Question

Evaluate the determinants 
(i) \(\begin{vmatrix}3 & -1 & -2\\ 0 & 0 & -1 \\ 3&-5&0\end{vmatrix}\)(iii) \(\begin{vmatrix} 3 & -4 & 5\\ 1 & 2 & -2 \\ 2&3&1\end{vmatrix}\)
(ii)\(\begin{vmatrix} 0 & 1 & 2\\ -1 & 0 & -3 \\ -2&3&0\end{vmatrix}\) (iv)\(\begin{vmatrix} 2 & -1 & -2\\ 0 & 2 & -1\\3&-5&0 \end{vmatrix}\)

Answer 1

(i) Let A=\(\begin{vmatrix}3 & -1 & -2\\ 0 & 0 & -1 \\ 3&-5&0\end{vmatrix}\)

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

IAI=-0\(\begin{vmatrix}-1 & -2 \\ -5&0\end{vmatrix}\)+0\(\begin{vmatrix}3&-2\\3&0\end{vmatrix}\)-(-1)\(\begin{vmatrix}3& -1 \\ 3&-5\end{vmatrix}\)= (-15+3)=-12 

(ii)Let A=\(\begin{vmatrix} 0 & 1 & 2\\ -1 & 0 & -3 \\ -2&3&0\end{vmatrix}\)

By expanding along the first row, we have:
 IAI=0\(\begin{vmatrix}0 & -3 \\ 3&0\end{vmatrix}\)-1\(\begin{vmatrix}-1 & -3 \\ -2&0\end{vmatrix}\)
=0-1(0-6)+2(-3-0) 
=6-6=0 

(iii) Let A=\(\begin{vmatrix} 3 & -4 & 5\\ 1 & 2 & -2 \\ 2&3&1\end{vmatrix}\)

By expanding along the first row, we have: 
IAI=3\(\begin{vmatrix}1 & -2 \\ 3&1\end{vmatrix}\)+4\(\begin{vmatrix}1 & -2 \\ 2&1\end{vmatrix}\)+5\(\begin{vmatrix}1 & 1 \\ 2&3\end{vmatrix}\) 
=3(1+6)+4(1+4)+5(3-2) 
=21+20+5
=46 

(iv) Let A=\(\begin{vmatrix} 2 & -1 & -2\\ 0 & 2 & -1\\3&-5&0 \end{vmatrix}\)

By expanding along the first column, we have: 
IAI=\(\begin{vmatrix}2 & -1 \\ -5&0\end{vmatrix}\)-0\(\begin{vmatrix}-1 & -2 \\ -5&0\end{vmatrix}\)+3\(\begin{vmatrix}-1 & -2 \\ 2&-2\end{vmatrix}\)
=2(0-5)-0+3(1+4) 
=-10+15
=5