Using Cofactors of elements of second row, evaluate = 5&3&8 2&0&1… - Vidyadip

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:
$M_{21} =\begin{vmatrix}3&8\\2&3\end{vmatrix}=9-16=-7$
$\therefore A_{21} =$ 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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