Q.Find minors and cofactors of all the elements of the determinant $\begin{vmatrix} 1 & -2 \\ 4 & 3 \end{vmatrix}$.

Concept: Minor and Cofactor definitions for a $2 \times 2$ determinant.

Step 1 — Minors

For element $a_{ij}$, the minor $M_{ij}$ is the determinant of the submatrix after removing row $i$ and column $j$.

• $M_{11} = |3| = 3$
• $M_{12} = |4| = 4$
• $M_{21} = |-2| = -2$
• $M_{22} = |1| = 1$

Step 2 — Cofactors

Cofactor $C_{ij} = (-1)^{i+j} M_{ij}$.

• $C_{11} = (-1)^{2} \cdot 3 = 3$
• $C_{12} = (-1)^{3} \cdot 4 = -4$
• $C_{21} = (-1)^{3} \cdot (-2) = 2$
• $C_{22} = (-1)^{4} \cdot 1 = 1$

The idea is simple: a minor is the determinant you get by deleting the row and column of that element. For a $2\times 2$ matrix, that means each minor is just a single number — the element that remains. The cofactor then adds a sign based on the position: $(-1)^{i+j}$ times the minor.

Let’s label the determinant as:

We’ll go element by element.

1. Element $a_{11} = 1$ (row 1, column 1)

   Delete row 1 and column 1. What’s left? The element at row 2, column 2, which is $3$.

   So the minor $M_{11} = 3$.

   The cofactor $C_{11} = (-1)^{1+1} \cdot M_{11} = (+1) \cdot 3 = 3$.

2. Element $a_{12} = -2$ (row 1, column 2)

   Delete row 1 and column 2. The remaining element is $a_{21} = 4$.

   So $M_{12} = 4$.

   Cofactor: $C_{12} = (-1)^{1+2} \cdot 4 = (-1) \cdot 4 = -4$.

3. Element $a_{21} = 4$ (row 2, column 1)

   Delete row 2 and column 1. The leftover is $a_{12} = -2$.

   So $M_{21} = -2$.

   Cofactor: $C_{21} = (-1)^{2+1} \cdot (-2) = (-1) \cdot (-2) = 2$.

4. Element $a_{22} = 3$ (row 2, column 2)

   Delete row 2 and column 2. The leftover is $a_{11} = 1$.

   So $M_{22} = 1$.

   Cofactor: $C_{22} = (-1)^{2+2} \cdot 1 = (+1) \cdot 1 = 1$.

Method: Finding All Minors and Cofactors of a Small Determinant

This method systematically computes the minor and cofactor of every element in a determinant — the standard first step before building an adjoint matrix.

Steps

Step 1: Go through the elements one at a time, in position order

Work through $a_{11}, a_{12}, a_{21}, a_{22}$ (and onward for larger matrices) in a fixed order so none is skipped.

Step 2: For each element, delete its row and column to get the minor

For a $2\times2$ matrix, deleting one row and one column leaves a single number — the opposite-diagonal entry.

Step 3: Attach the sign to get the cofactor

Use the checkerboard pattern to get the sign quickly: $(1,1)$ and $(2,2)$ positions are $+$; $(1,2)$ and $(2,1)$ are $-$.

Step 4: Tabulate all results together

List minors and cofactors side by side for each element — this makes it easy to spot a sign error, since minors and cofactors should only ever differ by a $\pm1$ factor.

Step 5: Use the quick $2\times2$ pattern as a cross-check

For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$: minors are $M_{11}=d,\ M_{12}=c,\ M_{21}=b,\ M_{22}=a$, and cofactors are $C_{11}=d,\ C_{12}=-c,\ C_{21}=-b,\ C_{22}=a$ — a fast way to check your row-by-row work.

This element-by-element method scales directly to $3\times3$ and larger determinants — only the size of each minor changes.

Common Mistakes

Mistake 1: Confusing the minor of $a_{12}$ with $a_{22}$ instead of $a_{21}$

Why it's wrong: deleting row 1 and column 2 (for the minor of $a_{12}$) leaves the element at the intersection of row 2 and column 1, i.e. $a_{21}$ — mistakenly picking $a_{22}$ gives the wrong minor. Correct approach: physically cross out the row and column of the element in question and read off whatever single entry remains, rather than guessing from the diagonal pattern.

Mistake 2: Forgetting the sign when converting a minor to a cofactor

Why it's wrong: leaving $C_{12}=+4$ instead of $C_{12}=(-1)^{1+2}\cdot4=-4$ silently drops the required sign flip for odd $i+j$ positions. Correct approach: always compute $(-1)^{i+j}$ explicitly for each position before finalizing a cofactor — never assume the minor's own sign carries over.