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Q : 3 Using Cofactors of elements of second row, evaluate . $\small \Delta =\begin{vmatrix} 5 &3 &8 \\ 2 & 0 & 1\\ 1 &2 &3 \end{vmatrix}$

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Given determinant : $\small \Delta =\begin{vmatrix} 5 &3 &8 \\ 2 & 0 & 1\\ 1 &2 &3 \end{vmatrix}$

First finding Minors of the second rows by the definition,

$M_{21} =$ minor of $a_{21} = \begin{vmatrix} 3 &8 \\ 2 &3 \end{vmatrix} =9-16 = -7$

$M_{22} =$ minor of $a_{22} = \begin{vmatrix} 5 &8 \\ 1 &3 \end{vmatrix} = 15-8=7$

$M_{23} =$ minor of $a_{23} = \begin{vmatrix} 5 &3 \\ 1 &2 \end{vmatrix} = 10-3 =7$

Finding the Cofactors of the second row:

$A_{21}=$ Cofactor of $a_{21} = (-1)^{2+1}M_{21} = 7$

$A_{22}=$ Cofactor of $a_{22} = (-1)^{2+2}M_{22} = 7$

$A_{23}=$ Cofactor of $a_{23} = (-1)^{2+3}M_{23} = -7$

Therefore we can calculate $\triangle$ by sum of the product of the elements of the second row with their corresponding cofactors.

Therefore we have,

$\triangle = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = 2(7) +0(7) +1(-7) =14-7=7$

Posted by

Divya Prakash Singh

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