Let $B$ be a $3\times 3$ matrix with $\det B=-2$. For which value of $k$ does the matrix $C=kB$ have determinant $54$?\n\nA) $-3$\nB) $3$\nC) $-2$\nD) $2$

Step-by-step solution

1. Recall the scaling property: $\det (kB)=k^n\det B$, where $n$ is the order of the matrix.
2. Here $n=3$, so $\det C=k^3\det B=k^3(-2)$.
3. Set this equal to the required determinant: $k^3(-2)=54$.
4. Divide both sides by $-2$: $k^3=-27$.
5. Take the real cube root of $-27$: $k=-3$.
6. Thus the correct option is A) $-3$.

Concept tested: Scaling property of determinants

Published 13 September 2026 · How this solution was prepared and checked