Express the following equations in matrix form and solve them by method of reduction:\(x+2y+z=8\)\(2x+3y-z=11\)\(3x-y-2z=5\)

Answer: \(x=3,\ y=2,\ z=1\)

Matrix form \(AX=B\): \(\begin{bmatrix}1 & 2 & 1\\2 & 3 & -1\\3 & -1 & -2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}8\\11\\5\end{bmatrix}\)

\(R_2\to R_2-2R_1\), \(R_3\to R_3-3R_1\):
\(\begin{bmatrix}1 & 2 & 1\\0 & -1 & -3\\0 & -7 & -5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}8\\-5\\-19\end{bmatrix}\)

\(R_3\to R_3-7R_2\):
\(\begin{bmatrix}1 & 2 & 1\\0 & -1 & -3\\0 & 0 & 16\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}8\\-5\\16\end{bmatrix}\)

Equating: \(x+2y+z=8\), \(-y-3z=-5\), \(16z=16\).
\(\therefore z=1\); \(-y-3=-5\Rightarrow y=2\); \(x+4+1=8\Rightarrow x=3\).

Hence \(x=3,\ y=2,\ z=1\).