If \(A=\begin{bmatrix}2&-1&0\\1&1&2\\-1&0&1\end{bmatrix}\), then which of the following statement(s) is/are correct?
(A) A is singular matrix
(B) \(|3A| = 135\)
(C) \(|\mathrm{adj}\,A| = 125\)
(D) \(|A^{-1}|=\displaystyle\frac{1}{5}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (B) and (D) only
Expanding along Row 1:
\( |A| = 2\begin{vmatrix}1&2\\0&1\end{vmatrix} - (-1)\begin{vmatrix}1&2\\-1&1\end{vmatrix} + 0 \)
\(|A| = 2(1 - 0) + (1 + 2)\)
\(|A| = 5\)
For a square matrix of order n = 3: \(|kA| = k^3|A|\), \(|\mathrm{adj}\,A| = |A|^{2}\) and \(|A^{-1}| = \frac{1}{|A|}\).
So |A| = 5, |3A| = 27 × 5 = 135, |adj A| = 25 and |A−1| = 1/5.
Checking the statements
(A) A is singular matrix — Incorrect. |A| = 5 ≠ 0, so A is non-singular.
(B) \(|3A| = 135\) — Correct. \(3^3 \times 5 = 135\).
(C) \(|\mathrm{adj}\,A| = 125\) — Incorrect. \(|\mathrm{adj}\,A| = 5^2 = 25\); 125 would be \(|A|^3\).
(D) \(|A^{-1}|=\frac{1}{5}\) — Correct. \(|A^{-1}| = \frac{1}{|A|} = \frac{1}{5}\).