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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS3 Marks
Question
Using Cofactors of elements of second row, evaluate  $\triangle=\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix}.$
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Answer

The given determinant is $\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix}.$
we have:
M21 $=\begin{vmatrix}3&8\\2&3\end{vmatrix}=9-16=-7$
$\therefore$ A21 = cofactor of a21 = (-1)2+1 M21 = 7
M22  $=\begin{vmatrix}5&8\\1&3\end{vmatrix}=15-8=7$
$\therefore$ A22 = cofactor of a22 = (-1)2+2 M22 = 7
M23  $=\begin{vmatrix}5&3\\1&2\end{vmatrix}=10-3=7$
$\therefore$ A23 = cofactor of a22 = (-1)2+3 M23 = -7
We know that $\triangle$ is equal to the sum of the product of the elements of the second row with their corresponding cofactor.
$\therefore\triangle$ = a21A21 + a22A22 + a23A23 = 2(7) + 0(7) + 1(-7) = 14 - 7 = 7

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