Solve the following system of equations by the method of reduction : \(x + y + z = 6\), \(y + 3z = 11\), \(x + z = 2y\).

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

The equations are \(x + y + z = 6\), \(0x + y + 3z = 11\), \(x - 2y + z = 0\). In matrix form \(AX = B\):

\(\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}\)

\(R_3 \to R_3 - R_1\): \(\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & -3 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 11 \\ -6 \end{bmatrix}\)

\(R_3 \to R_3 + 3R_2\): \(\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 9 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 11 \\ 27 \end{bmatrix}\)

So \(x + y + z = 6\), \(y + 3z = 11\), \(9z = 27\).

\(z = 3\); \(y = 11 - 9 = 2\); \(x = 6 - 2 - 3 = 1\).

Check: \(x + z = 4 = 2y\). Hence \(x = 1\), \(y = 2\), \(z = 3\).