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