Given that, $\mathrm{AB}=\left[\begin{array}{cc}4 & 11 \\ 4 & 5\end{array}\right]$ and $\mathrm{A}=\left[\begin{array}{ll}3 & 2 \\ 1 & 2\end{array}\right]$
Let $B=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}3 & 2 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right]=\left[\begin{array}{cc}4 & 11 \\ 4 & 5\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}3 a+2 c & 3 b+2 d \\ a+2 c & b+2 d\end{array}\right]=\left[\begin{array}{cc}4 & 11 \\ 4 & 5\end{array}\right]$
$\Rightarrow 3 \mathrm{a}+2 \mathrm{c}=4$ and $\mathrm{a}+2 \mathrm{c}=4$ ...(1)
and $3 b+2 d=11$ and $b+2 d=5$ ...(2)
From equation set $(1) \mathrm{a}=0$ and $\mathrm{c}=2$ and from equation $\operatorname{set}(2), \mathrm{b}=3$ and $\mathrm{d}=1$
$\Rightarrow B=\left[\begin{array}{ll}0 & 3 \\ 2 & 1\end{array}\right]$
Hence $|\mathrm{B}|=\left|\begin{array}{ll}0 & 3 \\ 2 & 1\end{array}\right|=0-6=-6$