Q63 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
If $A = [a_{ij}]_{3 \times 3}$ is a matrix such that $a_{ij} = |2i - 5j|$, where $|.|$ denotes the modulus function, then the element in the $2^{\text{nd}}$ row and $3^{\text{rd}}$ column of $A^{-1}$ is ...
- A$3$
- B$1$
- C$0$
- Correct answerD$-1$
Solution
Given $a_{ij} = |2i - 5j|$, we construct the matrix $A$ of order $3 \times 3$:
$A = \begin{bmatrix} |2(1)-5(1)| & |2(1)-5(2)| & |2(1)-5(3)| \\ |2(2)-5(1)| & |2(2)-5(2)| & |2(2)-5(3)| \\ |2(3)-5(1)| & |2(3)-5(2)| & |2(3)-5(3)| \end{bmatrix} = \begin{bmatrix} 3 & 8 & 13 \\ 1 & 6 & 11 \\ 1 & 4 & 9 \end{bmatrix}$
The determinant of matrix $A$ is:
$|A| = 3(54 - 44) - 8(9 - 11) + 13(4 - 6)$
$|A| = 3(10) - 8(-2) + 13(-2) = 30 + 16 - 26 = 20$
The element in the $2^{\text{nd}}$ row and $3^{\text{rd}}$ column of $A^{-1}$ is given by $\dfrac{C_{32}}{|A|}$, where $C_{32}$ is the cofactor of the element $a_{32}$ in matrix $A$.
Calculating the cofactor $C_{32}$:
$C_{32} = (-1)^{3+2} \begin{vmatrix} 3 & 13 \\ 1 & 11 \end{vmatrix} = -(33 - 13) = -20$
Therefore, the required element is:
$\dfrac{C_{32}}{|A|} = \dfrac{-20}{20} = -1$
Answer: $-1$
$A = \begin{bmatrix} |2(1)-5(1)| & |2(1)-5(2)| & |2(1)-5(3)| \\ |2(2)-5(1)| & |2(2)-5(2)| & |2(2)-5(3)| \\ |2(3)-5(1)| & |2(3)-5(2)| & |2(3)-5(3)| \end{bmatrix} = \begin{bmatrix} 3 & 8 & 13 \\ 1 & 6 & 11 \\ 1 & 4 & 9 \end{bmatrix}$
The determinant of matrix $A$ is:
$|A| = 3(54 - 44) - 8(9 - 11) + 13(4 - 6)$
$|A| = 3(10) - 8(-2) + 13(-2) = 30 + 16 - 26 = 20$
The element in the $2^{\text{nd}}$ row and $3^{\text{rd}}$ column of $A^{-1}$ is given by $\dfrac{C_{32}}{|A|}$, where $C_{32}$ is the cofactor of the element $a_{32}$ in matrix $A$.
Calculating the cofactor $C_{32}$:
$C_{32} = (-1)^{3+2} \begin{vmatrix} 3 & 13 \\ 1 & 11 \end{vmatrix} = -(33 - 13) = -20$
Therefore, the required element is:
$\dfrac{C_{32}}{|A|} = \dfrac{-20}{20} = -1$
Answer: $-1$
Hence, the correct answer is option: 'D'.