Q63 / 150 · MHT CET 2026 (13 April Shift 2) · Mathematics
The element in 1st row and 2nd column of the inverse of the matrix $\begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix}$ is ...
- Correct answerA$-\dfrac{2}{5}$
- B$-\dfrac{10}{3}$
- C$-\dfrac{2}{25}$
- D$\dfrac{1}{25}$
Solution
Let $A = \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix}$
The determinant of $A$ is given by:
$|A| = 1(0 - (-25)) - 3(0 - (-10)) - 2(-15 - 0)$
$|A| = 25 - 30 + 30 = 25$
The inverse of matrix $A$ is $A^{-1} = \dfrac{1}{|A|} \text{adj}(A)$.
The element in the 1st row and 2nd column of $A^{-1}$ is $\dfrac{1}{|A|} C_{21}$, where $C_{21}$ is the cofactor of the element $a_{21}$ in $A$.
$C_{21} = (-1)^{2+1} \begin{vmatrix} 3 & -2 \\ 5 & 0 \end{vmatrix} = -(0 - (-10)) = -10$
Therefore, the required element is $\dfrac{-10}{25} = -\dfrac{2}{5}$.
The determinant of $A$ is given by:
$|A| = 1(0 - (-25)) - 3(0 - (-10)) - 2(-15 - 0)$
$|A| = 25 - 30 + 30 = 25$
The inverse of matrix $A$ is $A^{-1} = \dfrac{1}{|A|} \text{adj}(A)$.
The element in the 1st row and 2nd column of $A^{-1}$ is $\dfrac{1}{|A|} C_{21}$, where $C_{21}$ is the cofactor of the element $a_{21}$ in $A$.
$C_{21} = (-1)^{2+1} \begin{vmatrix} 3 & -2 \\ 5 & 0 \end{vmatrix} = -(0 - (-10)) = -10$
Therefore, the required element is $\dfrac{-10}{25} = -\dfrac{2}{5}$.
Hence, the correct answer is option: 'A'.