\ \text{ and } \therefore \quad \mathrm{ s}=-5 \end{aligned}$ Hence, $A=\left[\begin{array}{ll}\mathrm{p}& \mathrm{q}\\ \mathrm{r}& \mathrm{s}\end{array}\right]=\left[\begin{array}{cc}-7 & 26 \\ 1 & -5\end{array}$\right]\therefore \quad Option (A) is correct.