Q.Evaluate $\begin{vmatrix} 2 & 0 & 1 \\ 3 & 1 & -2 \\ 1 & 4 & 5 \end{vmatrix}$ by expanding along the first row.

Expanding along the first row, each entry is multiplied by its minor with the $+,-,+$ signs and the results are added.

The first row's middle entry is $0$, which drops one term from the sum.

Expanding along the first row (entries $2, 0, 1$): $|A| = 2\begin{vmatrix} 1 & -2 \\ 4 & 5 \end{vmatrix} - 0\begin{vmatrix} 3 & -2 \\ 1 & 5 \end{vmatrix} + 1\begin{vmatrix} 3 & 1 \\ 1 & 4 \end{vmatrix}.$

Evaluating the needed minors: $\begin{vmatrix} 1 & -2 \\ 4 & 5 \end{vmatrix} = 5-(-8) = 13,\qquad \begin{vmatrix} 3 & 1 \\ 1 & 4 \end{vmatrix} = 12-1 = 11.$

So $|A| = 2(13) - 0 + 1(11) = 26 + 11 = 37.$

Verification by expanding along the first column (entries $2, 3, 1$): $C_{11}=+\begin{vmatrix} 1 & -2 \\ 4 & 5 \end{vmatrix}=13$, $C_{21}=-\begin{vmatrix} 0 & 1 \\ 4 & 5 \end{vmatrix}=-(0-4)=4$, $C_{31}=+\begin{vmatrix} 0 & 1 \\ 1 & -2 \end{vmatrix}=0-1=-1$. Then $|A| = 2(13)+3(4)+1(-1) = 26+12-1 = 37$ — the same value.

Because the first row's middle entry is 0, one could equally expand along the second column or elsewhere; the zero entry always removes its own term whichever line it lies on.

Students sometimes still compute the minor of the zero entry and waste effort, or forget the minus sign in front of the (already zero) middle term. Since that term is 0 either way it does not affect the answer, but the sign pattern must still be applied correctly to the non-zero terms.