Given find BA and use this to

Given,

$A=\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]$

Now,

$B A=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]=\left[\begin{array}{lll}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right]=6 I$

Thus,

$B^{-1}=\frac{A}{6}=\frac{1}{6}\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$

Given system of equations are:

$x-y=3,2 x+3 y+4 z=17$ and $y+2 z=7$

$\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}3 \\ 17 \\ 7\end{array}\right]$

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]^{-1}\left[\begin{array}{c}3 \\ 17 \\ 7\end{array}\right]=\frac{1}{6}\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]\left[\begin{array}{c}3 \\ 17 \\ 7\end{array}\right]$

$=\frac{1}{6}\left[\begin{array}{c}6+34-28 \\ -12+34-28 \\ 6-17+35\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}12 \\ -6 \\ 24\end{array}\right]=\left[\begin{array}{c}2 \\ -1 \\ 4\end{array}\right]$

Therefore,

$x=2, y=-1$ and $z=4$