Evaluate the determinants

(i) Let $A=\left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$.

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

$|A|=-0\left|\begin{array}{cc}-1 & -2 \\ -5 & 0\end{array}\right|+0\left|\begin{array}{cc}3 & -2 \\ 3 & 0\end{array}\right|-(-1)\left|\begin{array}{ll}3 & -1 \\ 3 & -5\end{array}\right|=(-15+3)=-12$

(ii) Let $A=\left[\begin{array}{ccc}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{array}\right]$.

By expanding along the first row, we have:

$\begin{aligned}|A| &=3\left|\begin{array}{cc}1 & -2 \\ 3 & 1\end{array}\right|+4\left|\begin{array}{cc}1 & -2 \\ 2 & 1\end{array}\right|+5\left|\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right| \\ &=3(1+6)+4(1+4)+5(3-2) \\ &=3(7)+4(5)+5(1) \\ &=21+20+5=46 \end{aligned}$

(iii) Let $A=\left[\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right]$.

By expanding along the first row, we have:

$\begin{aligned}|A| &=0\left|\begin{array}{cc}0 & -3 \\ 3 & 0\end{array}\right|-1\left|\begin{array}{cc}-1 & -3 \\ -2 & 0\end{array}\right|+2\left|\begin{array}{rr}-1 & 0 \\ -2 & 3\end{array}\right| \\ &=0-1(0-6)+2(-3-0) \\ &=-1(-6)+2(-3) \\ &=6-6=0 \end{aligned}$

(iv) Let $A=\left[\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right]$.

By expanding along the first column, we have:

$\begin{aligned}|A| &=2\left|\begin{array}{cc}2 & -1 \\ -5 & 0\end{array}\right|-0\left|\begin{array}{cc}-1 & -2 \\ -5 & 0\end{array}\right|+3\left|\begin{array}{cc}-1 & -2 \\ 2 & -1\end{array}\right| \\ &=2(0-5)-0+3(1+4) \\ &=-10+15=5 \end{aligned}$