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# Using Cofactors of elements of second row, evaluate.

Question:

Using Cofactors of elements of second row, evaluate $\Delta=\left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|$.

Solution:

The given determinant is $\left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|$.

We have:

$\mathrm{M}_{21}=\left|\begin{array}{ll}3 & 8 \\ 2 & 3\end{array}\right|=9-16=-7$

$\therefore \mathrm{A}_{21}=$ cofactor of $\mathrm{a}_{21}=(-1)^{2+1} \mathrm{M}_{21}=7$

$\mathrm{M}_{22}=\left|\begin{array}{ll}5 & 8 \\ 1 & 3\end{array}\right|=15-8=7$

$\therefore \mathrm{A}_{22}=$ cofactor of $\mathrm{a}_{22}=(-1)^{2+2} \mathrm{M}_{22}=7$

$\mathrm{M}_{23}=\left|\begin{array}{ll}5 & 3 \\ 1 & 2\end{array}\right|=10-3=7$

$\therefore \mathrm{A}_{23}=$ cofactor of $a_{23}=(-1)^{2+3} \mathrm{M}_{23}=-7$

We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

$\therefore \Delta=a_{21} \mathrm{~A}_{21}+a_{22} \mathrm{~A}_{22}+a_{23} \mathrm{~A}_{23}=2(7)+0(7)+1(-7)=14-7=7$