Q79 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
If $A=\left[\begin{array}{lll}2 & 1 & 3 \\ 3 & 1 & 2 \\ 1 & 2 & 3\end{array}\right]$ and I is an identity by matrix of order $3 \times 3$ then $A(\operatorname{adj} A)=$
- A$\left[\begin{array}{ccc}20 & 0 & 0 \\ 0 & -20 & 0 \\ 0 & 0 & 20\end{array}\right]$
- B$\text { }\left[\begin{array}{ccc} -1 & -7 & 5 \\ 3 & 3 & -3 \\ -1 & 5 & -1 \end{array}\right]$
- C20 I
- Correct answerD6 I
Solution
The property of the adjoint of a matrix states that $A(\operatorname{adj} A) = |A| I$, where $|A|$ is the determinant of matrix $A$ and $I$ is the identity matrix of the same order.
Given $A = \begin{bmatrix} 2 & 1 & 3 \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{bmatrix}$, we calculate the determinant $|A|$ by expanding along the first row:
$|A| = 2(1 \times 3 - 2 \times 2) - 1(3 \times 3 - 2 \times 1) + 3(3 \times 2 - 1 \times 1)$
$|A| = 2(3 - 4) - 1(9 - 2) + 3(6 - 1)$
$|A| = 2(-1) - 1(7) + 3(5)$
$|A| = -2 - 7 + 15 = 6$
Therefore, $A(\operatorname{adj} A) = 6I$.
Answer: 6 I
Given $A = \begin{bmatrix} 2 & 1 & 3 \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{bmatrix}$, we calculate the determinant $|A|$ by expanding along the first row:
$|A| = 2(1 \times 3 - 2 \times 2) - 1(3 \times 3 - 2 \times 1) + 3(3 \times 2 - 1 \times 1)$
$|A| = 2(3 - 4) - 1(9 - 2) + 3(6 - 1)$
$|A| = 2(-1) - 1(7) + 3(5)$
$|A| = -2 - 7 + 15 = 6$
Therefore, $A(\operatorname{adj} A) = 6I$.
Answer: 6 I
Hence, the correct answer is option: 'D'.