(i) If $\mathrm{A}=\left[\begin{array}{ccc}5& 6& -3\\ -4& 3& 2\\ -4& -7& 3\end{array}\right], \mathrm{then}$
a) The cofactor of the element 𝑎₂₁ of its 2^nd row.
b) Determinant of A

(ii) In the interval it $\mathrm{\pi }/2<\mathrm{x}<\mathrm{\pi }$, find the value of x for which the matrix, $\left[\begin{array}{cc}2\sin \mathrm{x}& 3\\ 1& 2\sin \mathrm{x}\end{array}\right]$ is singular.

(i)- (a) Given, $\mathrm{A}=\left[\begin{array}{ccc}5& 6& -3\\ -4& 3& 2\\ -4& -7& 3\end{array}\right]$
Now, cofactors of ${\mathrm{a}}_{21}=-1\left|\begin{array}{cc}6& -3\\ -7& 3\end{array}\right|$
$\begin{array}{l}=-(18-21)\\ =3\end{array}$

(b) Determinant of $\mathrm{A}=\left|\begin{array}{ccc}5& 6& -3\\ -4& 3& 2\\ -4& -7& 3\end{array}\right|$
$\begin{array}{l}=5(9+14)-6(-12+8)-3(28+12)\\ =115+24-120\\ =19\end{array}$

ii-
$\mathrm{Let} \mathrm{A}=\left[\begin{array}{cc}2\mathrm{sinx}& 3\\ 1& 2\mathrm{sinx}\end{array}\right]$
$\therefore$A is a singular matrix.
$\begin{array}{l}\therefore \left|\mathrm{A}\right|=0\Rightarrow \begin{array}{cc}2\sin \mathrm{x}& 3\\ 1& 2\sin \mathrm{x}\end{array}\mid =0\\ \Rightarrow 4{\sin }^2 \mathrm{x}-3=0\Rightarrow {\sin }^2 \mathrm{x}=\frac{3}{4}\\ \Rightarrow \sin \mathrm{x}=\frac{\sqrt{3}}{2}\left[\begin{array}{c}\text{ taking positive square }\\ \text{ root because }\frac{\mathrm{\pi }}{2}<\mathrm{x}<\mathrm{\pi }\end{array}\right]\\ \therefore \mathrm{x}=\frac{2\mathrm{\pi }}{3}\end{array}$