Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS3 Marks
Question
Using Cofactors of elements of second row, evaluate $\triangle=\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix}.$
✓
Answer
The given determinant is $\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix}.$
we have:
$M_{21}$ $=\begin{vmatrix}3&8\\2&3\end{vmatrix}=9-16=-7$
$\therefore$ $A_{21}$ = cofactor of $a_{21} = (-1)^{2+1}M_{21} = 7$
$M_{22}$ $=\begin{vmatrix}5&8\\1&3\end{vmatrix}=15-8=7$
$\therefore$ $A_{22}$ = cofactor of $a_{22} = (-1)^{2+2} M_{22} = 7$
$M_{23}$ $=\begin{vmatrix}5&3\\1&2\end{vmatrix}=10-3=7$
$\therefore$ $A_{23}$ = cofactor of $a_{22} = (-1)^{2+3} M_{23}= -7$
We know that $\triangle$ is equal to the sum of the product of the elements of the second row with their corresponding cofactor.
$\therefore\triangle$ $= a_{21}A_{21} + a_{22}A_{22}+ a_{23}A_{23}= 2(7) + 0(7) + 1(-7) = 14 - 7 = 7$
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