# Simultaneous Linear Equations/Examples/Arbitrary System 11

## Example of Simultaneous Linear Equations

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

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

$S$ has the single solution:

 $\ds x$ $=$ $\ds 0$ $\ds y$ $=$ $\ds \dfrac 3 2$ $\ds z$ $=$ $\ds \dfrac 1 2$

## Proof

We express $S$ in matrix representation:

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

and consider the augmented matrix:

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

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 $\begin {pmatrix} \mathbf A_k & \mathbf b_k \end {pmatrix}$.

$e_1 := r_3 \to r_3 - r_1$

$e_2 := r_4 \to r_4 - r_1$

Hence:

$\begin {pmatrix} \mathbf A_2 & \mathbf b_2 \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 1 & -1 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & -1 & 3 & 0 \\ 0 & -2 & 6 & 0 \end {array} }$

$e_3 := r_1 \to r_1 - r_2$

$e_4 := r_3 \to r_3 + r_2$

$e_5 := r_4 \to r_4 + 2 r_2$

Hence:

$\begin {pmatrix} \mathbf A_5 & \mathbf b_5 \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 0 & -2 & -1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 4 & 2 \\ 0 & 0 & 8 & 4 \end {array} }$

$e_6 := r_3 \to \dfrac 1 4 r_3$

Hence:

$\begin {pmatrix} \mathbf A_6 & \mathbf b_6 \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 0 & -2 & -1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & 1/2 \\ 0 & 0 & 8 & 4 \end {array} }$

$e_7 := r_1 \to r_1 + 2 r_3$

$e_8 := r_2 \to r_2 - r_3$

$e_9 := r_4 \to r_4 - 8 r_3$

Hence:

$\begin {pmatrix} \mathbf A_9 & \mathbf b_9 \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 3/2 \\ 0 & 0 & 1 & 1/2 \\ 0 & 0 & 0 & 0 \end {array} }$

Thus:

 $\ds x$ $=$ $\ds 0$ $\ds y$ $=$ $\ds \dfrac 3 2$ $\ds z$ $=$ $\ds \dfrac 1 2$

$\blacksquare$