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