Question

If B is a non-singular matrix and A is a square matrix, then det (B⁻¹ AB) is equal to
(a) Det (A⁻¹)
(b) Det (B⁻¹)
(c) Det (A)
(d) Det (B)

Answer 1

(c) Det (A)

$$\begin{gathered} B\mathrm{is}\mathrm{non}-\mathrm{singular}. \\ \mathrm{This}\mathrm{implies}\mathrm{that}\left|B\right|\ne 0,\mathrm{that}B\mathrm{is}\mathrm{invertible}\mathrm{and}\mathrm{that}{B}^{-1}\mathrm{exists}. \\ \mathrm{Here},B\mathrm{is}\mathrm{invertible}. \\ \therefore \left|B^{-1}\right|={\left|B\right|}^{-1}=\frac{1}{\left|B\right|} \\ \Rightarrow \left|B^{-1}AB\right|=\left|B^{-1}\right|\left|AB\right| \\ \Rightarrow \left|B^{-1}AB\right|={\left|B\right|}^{-1}\left|A\right|\left|B\right| \\ \Rightarrow \left|B^{-1}AB\right|=\frac{1}{\left|B\right|}\left|A\right|\left|B\right| \\ \Rightarrow \left|B^{-1}AB\right|=\left|A\right| \end{gathered}$$