If \(A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}\) and \(A^2 = pA + qI\), where I is the identity matrix of order 3, then \((p, q)\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \((1, 2)\)
Each diagonal element of \(A^2\) is a row of A times the same column: \(0 + 1 + 1 = 2\). Each off-diagonal element is a row times a different column, for example row 1 times column 2: \(0 + 0 + 1 = 1\).
\( \displaystyle A^2 = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} \)
On the other side:
\( \displaystyle pA + qI = \begin{bmatrix} q & p & p \\ p & q & p \\ p & p & q \end{bmatrix} \)
By equality of matrices, comparing the off-diagonal and the diagonal elements:
\(p = 1,\quad q = 2\)
So \(A^2 = A + 2I\) and \((p, q) = (1, 2)\).