Let \(A = \displaystyle\frac{1}{3}\begin{bmatrix}2 & -2 & 1\\ 1 & 2 & 2\\ 2 & 1 & -2\end{bmatrix}\). Consider the following statements:
(A) \(AA^{T} = I\)
(B) \(|A| = -1\)
(C) \(A^{-1} = A^{T}\)
(D) \(\mathrm{adj}\,A = A^{T}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
Let \(M = \begin{bmatrix}2 & -2 & 1\\ 1 & 2 & 2\\ 2 & 1 & -2\end{bmatrix}\), so \(A = \frac{1}{3}M\).
Each row of M has \(2^{2} + 2^{2} + 1^{2} = 9\), and the products of different rows are
\(2 - 4 + 2 = 0,\ \ 4 - 2 - 2 = 0,\ \ 2 + 2 - 4 = 0\)
\(MM^{T} = 9I\), so \(\displaystyle AA^{T} = \frac{1}{9}MM^{T} = I\)
\(|M| = 2(-4 - 2) + 2(-2 - 4) + 1(1 - 4) = -27\)
\(\displaystyle |A| = \frac{1}{27}|M| = -1\)
From \(AA^{T} = I\), the inverse is the transpose. Then \(\mathrm{adj}\,A = |A|\,A^{-1} = -A^{T}\).
So \(AA^{T} = I\), \(|A| = -1\), \(A^{-1} = A^{T}\) and \(\mathrm{adj}\,A = -A^{T}\).
Checking the statements
(A) \(AA^{T} = I\) — Correct. \(MM^{T} = 9I\) and the factor is \(\frac{1}{9}\).
(B) \(|A| = -1\) — Correct. \(|M| = -27\) and \(\left(\frac{1}{3}\right)^{3}(-27) = -1\).
(C) \(A^{-1} = A^{T}\) — Correct. \(AA^{T} = I\).
(D) \(\mathrm{adj}\,A = A^{T}\) — Incorrect. \(\mathrm{adj}\,A = |A|\,A^{-1} = -A^{T}\).