Question

If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, then $\text{adj}(A)$ is:

• A. \( \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} \)
• B. \( \begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix} \)
• C. \( \begin{bmatrix} 4 & 2 \\ 3 & 1 \end{bmatrix} \)
• D. \( \begin{bmatrix} -4 & 2 \\ 3 & -1 \end{bmatrix} \)

Hint:

For any \(2 \times 2\) matrix \(\begin{bmatrix} a & b c & d \end{bmatrix}\), you can find the adjugate instantly using a simple shortcut rule: Swap the elements on the main diagonal (\(a\) and \(d\)), and change the signs of the off-diagonal elements (\(b\) and \(c\)). Applying this rule to \(\begin{bmatrix} 1 & 2 3 & 4 \end{bmatrix}\) immediately yields \(\begin{bmatrix} 4 & -2 -3 & 1 \end{bmatrix}\).

Answer 1

The Correct Option is A

Concept: The adjugate matrix (\(\text{adj}(A)\)) of a matrix is the transpose of its cofactor matrix: \[ \text{adj}(A) = (C)^\text{T} \] Where each entry \(C_{ij}\) is the cofactor of element \(a_{ij}\), calculated using the formula: \[ C_{ij} = (-1)^{i+j} M_{ij} \quad (M_{ij} \text{ is the minor matrix determinant}) \]

Step 1: Calculating the cofactors for each element of matrix A.
Given matrix: \[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \] Let's find the cofactors for all 4 positions:

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

Constructing the cofactor matrix \(C\): \[ C = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix} \]

Step 2: Transposing the cofactor matrix to determine the adjugate.
\[ \text{adj}(A) = C^\text{T} = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}^\text{T} = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} \]