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 found by computing the matrix of minors, taking their cofactors (with appropriate signs), and then transposing. Each element of adj(A) at position (i,j) equals the cofactor of element a(j,i) in the original matrix.
<p><strong>Step 1: Set up the cofactor matrix</strong></p><p>For a 3×3 matrix, we need cofactors C(i,j) = (-1)^(i+j) × M(i,j), where M(i,j) is the minor (determinant of 2×2 submatrix).</p><p><strong>Step 2: Calculate each cofactor</strong></p><p>C(1,1) = (-1)^(1+1)|5 1; 1 3| = +(15-1) = 14</p><p>C(1,2) = (-1)^(1+2)|3 1; 5 3| = -(9-5) = -4</p><p>C(1,3) = (-1)^(1+3)|3 5; 5 1| = +(3-25) = -22</p><p>C(2,1) = (-1)^(2+1)|3 5; 1 3| = -(9-5) = -4</p><p>C(2,2) = (-1)^(2+2)|1 5; 5 3| = +(3-25) = -22</p><p>C(2,3) = (-1)^(2+3)|1 3; 5 1| = -(1-15) = 14</p><p>C(3,1) = (-1)^(3+1)|3 5; 5 1| = +(3-25) = -22</p><p>C(3,2) = (-1)^(3+2)|1 5; 3 1| = -(1-15) = 14</p><p>C(3,3) = (-1)^(3+3)|1 3; 3 5| = +(5-9) = -4</p><p><strong>Step 3: Form the cofactor matrix</strong></p><p>Cofactor matrix = [14, -4, -22; -4, -22, 14; -22, 14, -4]</p><p><strong>Step 4: Transpose to get adjugate</strong></p><p>adj(A) = [14, -4, -22; -4, -22, 14; -22, 14, -4]^T = [14, -4, -22; -4, -22, 14; -22, 14, -4]</p><p><strong>Note:</strong> This matrix is symmetric, so transposing doesn't change it.</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A