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

Concept: Determinant evaluation using cofactor expansion along a chosen row.

We expand $\Delta$ along the second row (elements $a_{21}=2$, $a_{22}=0$, $a_{23}=1$). The formula is:

where $C_{ij} = (-1)^{i+j} M_{ij}$, and $M_{ij}$ is the minor (determinant of the submatrix after removing row $i$, column $j$).

Step 1: Compute $C_{21}$.

$M_{21} = \begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = (3)(3) - (8)(2) = 9 - 16 = -7$.

$C_{21} = (-1)^{2+1}(-7) = (-1)(-7) = 7$.

Step 2: Compute $C_{22}$.

$M_{22} = \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = (5)(3) - (8)(1) = 15 - 8 = 7$.

$C_{22} = (-1)^{2+2}(7) = (1)(7) = 7$. …

We expand $\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$ along row 2, whose entries are $a_{21}=2,\ a_{22}=0,\ a_{23}=1$. The cofactor is $A_{ij}=(-1)^{i+j}M_{ij}$.

Cofactor $A_{21}$ (delete row 2, column 1):

Middle term vanishes since $a_{22}=0$.

Cofactor $A_{23}$ (delete row 2, column 3): …

Method: Cofactor Expansion Along a Specified Row

Use this whenever the question explicitly directs which row (or column) to expand along.

Steps

Step 1: Identify the specified row's entries

Read off $a_{i1},a_{i2},a_{i3}$ for the required row $i$.

Step 2: Form each minor by deleting that row and the entry's own column

$M_{ij}$ is the determinant of the $2\times2$ submatrix left after removing row $i$ and column $j$.

Step 3: Attach the correct sign to each cofactor

$C_{ij}=(-1)^{i+j}M_{ij}.$ …

Common Mistakes

Mistake 1: Deleting the wrong row or column when forming a minor

Why it's wrong: the minor $M_{ij}$ must come from deleting row $i$ AND column $j$ specifically — deleting a different row/column produces a submatrix that has nothing to do with the required cofactor. Correct approach: for each cofactor, explicitly write out which row and column are being removed before writing the remaining $2\times2$ determinant.

Mistake 2: Applying the row-$1$ sign pattern ($+,-,+$) automatically instead of recomputing $(-1)^{i+j}$ for the actual row being expanded …