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Question

Using cofactors of elements of second row, evaluate=∣ ∣538201123∣ ∣

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Solution

We have to evaluate Δ=∣ ∣538201123∣ ∣ using the cofactors of elements of second row.

Δ=a21A21+a22A22+a23A23 where A21,A22,A23 are the cofactors of second row.

A21=(1)(916)=(7)=7

A22=(+1)(158)=7

A23=(1)(103)=7

Δ=2×7+0×7+1×(7)

=14+07

=7

Δ=7

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