Question

Using Cofactors of elements of second row, evaluate △=$\begin{vmatrix}5&3&8\\\2&0&1\\\1&2&3\end{vmatrix}$

Answer 1

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
∴A21=cofactor of a21=(−1)2+1 M21=7
M22 =$\begin{vmatrix}5&8\\\1&3\end{vmatrix}$=15-8=7
∴A22=cofactor of a22=(−1)2+2 M22=7
M23=$\begin{vmatrix}5&3\\\1&2\end{vmatrix}$=10-3=7
∴A23 = cofactor of a23 = (−1)2+3 M23 = −7
We know that ∆ is equal to the sum of the product of the elements of the second row
with their corresponding cofactors.

∴ ∆ = a21A21 + a22A22 + a23A23 = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7