Q56 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics
The element in the third row and the second column in the inverse matrix of a matrix $\begin{bmatrix} 1 & 3 & 3 \\ 3 & 1 & 3 \\ 3 & 3 & 4 \end{bmatrix}$ is
- Correct answerA$\dfrac{3}{2}$
- B$\dfrac{2}{3}$
- C$\dfrac{-3}{2}$
- D$\dfrac{-2}{3}$
Solution
Let $A = \begin{bmatrix} 1 & 3 & 3 \\ 3 & 1 & 3 \\ 3 & 3 & 4 \end{bmatrix}$.
The element in the third row and second column of $A^{-1}$ is given by $\dfrac{C_{23}}{|A|}$, where $C_{23}$ is the cofactor of the element $a_{23}$ in $A$.
First, we calculate the determinant of $A$:
$|A| = 1(1 \times 4 - 3 \times 3) - 3(3 \times 4 - 3 \times 3) + 3(3 \times 3 - 1 \times 3)$
$|A| = 1(-5) - 3(3) + 3(6) = -5 - 9 + 18 = 4$
Next, we find the cofactor $C_{23}$:
$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 3 \\ 3 & 3 \end{vmatrix} = -(1 \times 3 - 3 \times 3) = -(3 - 9) = 6$
Therefore, the required element is $\dfrac{C_{23}}{|A|} = \dfrac{6}{4} = \dfrac{3}{2}$.
Answer: $\dfrac{3}{2}$
The element in the third row and second column of $A^{-1}$ is given by $\dfrac{C_{23}}{|A|}$, where $C_{23}$ is the cofactor of the element $a_{23}$ in $A$.
First, we calculate the determinant of $A$:
$|A| = 1(1 \times 4 - 3 \times 3) - 3(3 \times 4 - 3 \times 3) + 3(3 \times 3 - 1 \times 3)$
$|A| = 1(-5) - 3(3) + 3(6) = -5 - 9 + 18 = 4$
Next, we find the cofactor $C_{23}$:
$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 3 \\ 3 & 3 \end{vmatrix} = -(1 \times 3 - 3 \times 3) = -(3 - 9) = 6$
Therefore, the required element is $\dfrac{C_{23}}{|A|} = \dfrac{6}{4} = \dfrac{3}{2}$.
Answer: $\dfrac{3}{2}$
Hence, the correct answer is option: 'A'.