Solve the following equation by method of reduction:\(x-y+z=4\)\(2x+y-3z=0\)\(x+y+z=2\)

Answer: \(x=2,\ y=-1,\ z=1\)

In matrix form \(AX=B\): \(\begin{bmatrix}1 & -1 & 1\\2 & 1 & -3\\1 & 1 & 1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\0\\2\end{bmatrix}\).

\(R_2\to R_2-2R_1,\ R_3\to R_3-R_1\): \(\begin{bmatrix}1 & -1 & 1\\0 & 3 & -5\\0 & 2 & 0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\-8\\-2\end{bmatrix}\).

\(R_3\to 3R_3-2R_2\): \(\begin{bmatrix}1 & -1 & 1\\0 & 3 & -5\\0 & 0 & 10\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\-8\\10\end{bmatrix}\).

So \(x-y+z=4,\ 3y-5z=-8,\ 10z=10\).

\(z=1\); \(3y-5=-8\Rightarrow y=-1\); \(x+1+1=4\Rightarrow x=2\).