Question

Evaluate the determinants..

$\left|\begin{array}{ccc}3& -1& -2\\ 0& 0& -1\\ 3& -5& 0\end{array}\right|$

$\left|\begin{array}{ccc}3& -4& 5\\ 1& 1& -2\\ 2& 3& 1\end{array}\right|$

$\left|\begin{array}{ccc}0& 1& 2\\ -1& 0& -3\\ -2& 3& 0\end{array}\right|$

$\left|\begin{array}{ccc}2& -1& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}\right|$

Answer 1

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 the determinant along the second row for easier calculation.
$|A|=-0\left|\begin{array}{cc}-1& -2\\ -5& 0\end{array}\right|+0\left|\begin{array}{cc}3& -2\\ 3& 0\end{array}\right|-(-1)\left|\begin{array}{cc}3& -1\\ 3& -5\end{array}\right|=(3\times (-5)-3\times (-1\left)\right)=-15+3=-12$

Let $A=\left|\begin{array}{ccc}3& -4& 5\\ 1& 1& -2\\ 2& 3& 1\end{array}\right|$
By expanding along $R_1$ (first row), we get
$|A|=3\left|\begin{array}{cc}1& -2\\ 3& 1\end{array}\right|-(-4)\left|\begin{array}{cc}1& -2\\ 2& 1\end{array}\right|+5\left|\begin{array}{cc}1& 1\\ 2& 3\end{array}\right|=3(1+6)+4(1+4)+5(3-2)=3\left(7\right)+4\left(5\right)+5\left(1\right)=21+20+5=46$

Let $A=\left|\begin{array}{ccc}0& 1& 2\\ -1& 0& -3\\ -2& 3& 0\end{array}\right|$
By expanding along $R_1$ (first row), we get
$|A|=0\left|\begin{array}{cc}0& -3\\ 3& 0\end{array}\right|-1\left|\begin{array}{cc}-1& -3\\ -2& 0\end{array}\right|+2\left|\begin{array}{cc}-1& 0\\ -2& 3\end{array}\right|=-1(0-6)+2(-3-0)=-1(-6)+2(-3)=6-6=0$

Let $A=\left|\begin{array}{ccc}2& -2& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}\right|$
By expanding along $R_2$ (second row), we get
$|A|=0\left|\begin{array}{cc}-1& -2\\ -5& 0\end{array}\right|-2\left|\begin{array}{cc}2& -2\\ 3& 0\end{array}\right|-(-1)\left|\begin{array}{cc}2& -1\\ 3& -5\end{array}\right|=0+2(0+6)+(-10+3)=12-7=5$