We have,
$$
\begin{aligned}
&\mathrm{A}=\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right] \\
&\therefore \mathrm{BA}=\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right]=\left[\begin{array}{lll}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 6
\end{array}\right]=6 \mathrm{I} \\
&\therefore \quad \mathrm{B}^{-1}=\frac{\mathrm{A}}{6}=\frac{1}{6} \mathrm{~A}=\frac{1}{6}\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right] \quad \ldots \text { (i) }
\end{aligned}
$$

Also, $x-y=3,2 x+3 y+4 z=17$ and $y+2 z=7$
$$
\begin{aligned}
&\Rightarrow\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
3 \\
17 \\
7
\end{array}\right] \therefore\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]^{-1}\left[\begin{array}{c}
3 \\
17 \\
7
\end{array}\right] \\
&=\frac{1}{6}\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right]\left[\begin{array}{c}
3 \\
17 \\
7
\end{array}\right] \\
&\therefore x=2, y=-1 \text { and } z=4
\end{aligned}
$$