# Simultaneous Linear Equations/Examples/Arbitrary System 7

## Example of Simultaneous Linear Equations

Let $S$ denote the system of simultaneous linear equations:

 $\ds x + 2 y + z$ $=$ $\ds 2$ $\ds -x + 2 y$ $=$ $\ds -1$ $\ds 5 x - 2 y + 2 z$ $=$ $\ds 7$

$S$ has as its solution set:

 $\ds 2 x + z$ $=$ $\ds 3$ $\ds 4 y + z$ $=$ $\ds 1$

where $z$ can be any number.

## Proof

We express $S$ in matrix representation:

$\begin {pmatrix} 1 & 2 & 1 \\ -1 & 2 & 0 \\ 5 & -2 & 2 \end {pmatrix} \begin {pmatrix} x \\ y \\ z \end {pmatrix} = \begin {pmatrix} 2 \\ -1 \\ 7 \end {pmatrix}$

and consider the augmented matrix:

$\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 2 & 1 & 2 \\ -1 & 2 & 0 & -1 \\ 5 & -2 & 2 & 7 \end {array} }$

In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$.

The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf b_k \end {pmatrix}$.

$e_1 := r_2 \to r_2 + r_1$

$e_2 := r_3 \to r_3 - 5 r_1$

Hence:

$\begin {pmatrix} \mathbf A_2 & \mathbf b_2 \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 2 & 1 & 2 \\ 0 & 4 & 1 & 1 \\ 0 & -12 & -3 & -3 \\ \end {array} }$

$e_3 := r_3 \to r_3 + 3 r_2$

Hence:

$\begin {pmatrix} \mathbf A_3 & \mathbf b_3 \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 2 & 1 & 2 \\ 0 & 4 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end {array} }$

$e_4 := r_2 \to \dfrac 1 4 r_2$

Hence:

$\begin {pmatrix} \mathbf A_4 & \mathbf b_4 \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 2 & 1 & 2 \\ 0 & 1 & 1/4 & 1/4 \\ 0 & 0 & 0 & 0 \\ \end {array} }$

$e_5 := r_1 \to r_1 - 2 r_2$

Hence:

$\begin {pmatrix} \mathbf A_5 & \mathbf b_5 \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 0 & 1/2 & 3/2 \\ 0 & 1 & 1/4 & 1/4 \\ 0 & 0 & 0 & 0 \\ \end {array} }$

$e_6 := r_1 \to 2 r_1$

$e_7 := r_5 \to 4 r_2$

Hence:

$\begin {pmatrix} \mathbf A_7 & \mathbf b_7 \end {pmatrix} = \paren {\begin {array} {ccc|c} 2 & 0 & 1 & 3 \\ 0 & 4 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end {array} }$

Thus $z$ is not restricted, and can be anything, leaving us with:

 $\ds 2 x + z$ $=$ $\ds 3$ $\ds 4 y + z$ $=$ $\ds 1$

as the solution set:

$\blacksquare$