If \(A\) is a square matrix of order 3 such that \(A (\mathrm{adj}\, A) = \begin{bmatrix}-2 & 0 & 0\\0 & -2 & 0\\0 & 0 & -2\end{bmatrix}\) then \(|A|\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → −2
For every square matrix, \(A(\mathrm{adj}\, A) = |A|\, I\). The given matrix is
\( \displaystyle \begin{bmatrix}-2 & 0 & 0\\0 & -2 & 0\\0 & 0 & -2\end{bmatrix} = -2\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix} = -2I \)
So \(|A|\, I = -2I\), which gives \(|A| = -2\).
Why the other options are incorrect
Option (1) 2 — Incorrect. The diagonal entries are \(-2\), so \(|A| = -2\), not 2.
Option (2) −8 — Incorrect. \(-8\) is the determinant of the matrix \(A(\mathrm{adj}\, A)\) itself, which equals \(|A|^3 = (-2)^3\). The question asks for \(|A|\), the number multiplying I.
Option (3) 8 — Incorrect. \(|A| = -2\); 8 is neither \(|A|\) nor \(|A|^3\).
CBSE Applied Mathematics (Class XII):
Adjoint of a Matrix
