Answer
0
Work Step by Step
We are given:
$C=\begin{bmatrix}
1 & 0 & 5\\
3& -1 & 4\\
2 & -2 & 6
\end{bmatrix}$
$\rightarrow C^{-1}=\begin{bmatrix}
\frac{5}{9} &\frac{2}{9} &\frac{-5}{18} \\
\frac{-1}{9} &\frac{5}{9} &\frac{-11}{18} \\
\frac{2}{9} &\frac{-1}{9} &\frac{1}{18}
\end{bmatrix}$
$\det(C^{-1})=\begin{vmatrix}
\frac{1}{3} & \frac{1}{54}\\
\frac{-1}{9} & \frac{-1}{27}
\end{vmatrix}=\frac{-5}{486}$
We know that:
$\det(AB)=\det(A)\det(B)$
Hence here:
$\det(C^{-1}BA)=\det(C^{-1})\det(BA)=\frac{-5}{486}.0=0$
