Exercise 13

Suppose that $\mathchoice{\left|A\right|}{|A|}{|A|}{|A|}=3$ and that $\mathchoice{\left|B\right|}{|B|}{|B|}{|B|}=2$. Find $\mathchoice{\left|A^{2}\cdot{{B}^{\mathsf{T}}}\cdot B^{-2}\cdot{{A}^{\mathsf{T}}}\right|}{|A^{2}\cdot{{B}^{\mathsf{T}}}\cdot B^{-2}\cdot{{A}^{\mathsf{T}}}|}{|A^{2}\cdot{{B}^{\mathsf{T}}}\cdot B^{-2}\cdot{{A}^{\mathsf{T}}}|}{|A^{2}\cdot{{B}^{\mathsf{T}}}\cdot B^{-2}\cdot{{A}^{\mathsf{T}}}|}$.

Assume that $\mathchoice{\left|A\right|}{|A|}{|A|}{|A|}=0$. Prove that $\mathchoice{\left|6A^{3}+5A^{2}+2A\right|}{|6A^{3}+5A^{2}+2A|}{|6A^{3}+5A^{2}+2A|}{|6A^{3}+5A^{2}+2A|}=0$.