Question

If \(A = [a_{ij}]_{3\times 3}\), where \(a_{ij} = \{\begin{array}{cc}1, & \text{if }i+j\text{ is even} \\ 0, & \text{if }i+j\text{ is odd}\end{array}\), then \(\text{adj}(A) = \ldots\) ..

• A. \(\left[ \begin{array}{ccc}0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array} \right]\)
• B. \(\left[ \begin{array}{ccc}1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1\end{array} \right]\)
• C. \(\left[ \begin{array}{ccc}0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0\end{array} \right]\)
• D. \(\left[ \begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1\end{array} \right]\)

Hint:

Build the matrix from the rule, then find the cofactors.

Answer 1

The Correct Option is B

Step 1: Observation:
Rows 1 and 3 of $A$ are equal, so $\det A=0$. For any square matrix, $A\cdot\text{adj}(A)=\det(A)\,I=0$ here.

Step 2: Test Option (B):
Let $N$ have rows $(1,0,-1),(0,0,0),(-1,0,1)$. Row 1 of $A$ times $N$ is $1\cdot(1,0,-1)+1\cdot(-1,0,1)=(0,0,0)$. Row 2 of $A$ picks the zero row of $N$. Row 3 equals row 1. So $AN=0$, as required.

Step 3: Test Option (A):
Option (A) has rows $(0,1,0),(1,0,1),(0,1,0)$. Row 1 of $A$ times it gives $(0,1,0)+(0,1,0)=(0,2,0)$, which is not zero. Options (C) and (D) also give a nonzero first row ($(0,2,0)$ and $(1,1,2)$). Only (B) works.

Final Answer:
Option (B). \[ \boxed{\text{(B)}} \]