Q.Find $\operatorname{adj} A$ for $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$.

Concept: Adjoint Matrix Property — For a $2 \times 2$ matrix, the adjoint is the transpose of the cofactor matrix.

Step 1: Compute the cofactor matrix.

For $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$:

• $C_{11} = +4$
• $C_{12} = -1$
• $C_{21} = -3$
• $C_{22} = +2$

So the cofactor matrix is $\begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}$.

Step 2: Transpose the cofactor matrix to get the adjoint:

$\operatorname{adj} A = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$.

The adjoint of a matrix is the transpose of its cofactor matrix. For a $2 \times 2$ matrix, this simplifies to a neat pattern that saves you from computing cofactors individually every time.

Why this works: The cofactor of an entry $a_{ij}$ is $(-1)^{i+j}$ times the determinant of the submatrix obtained by deleting row $i$ and column $j$. For a $2 \times 2$ matrix, each cofactor is just a single number (the other entry, with a possible sign change). Transposing the cofactor matrix then gives the adjoint.

Let’s apply this step by step.

1. Write down the matrix.

   $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$.

2. Find the cofactor of each entry.

   • For $a_{11} = 2$: delete row 1, column 1 → submatrix is $[4]$. Cofactor $C_{11} = (+1) \cdot 4 = 4$.
   • For $a_{12} = 3$: delete row 1, column 2 → submatrix is $[1]$. Cofactor $C_{12} = (-1) \cdot 1 = -1$.
   • For $a_{21} = 1$: delete row 2, column 1 → submatrix is $[3]$. Cofactor $C_{21} = (-1) \cdot 3 = -3$.
   • For $a_{22} = 4$: delete row 2, column 2 → submatrix is $[2]$. Cofactor $C_{22} = (+1) \cdot 2 = 2$.

   So the cofactor matrix is $\begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}$.

3. Transpose the cofactor matrix to get the adjoint.

   The adjoint is the transpose: swap rows and columns.

   $\operatorname{adj} A = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$.

Method: Finding the Adjoint of a 2×2 Matrix

This method computes the adjoint of a $2\times2$ matrix using the fast shortcut pattern, rather than computing four separate cofactors from scratch.

Steps

Step 1: Write down the matrix in standard form

Step 2: Apply the $2\times2$ adjoint shortcut

Swap the two diagonal entries ($a\leftrightarrow d$), and flip the sign of the two off-diagonal entries ($b\to -b$, $c\to -c$).

Step 3 (to see why the shortcut works): Compute the cofactors explicitly

$C_{11}=+d$, $C_{12}=-c$, $C_{21}=-b$, $C_{22}=+a$ — the cofactor matrix is $\begin{pmatrix}d&-c\\-b&a\end{pmatrix}$.

Step 4: Transpose the cofactor matrix

The adjoint is defined as the transpose of the cofactor matrix, which swaps the off-diagonal entries $-c$ and $-b$, giving exactly $\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ — matching the shortcut.

This $2\times2$ shortcut is safe to use directly on any exam question — just be careful it does NOT generalize the same way to $3\times3$ or larger, where the full cofactor-then-transpose procedure is required.

Common Mistakes

Mistake 1: Forgetting to transpose the cofactor matrix

Why it's wrong: the adjoint is the transpose of the cofactor matrix, not the cofactor matrix itself — for a $2\times2$ matrix, submitting the cofactor matrix directly instead of its transpose swaps the off-diagonal entries and gives the wrong adjoint. Correct approach: after computing all four cofactors, always transpose the resulting matrix as a separate, explicit final step.