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Question Number 54903 by Meritguide1234 last updated on 14/Feb/19
Answered by kaivan.ahmadi last updated on 14/Feb/19
$$\mathrm{det}\left(\mathrm{ABA}^{\mathrm{T}} \right)=\mathrm{det}\left(\mathrm{A}^{\mathrm{2}} \mathrm{B}\right)=\mathrm{detA}^{\mathrm{2}} \mathrm{detB}=\mathrm{8}\:\:\left(\mathrm{i}\right) \\ $$$$\mathrm{det}\left(\mathrm{AB}^{−\mathrm{1}} \right)=\frac{\mathrm{detA}}{\mathrm{detB}}=\mathrm{8}\Rightarrow\mathrm{detA}=\mathrm{8detB}\:\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{replace}\:\left(\mathrm{ii}\right)\:\mathrm{in}\:\left(\mathrm{i}\right)\:\mathrm{deduce} \\ $$$$\mathrm{64}\left(\mathrm{detB}\right)^{\mathrm{3}} =\mathrm{8}\Rightarrow\mathrm{detB}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\left(\mathrm{iii}\right) \\ $$$$\mathrm{replace}\:\left(\mathrm{iii}\right)\:\mathrm{in}\:\left(\mathrm{ii}\right)\:\mathrm{deduce} \\ $$$$\mathrm{detA}=\mathrm{4} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{det}\left(\mathrm{BA}^{−\mathrm{1}} \mathrm{B}^{\mathrm{T}} \right)=\frac{\left(\mathrm{detB}\right)^{\left.\right)\mathrm{2}} }{\mathrm{detA}}=\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$ \\ $$$$ \\ $$