If \(A =\begin{bmatrix}2&0&3\\-1&1&3\\0&-4&0\end{bmatrix}\), then the value of det (2A) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 288
For a square matrix of order n, \(|kA| = k^n|A|\), since each of the n rows is multiplied by k. Here \(n = 3\):
\(|2A| = 2^3|A| = 8|A|\)
Expanding \(|A|\) along the first row:
\(\displaystyle |A| = 2\begin{vmatrix}1&3\\-4&0\end{vmatrix} - 0 + 3\begin{vmatrix}-1&1\\0&-4\end{vmatrix}\)
\(= 2(0 + 12) + 3(4 - 0)\)
\(= 24 + 12 = 36\)
\(|2A| = 8 \times 36 = 288\)
Why the other options are incorrect
Option (1) 72 — Incorrect. This is \(2|A|\): the factor 2 is taken out only once, but it comes out of each of the 3 rows.
Option (2) -144 — Incorrect. \(|A| = 36\) is positive, so \(|2A| = 8|A|\) is positive too.
Option (4) 144 — Incorrect. This is \(2^2|A|\), the rule for a matrix of order 2; A is of order 3.