If \(\mathrm{AX}=\mathrm{B}\), where \(\mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]\) and \(\mathrm{X}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]\), then \(x+y+z=\)

Given \(A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right]\) and \(B=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]\) and \(A X=B \therefore AA ^{-1} X = A ^{-1} B\)
\(\Rightarrow IX = A ^{-1} B \Rightarrow X = A ^{-1} B\)
Now \(| A |=4+(-8)+9=5\)
\(\therefore A^{-1}=\left[\begin{array}{ccc}4 & -1 & 1 \\ 8 & -7 & 2 \\ 9 & -6 & 1\end{array}\right] \times \frac{1}{5}\)
\(\therefore\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{5}\left[\begin{array}{lll}4 & -1 & 1 \\ 8 & -7 & 2 \\ 9 & -6 & 1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]\)
\(=\frac{1}{5}\left[\begin{array}{l}8-1+2 \\ 8-7+4 \\ 9-6+2\end{array}\right]=\frac{1}{5}\left[\begin{array}{l}5 \\ 5 \\ 5\end{array}\right]\)
\(y=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]\)
On comparing both side, we get
\(x=y=z=1 \Rightarrow x+y+z=1+1+1=3\)