Q54 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics
If $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}_{3 \times 3}$, and $A^{-1} = \begin{bmatrix} \gamma & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix}_{3 \times 3}$, then $|\alpha \cdot \beta \cdot \gamma| = $ _____ (where $|\cdot|$ denotes the absolute value)
- A$125$
- B$220$
- Correct answerC$225$
- D$-225$
Solution
The determinant of matrix $A$ is:
$|A| = 2(1 \times 3 - 0 \times 1) - 0 + (-1)(5 \times 1 - 1 \times 0) = 6 - 5 = 1$
Since $|A| = 1$, the inverse of $A$ is simply its adjugate matrix, $A^{-1} = \text{adj}(A)$.
The elements $\gamma, \alpha, \beta$ form the first column of $A^{-1}$, which corresponds to the cofactors $C_{11}, C_{12}, C_{13}$ of the first row of matrix $A$.
Calculating these cofactors:
$C_{11} = (1)(3) - (0)(1) = 3 \Rightarrow \gamma = 3$
$C_{12} = -((5)(3) - (0)(0)) = -15 \Rightarrow \alpha = -15$
$C_{13} = (5)(1) - (1)(0) = 5 \Rightarrow \beta = 5$
The product of these values is:
$\alpha \cdot \beta \cdot \gamma = (-15) \cdot 5 \cdot 3 = -225$
Taking the absolute value:
$|\alpha \cdot \beta \cdot \gamma| = |-225| = 225$
Answer: $225$
$|A| = 2(1 \times 3 - 0 \times 1) - 0 + (-1)(5 \times 1 - 1 \times 0) = 6 - 5 = 1$
Since $|A| = 1$, the inverse of $A$ is simply its adjugate matrix, $A^{-1} = \text{adj}(A)$.
The elements $\gamma, \alpha, \beta$ form the first column of $A^{-1}$, which corresponds to the cofactors $C_{11}, C_{12}, C_{13}$ of the first row of matrix $A$.
Calculating these cofactors:
$C_{11} = (1)(3) - (0)(1) = 3 \Rightarrow \gamma = 3$
$C_{12} = -((5)(3) - (0)(0)) = -15 \Rightarrow \alpha = -15$
$C_{13} = (5)(1) - (1)(0) = 5 \Rightarrow \beta = 5$
The product of these values is:
$\alpha \cdot \beta \cdot \gamma = (-15) \cdot 5 \cdot 3 = -225$
Taking the absolute value:
$|\alpha \cdot \beta \cdot \gamma| = |-225| = 225$
Answer: $225$
Hence, the correct answer is option: 'C'.