Matrices & Determinants
Grade 12
Question:
If $A = \begin{bmatrix} 1 & 3 & 5 \\ 3 & 5 & 1 \\ 5 & 1 & 3 \end{bmatrix}$, then $\text{adj } A$ is equal to -
$\begin{bmatrix} 14 & -4 & -22 \\ -4 & -22 & 14 \\ -22 & 14 & -4 \end{bmatrix}$
$\begin{bmatrix} -14 & 4 & 22 \\ 4 & 22 & -14 \\ 22 & -14 & 4 \end{bmatrix}$
$\begin{bmatrix} 14 & 4 & -22 \\ 4 & -22 & -14 \\ -22 & -14 & -4 \end{bmatrix}$
none of these
Step-by-Step Solution
Key Concept: The adjugate matrix is the transpose of the cofactor matrix. Each element of adj(A) is the cofactor of the corresponding element in A, and we must find all 9 cofactors, then transpose.
<p><strong>Step 1: Find the Cofactor Matrix</strong></p><p>For each element a_ij, compute cofactor C_ij = (-1)^(i+j) × (minor M_ij)</p><p><strong>C₁₁:</strong> (-1)^(1+1) × |5 1; 1 3| = (15-1) = 14</p><p><strong>C₁₂:</strong> (-1)^(1+2) × |3 1; 5 3| = -(9-5) = -4</p><p><strong>C₁₃:</strong> (-1)^(1+3) × |3 5; 5 1| = (3-25) = -22</p><p><strong>C₂₁:</strong> (-1)^(2+1) × |3 5; 1 3| = -(9-5) = -4</p><p><strong>C₂₂:</strong> (-1)^(2+2) × |1 5; 5 3| = (3-25) = -22</p><p><strong>C₂₃:</strong> (-1)^(2+3) × |1 3; 5 1| = -(1-15) = 14</p><p><strong>C₃₁:</strong> (-1)^(3+1) × |3 5; 5 1| = (3-25) = -22</p><p><strong>C₃₂:</strong> (-1)^(3+2) × |1 5; 3 1| = -(1-15) = 14</p><p><strong>C₃₃:</strong> (-1)^(3+3) × |1 3; 3 5| = (5-9) = -4</p><p><strong>Step 2: Form Cofactor Matrix</strong></p><p>Cofactor matrix = [14 -4 -22; -4 -22 14; -22 14 -4]</p><p><strong>Step 3: Transpose to get adj(A)</strong></p><p>adj(A) = (Cofactor matrix)^T = [14 -4 -22; -4 -22 14; -22 14 -4]</p><p>Note: The matrix is symmetric, so transpose gives the same matrix.</p><p><strong>∴ Answer:</strong> A</p>
Correct Answer: A