Q69 / 180 · COMEDK 2025 (10 May Shift 1) · Mathematics
If $A=\left[\begin{array}{ccc}0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0\end{array}\right]$ then $A^{-1}$
- Aequal to $-\dfrac{1}{12}(\operatorname{adj} A)$
- Bequal to -12
- Correct answerCdoesn't exist
- Dequal to $\dfrac{1}{12}(\operatorname{adj} A)$
Solution
The matrix $A$ is given by $A = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix}$.
Calculate the determinant of $A$ by expanding along the first row:
$\det(A) = 0(0 - (-9)) - 1(0 - 6) + (-2)(3 - 0)$
$\det(A) = 0 - 1(-6) - 2(3) = 6 - 6 = 0$.
Since $\det(A) = 0$, the matrix $A$ is a singular matrix.
A matrix has an inverse if and only if its determinant is non-zero. Therefore, $A^{-1}$ does not exist.
Answer: doesn't exist
Calculate the determinant of $A$ by expanding along the first row:
$\det(A) = 0(0 - (-9)) - 1(0 - 6) + (-2)(3 - 0)$
$\det(A) = 0 - 1(-6) - 2(3) = 6 - 6 = 0$.
Since $\det(A) = 0$, the matrix $A$ is a singular matrix.
A matrix has an inverse if and only if its determinant is non-zero. Therefore, $A^{-1}$ does not exist.
Answer: doesn't exist
Hence, the correct answer is option: 'C'.