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: Build the Matrix:
Entry is 1 when \(i+j\) is even and 0 when odd. So row 1 is \((1,0,1)\), row 2 is \((0,1,0)\) and row 3 is \((1,0,1)\).

Step 2: Cofactors:
\(C_{11}=\begin{vmatrix}1&0\\0&1\end{vmatrix}=1\), \(C_{12}=-\begin{vmatrix}0&0\\1&1\end{vmatrix}=0\), \(C_{13}=\begin{vmatrix}0&1\\1&0\end{vmatrix}=-1\).
\(C_{21}=-\begin{vmatrix}0&1\\0&1\end{vmatrix}=0\), \(C_{22}=\begin{vmatrix}1&1\\1&1\end{vmatrix}=0\), \(C_{23}=-\begin{vmatrix}1&0\\1&0\end{vmatrix}=0\).
\(C_{31}=\begin{vmatrix}0&1\\1&0\end{vmatrix}=-1\), \(C_{32}=-\begin{vmatrix}1&1\\0&0\end{vmatrix}=0\), \(C_{33}=\begin{vmatrix}1&0\\0&1\end{vmatrix}=1\).

Step 3: Adjoint:
The adjoint is the transpose of the cofactor matrix. The cofactor matrix has rows \((1,0,-1)\), \((0,0,0)\), \((-1,0,1)\). It is symmetric, so the adjoint is the same matrix.

Step 4: Check the Options:
Option (B) has exactly these rows. Options (A), (C) and (D) have only 0 and 1 as entries, but the cofactors \(C_{13}\) and \(C_{31}\) are \(-1\), so they cannot be the adjoint.

Final Answer:
\(\text{adj}(A)\) has rows \((1,0,-1),(0,0,0),(-1,0,1)\), option (B). \[ \boxed{\text{(B)}} \]