Let \(\mathrm{A} = [a_{ij}]_{3 \times 3}\) such that \(|\mathrm{A}| = -5\). Then the value of det (5A) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → -625
Multiplying a matrix by k multiplies every row by k, and each row brings out a factor k from the determinant. So for a matrix of order n:
\(|kA| = k^n |A|\)
Here k = 5 and n = 3:
\(|5A| = 5^3 \times (-5)\)
\(= 125 \times (-5) = -625\)