Q.Find minors and cofactors of the elements $a_{11}$, $a_{21}$ in the determinant $\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$.

Concept: Minor and Cofactor Definition

The minor $M_{ij}$ of element $a_{ij}$ is the determinant obtained by deleting the $i$-th row and $j$-th column. The cofactor $C_{ij}$ is $(-1)^{i+j} M_{ij}$.

Step 1: Minor of $a_{11}$

Delete row 1 and column 1:

Step 2: Cofactor of $a_{11}$

$C_{11} = (-1)^{1+1} M_{11} = M_{11}$.

Step 3: Minor of $a_{21}$

Delete row 2 and column 1: …

The idea is simple: every element in a determinant has a "shadow" — the smaller determinant you get by removing its row and column. That shadow is the minor. But the sign matters when you use it to expand the determinant, so we attach a sign based on the element's position: $(-1)^{\text{row}+\text{column}}$. That signed minor is the cofactor.

Let’s walk through it for the two elements asked.

1. For $a_{11}$ (first row, first column): Delete row 1 and column 1. What remains is the $2 \times 2$ determinant formed by the other rows (2 and 3) and other columns (2 and 3):

The cofactor $A_{11}$ is $(-1)^{1+1} M_{11} = (+1) \cdot M_{11} = M_{11}$.

So $A_{11} = \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}$.

2. For $a_{21}$ (second row, first column): Delete row 2 and column 1. The remaining rows are 1 and 3, remaining columns are 2 and 3:

The cofactor $A_{21} = (-1)^{2+1} M_{21} = (-1)^3 M_{21} = -M_{21}$.

So $A_{21} = -\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix}$. …

Method: Finding Minors and Cofactors in a General (Symbolic) Determinant

This method finds the minor and cofactor of a named element when the determinant's entries are symbols ($a_{ij}$) rather than numbers — exactly the same deletion rule applies, just carried out algebraically.

Steps

Step 1: Identify the element's exact position

Read off the row $i$ and column $j$ from the subscript of the element itself — e.g. $a_{21}$ sits in row 2, column 1.

Step 2: Delete that row and column symbolically

Cross out every symbol in row $i$ and every symbol in column $j$ from the full $3\times3$ array, keeping the remaining symbols in their original relative positions.

Step 3: Write the resulting $2\times2$ symbolic minor

where $p,s$ are the two surviving row indices and $q,r$ are the two surviving column indices, in their original order.

Step 4: Attach the sign for the cofactor

$C_{ij}=(-1)^{i+j}M_{ij}$ …

Common Mistakes

Mistake 1: Forgetting the negative sign on the cofactor of $a_{21}$

Why it's wrong: since $2+1=3$ is odd, $(-1)^{2+1}=-1$, so $A_{21}=-M_{21}$ — writing $A_{21}=M_{21}$ (without the negative) is a very common slip, especially since $a_{11}$'s cofactor (an even position) does NOT need the sign flip, and it's easy to apply that same "no flip" habit to every element. Correct approach: check the parity of $i+j$ for every single element individually — never assume the pattern from a neighboring element.

Mistake 2: Deleting the wrong row/column pair symbolically …