Question

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. \(0\)
• D. \(-1\)

Hint:

Build the matrix, find its determinant, and use the cofactor C32 for the (2,3) entry of the inverse.

Answer 1

The Correct Option is D

Step 1: Adjoint rule
Entry $(A^{-1})_{ij} = C_{ji}/\det A$, so entry (2,3) uses the cofactor of position (3,2).

Step 2: Determinant by row operations
Rows: $R_1=(3,8,13), R_2=(1,6,11), R_3=(1,4,9)$. Replace $R_1\to R_1-R_2$ and $R_2\to R_2-R_3$ to get rows $(2,2,2), (0,2,2), (1,4,9)$. These row operations do not change the determinant. Expanding along the first row: $2(18-8) - 2(0-2) + 2(0-2) = 20+4-4 = 20$.

Step 3: Compute
Cofactor $C_{32} = -(3\cdot11-13\cdot1) = -20$, so the entry is $-1$. Option (D).

Final Answer:
-1. \[ \boxed{\text{(D)}\ -1} \]