Let A be a square matrix of order 3 such that \(A^{2} = I\). Consider the following statements:
(A) \(|A| = 1\) or \(|A| = -1\)
(B) \(A^{-1} = A\)
(C) A must be either I or \(-I\)
(D) \(\mathrm{adj}\,A = |A|\,A\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
Taking the determinant of both sides of \(A^{2} = I\):
\(|A|^{2} = |I| = 1\)
\(|A| = \pm 1\)
So A is non-singular, and \(A \cdot A = I\) shows that A is its own inverse.
From \(A^{-1} = \displaystyle\frac{\mathrm{adj}\,A}{|A|}\):
\(\mathrm{adj}\,A = |A|\,A^{-1} = |A|\,A\)
The matrix \(D = \begin{bmatrix}1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1\end{bmatrix}\) satisfies \(D^{2} = I\) but is neither I nor \(-I\).
So \(|A| = \pm 1\), \(A^{-1} = A\) and \(\mathrm{adj}\,A = |A|\,A\), while A need not be \(\pm I\).
Checking the statements
(A) \(|A| = 1\) or \(-1\) — Correct. \(|A|^{2} = 1\).
(B) \(A^{-1} = A\) — Correct. \(A \cdot A = I\).
(C) A must be I or \(-I\) — Incorrect. D = diag(1, −1, 1) has \(D^{2} = I\).
(D) \(\mathrm{adj}\,A = |A|\,A\) — Correct. \(\mathrm{adj}\,A = |A|\,A^{-1}\) and \(A^{-1} = A\).