Exercises: Gaussian elimination

Solve the system of equations using the Gaussian elimination method, and give the solution in general form. (Use parameters in the solution if necessary).

$\begin{array}{ccccc} x & + & 2 y & = & 0 \\ x & + & y & + & z & = & 0 \\ 2 x & + & 3 y & + & z & = & 0 \end{array}$

$\begin{array}{ccccc} x & + & 2 y & = & 0 \\ x & + & y & + & z & = & 0 \\ 2 x & + & 3 y & + & z & = & 0 \end{array}$
First, construct the extended coefficient matrix and transform it according to the Gaussian method.
$(\begin{array}{cccc} 1 & 2 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 2 & 3 & 1 & 0 \end{array}) \overset{1 \cdot I - 1 \cdot I I}{\underset{2 \cdot I - 1 \cdot I I I}{⟶}} (\begin{array}{cccc} 1 & 2 & 0 & 0 \\ 0 & 1 & - 1 & 0 \\ 0 & 1 & - 1 & 0 \end{array}) \overset{1 \cdot I I - 1 \cdot I I I}{⟶} (\begin{array}{cccc} 1 & 2 & 0 & 0 \\ 0 & 1 & - 1 & 0 \\ 0 & 0 & 0 & 0 \end{array})$
The third line is a true statement. So you can choose $z$ , as you like. For instance, $z = r$ , with $r$ being arbitrary.
Plug $z$ into the second line.
$y - r$ $=$ $0$ $+ r$
$y$ $=$ $r$
Plug $y$ and $z$ into the first line.
$x + 2 r$ $=$ $0$ $- 2 r$
$x$ $=$ $- 2 r$
$\Rightarrow x = - 2 r; y = r; z = r$
$\begin{array}{rcrcrcc} x & - & 2 y & + & 3 z & = & 0 \\ - x & + & 2 y & - & 3 z & = & 0 \\ 2 x & - & 4 y & + & 6 z & = & 0 \end{array}$

$\begin{array}{rcrcrcc} x & - & 2 y & + & 3 z & = & 0 \\ - x & + & 2 y & - & 3 z & = & 0 \\ 2 x & - & 4 y & + & 6 z & = & 0 \end{array}$
First, construct the extended coefficient matrix and transform it according to the Gaussian method.
$(\begin{array}{rrrc} 1 & - 2 & 3 & 0 \\ - 1 & 2 & - 3 & 0 \\ 2 & - 4 & 6 & 0 \end{array}) \overset{1 \cdot I + 1 \cdot I I}{\underset{2 \cdot I - 1 \cdot I I I}{⟶}} (\begin{array}{rrrc} 1 & - 2 & 3 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array})$
The two lower lines are true statements. So you can freely choose two of the three variables, indicated by parameters in the solution.
Set for example
$y = r; z = s$
And plug both into the first line.
$x - 2 r + 3 s = 0$
Solve for $x$ .
$x - 2 r + 3 s$ $=$ $0$ $+ 2 r - 3 s$
$x$ $=$ $2 r - 3 s$
$\Rightarrow x = 2 r - 3 s; y = r; z = s$
$\begin{array}{rcrcrcc} 2 x & - & y & + & 3 z & = & 1 \\ x & + & 3 y & - & 2 z & = & 1 \\ 3 x & - & 2 y & + & 5 z & = & 1 \end{array}$

$\begin{array}{rcrcrcc} 2 x & - & y & + & 3 z & = & 1 \\ x & + & 3 y & - & 2 z & = & 1 \\ 3 x & - & 2 y & + & 5 z & = & 1 \end{array}$
First, construct the extended coefficient matrix and transform it according to the Gaussian method.
$(\begin{array}{crrc} 2 & - 1 & 3 & 1 \\ 1 & 3 & - 2 & 1 \\ 3 & - 2 & 5 & 1 \end{array}) \overset{1 \cdot I - 2 \cdot I I}{\underset{3 \cdot I - 2 \cdot I I I}{⟶}} (\begin{array}{crrc} 2 & - 1 & 3 & 1 \\ 0 & - 7 & 7 & - 1 \\ 0 & 1 & - 1 & 1 \end{array}) \overset{1 \cdot I I + 7 \cdot I I I}{⟶} (\begin{array}{crrc} 2 & - 1 & 3 & 1 \\ 0 & - 7 & 7 & - 1 \\ 0 & 0 & 0 & 6 \end{array})$
The third line implies:
$0 \cdot z = 6$
$\Rightarrow$ This has no solution.
$\begin{array}{rcrcrcr} x & - & 2 y & + & z & = & - 1 \\ - 2 x & + & y & + & 2 z & = & - 5 \\ 3 x & - & y & + & 2 z & = & 3 \\ x & - & 3 y & + & 8 z & = & - 9 \end{array}$

For this task you need the following basic knowledge: Gaussian elimination
$\begin{array}{rcrcrcr} x & - & 2 y & + & z & = & - 1 \\ - 2 x & + & y & + & 2 z & = & - 5 \\ 3 x & - & y & + & 2 z & = & 3 \\ x & - & 3 y & + & 8 z & = & - 9 \end{array}$
First, construct the extended coefficient matrix and transform it according to the Gaussian method.
$\begin{aligned} (\begin{array}{rrcr} 1 & - 2 & 1 & - 1 \\ - 2 & 1 & 2 & - 5 \\ 3 & - 1 & 2 & 3 \\ 1 & - 3 & 8 & - 9 \end{array}) \\ \overset{2 \cdot I + (- 2) \cdot I I, 3 \cdot I - 1 \cdot I I I}{\underset{1 \cdot I - 1 \cdot I V}{⟶}} & (\begin{array}{rrcr} 1 & - 2 & 1 & - 1 \\ 0 & - 3 & 4 & - 7 \\ 0 & - 5 & 1 & - 6 \\ 0 & 1 & - 7 & 8 \end{array}) \\ \overset{5 \cdot I I - 3 \cdot I I I}{\underset{1 \cdot I I + 3 \cdot I V}{⟶}} & (\begin{array}{rrcr} 1 & - 2 & 1 & - 1 \\ 0 & - 3 & 4 & - 7 \\ 0 & 0 & 17 & - 17 \\ 0 & 0 & - 17 & 17 \end{array}) \end{aligned}$
The third (or fourth) line implies:
$17 z$ $=$ $- 17$ $: 17$
$z$ $=$ $- 1$
Plug this result into the second line.
$- 3 y + 4 \cdot (- 1)$ $=$ $- 7$
↓
Multiply out.
$- 3 y - 4$ $=$ $- 7$ $+ 4$
$- 3 y$ $=$ $- 3$ $: (- 3)$
$y$ $=$ $- 1$
Plug $y$ and $z$ into the first line and solve for $x$ .
$x - 2 - 1_{}$ $=$ $- 1$
$x - 3$ $=$ $- 1$ $+ 3$
$x$ $=$ $2$
$\Rightarrow x = 2; y = 1; z = - 1$
$\begin{array}{rcrcrcr} x & - & 3 y & + & z & = & 4 \\ - 2 x & + & 4 y & - & 3 z & = & - 9 \end{array}$

$\begin{array}{rcrcrcr} x & - & 3 y & + & z & = & 4 \\ - 2 x & + & 4 y & - & 3 z & = & - 9 \end{array}$
First, construct the extended coefficient matrix and transform it according to the Gaussian method.
$(\begin{array}{rrrr} 1 & - 3 & 1 & 4 \\ - 2 & 4 & - 3 & - 9 \end{array}) \overset{2 \cdot I + 1 \cdot I I}{⟶} (\begin{array}{rrrr} 1 & - 3 & 1 & 4 \\ 0 & - 2 & - 1 & - 1 \end{array})$
The second line implies:
$- 2 y - 1 z = - 1$
Parametrize $z = r$ .
$- 2 y - r$ $=$ $- 1$ $+ r$
$- 2 y$ $=$ $- 1 + r$ $: (- 2)$
$y$ $=$ $\frac{1 - r}{2}$
Plug $y$ and $z$ into the first line.
$x - 3 \cdot \frac{1 - r}{2} + r$ $=$ $4$
↓
Extend the fraction.
$x - \frac{3 - 3 r}{2} + r$ $=$ $4$ $+ \frac{3 - 3 r}{2} - r$
$x$ $=$ $4 + \frac{3 - 3 r}{2} - r$
↓
Conclude.
$x$ $=$ $\frac{8}{2} + \frac{3 - 3 r}{2} - \frac{2 r}{2}$
$x$ $=$ $\frac{- 5 r + 11}{2}$
$\Rightarrow x = \frac{- 5 r + 11}{2}; y = \frac{1 - r}{2}; z = r$

This content is licensed under
CC BY-SA 4.0 → What does that mean? serlo.org

$- 3 y + 4 \cdot (- 1)$	$=$	$- 7$
	↓	Multiply out.
$- 3 y - 4$	$=$	$- 7$	$+ 4$
$- 3 y$	$=$	$- 3$	$: (- 3)$
$y$	$=$	$- 1$

$- 2 y - r$	$=$	$- 1$	$+ r$
$- 2 y$	$=$	$- 1 + r$	$: (- 2)$
$y$	$=$	$\frac{1 - r}{2}$

$x - 3 \cdot \frac{1 - r}{2} + r$	$=$	$4$
	↓	Extend the fraction.
$x - \frac{3 - 3 r}{2} + r$	$=$	$4$	$+ \frac{3 - 3 r}{2} - r$
$x$	$=$	$4 + \frac{3 - 3 r}{2} - r$
	↓	Conclude.
$x$	$=$	$\frac{8}{2} + \frac{3 - 3 r}{2} - \frac{2 r}{2}$
$x$	$=$	$\frac{- 5 r + 11}{2}$

$17 z$	$=$	$- 17$	$: 17$
$z$	$=$	$- 1$