Let $B$ be a $4\times 4$ matrix with $\det B=3$. Define $C=(2B^{-1}{)}^T\cdot (5B)$. Find $\det C$.

Step-by-step solution

1. Use $\det (X^T)=\det X$ for any square matrix.
2. Hence $\det ((2B^{-1}{)}^T)=\det (2B^{-1})$.
3. For a scalar $k$ and $n\times n$ matrix $M$, $\det (kM)=k^n\det M$; here $n=4$.
4. Thus $\det (2B^{-1})=2^4\det (B^{-1})=16\cdot \frac{1}{\det B}=16\cdot \frac{1}{3}=\frac{16}{3}$.
5. Similarly $\det (5B)=5^4\det B=625\cdot 3=1875$.
6. The determinant of a product is the product of determinants: $\det C=\det ((2B^{-1}{)}^T)\cdot \det (5B)=\frac{16}{3}\times 1875=\frac{30000}{3}=10000$.

Concept tested: Determinant properties for transpose, inverse, scalar multiplication, and product

Published 16 October 2026 · How this solution was prepared and checked