Using detA and detB to calculate the determinant of matrix C

Question

If we have C=($A^t$)$^2$BA$^3$B$^-$$^1$A$^-$$^3$ and detA=-2 and detB doesnt equal 0, how do we calculate det C?

I know that the transpose of a matrix does not affect the determinant. Does this mean that ($A^t$)$^2$=(-2)$^2$=4?

And then how is A$^-$$^3$ affected? Does this mean the inverse of A cubed? And how does the inverse affect the determinant? Thanks

Answer 1

One may recall that, if $\det A \neq 0$, then $$ \det A{}^t=\det A, \qquad \det (A{}^{p})=(\det A)^p,\quad p=0,\pm 1,\pm2, \ldots. $$and $$ \det (AB)=(\det A)(\det B)=(\det B)(\det A)=\det (BA) $$ Thus here: $$ \det C=\det ( (A^t)^2BA^3B^{-1}A^{-3})=(\det A)^{2+3-3}(\det B)^{1-1}=(\det A)^{2}=4. $$

Answer 2

\begin{equation*} \det(A^k) = \det(A)^k~\text{and}~\det(A^t) = \det(A) \end{equation*}

Answer 3

Remember the following

$$\operatorname{det}(AB)=\operatorname{det}{A}\operatorname{det}{B}$$ $$\operatorname{det}{A^n}=(\operatorname{det}{A})^n$$ $$\operatorname{det}{A^T}=\operatorname{det}{A}$$

So in our case we have $B$ and $B^{-1}$ whose determinant cancel out because they are not $0$ (otherwise $B^{-1}$ would not exist) $A^3$ and $A^{-3}$ whose determinant cancel out and we are left with $(\operatorname{det}{A^T})^2=4$