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Mathematics Question on Determinants

Evaluate the determinants
(i) ∣3−1−2 00−1 3−50∣\begin{vmatrix}3 & -1 & -2\\\ 0 & 0 & -1 \\\ 3&-5&0\end{vmatrix}(iii) ∣3−45 12−2 231∣\begin{vmatrix} 3 & -4 & 5\\\ 1 & 2 & -2 \\\ 2&3&1\end{vmatrix}
(ii)∣012 −10−3 −230∣\begin{vmatrix} 0 & 1 & 2\\\ -1 & 0 & -3 \\\ -2&3&0\end{vmatrix} (iv)∣2−1−2 02−1\3−50∣\begin{vmatrix} 2 & -1 & -2\\\ 0 & 2 & -1\\\3&-5&0 \end{vmatrix}

Answer

(i) Let A=∣3−1−2 00−1 3−50∣\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∣−1−2 −50∣\begin{vmatrix}-1 & -2 \\\ -5&0\end{vmatrix}+0∣3−2\30∣\begin{vmatrix}3&-2\\\3&0\end{vmatrix}-(-1)∣3−1 3−5∣\begin{vmatrix}3& -1 \\\ 3&-5\end{vmatrix}= (-15+3)=-12

(ii)Let A=∣012 −10−3 −230∣\begin{vmatrix} 0 & 1 & 2\\\ -1 & 0 & -3 \\\ -2&3&0\end{vmatrix}

By expanding along the first row, we have:
IAI=0∣0−3 30∣\begin{vmatrix}0 & -3 \\\ 3&0\end{vmatrix}-1∣−1−3 −20∣\begin{vmatrix}-1 & -3 \\\ -2&0\end{vmatrix}
=0-1(0-6)+2(-3-0)
=6-6=0

(iii) Let A=∣3−45 12−2 231∣\begin{vmatrix} 3 & -4 & 5\\\ 1 & 2 & -2 \\\ 2&3&1\end{vmatrix}

By expanding along the first row, we have:
IAI=3∣1−2 31∣\begin{vmatrix}1 & -2 \\\ 3&1\end{vmatrix}+4∣1−2 21∣\begin{vmatrix}1 & -2 \\\ 2&1\end{vmatrix}+5∣11 23∣\begin{vmatrix}1 & 1 \\\ 2&3\end{vmatrix}
=3(1+6)+4(1+4)+5(3-2)
=21+20+5
=46

(iv) Let A=∣2−1−2 02−1\3−50∣\begin{vmatrix} 2 & -1 & -2\\\ 0 & 2 & -1\\\3&-5&0 \end{vmatrix}

By expanding along the first column, we have:
IAI=∣2−1 −50∣\begin{vmatrix}2 & -1 \\\ -5&0\end{vmatrix}-0∣−1−2 −50∣\begin{vmatrix}-1 & -2 \\\ -5&0\end{vmatrix}+3∣−1−2 2−2∣\begin{vmatrix}-1 & -2 \\\ 2&-2\end{vmatrix}
=2(0-5)-0+3(1+4)
=-10+15
=5