Exercises: Gaussian elimination

Solve the following systems of equations using the Gaussian elimination method.

$ \begin{array}{ccrcr} 3 x & + & 2 y & = & - 1 \\ 4 x & + & y & = & - 2 \\ 6 x & + & 4 y & = & 3 \end{array}$

For this task you need the following basic knowledge: Gaussian elimination
$\begin{array}{ccrcr} 3 x & + & 2 y & = & - 1 \\ 4 x & + & y & = & - 2 \\ 6 x & + & 4 y & = & 3 \end{array}$
First, you construct the extended coefficient matrix.
$(\begin{array}{ccr} 3 & 2 & - 1 \\ 4 & 1 & - 2 \\ 6 & 4 & 3 \end{array})$
Then divide the first line by 3, the second by 4, and the third by 6.
$\underset{I I I : 6}{\overset{\overset{}{I : 3, I I : 4}}{\to}} (\begin{array}{ccr} 1 & \frac{2}{3} & - \frac{1}{3} \\ 1 & \frac{1}{4} & - \frac{2}{4} \\ 1 & \frac{2}{3} & \frac{1}{2} \end{array})$
Now subtract the first from the second line and from the third.
$\overset{\overset{}{I I - I, I I I - I}}{\to} (\begin{array}{ccr} 1 & \frac{2}{3} & - \frac{1}{3} \\ 0 & - \frac{5}{12} & - \frac{1}{6} \\ 0 & 0 & \frac{5}{6} \end{array})$
According to the criterion for solvability of linear systems, it already follows that the system of equations has no solution, because:
$rank (\begin{array}{cc} 1 & \frac{2}{3} \\ 0 & - \frac{5}{12} \\ 0 & 0 \end{array}) = 2 < 3 = rank (\begin{array}{ccr} 1 & \frac{2}{3} & - \frac{1}{3} \\ 0 & - \frac{5}{12} & - \frac{1}{3} \\ 0 & 0 & \frac{5}{6} \end{array})$
$\begin{array}{rcrcc} x & - & 3 y & = & 4 \\ 2 x & + & y & = & 1 \\ 4 x & + & 5 y & = & 9 \end{array}$

For this task you need the following basic knowledge: Gaussian elimination
$\begin{array}{rcrcc} x & - & 3 y & = & 4 \\ 2 x & + & y & = & 1 \\ 4 x & + & 5 y & = & 9 \end{array}$
First, you construct the extended coefficient matrix.
$(\begin{array}{ccc} 1 & - 3 & 4 \\ 2 & 1 & 1 \\ 4 & 5 & 9 \end{array})$
Then divide the second line by 2 and the third by 4.
$\overset{\overset{}{I I : 2, I I I : 4}}{\to} (\begin{array}{ccc} 1 & - 3 & 4 \\ 1 & \frac{1}{2} & \frac{1}{2} \\ 1 & \frac{5}{4} & \frac{9}{4} \end{array})$
Now subtract the first from the second line and from the third.
$\overset{\overset{}{I I - I, I I I - I}}{\to} (\begin{array}{ccc} 1 & - 3 & 4 \\ 0 & \frac{7}{2} & - \frac{7}{2} \\ 0 & \frac{17}{4} & - \frac{7}{4} \end{array})$
Then divide the second line by $\frac{7}{2}$ and the third by $\frac{17}{4}$ .
$\overset{\overset{}{I I : \frac{7}{2}, I I I : \frac{17}{4}}}{\to} (\begin{array}{ccc} 1 & - 3 & 4 \\ 0 & 1 & - 1 \\ 0 & 1 & - \frac{7}{17} \end{array})$
Now subtract the second line from the third.
$\overset{\overset{}{I I I - I I}}{\to} (\begin{array}{ccc} 1 & - 3 & 4 \\ 0 & 1 & - 1 \\ 0 & 0 & \frac{10}{17} \end{array})$
According to the criterion for solvability of linear systems, it already follows that the system of equations has no solution, because:
$rank (\begin{array}{cc} 1 & - 3 \\ 0 & 1 \\ 0 & 0 \end{array}) = 2 < 3 = rank (\begin{array}{ccc} 1 & - 3 & 4 \\ 0 & 1 & - 1 \\ 0 & 0 & \frac{10}{17} \end{array})$
$\begin{array}{rcrcr} x & - & 2 y & = & 3 \\ - 2 x & + & 4 y & = & - 6 \\ - x & + & 2 y & = & - 3 \end{array}$

For this task you need the following basic knowledge: Gaussian elimination
$\begin{array}{rcrcr} x & - & 2 y & = & 3 \\ - 2 x & + & 4 y & = & - 6 \\ - x & + & 2 y & = & - 3 \end{array}$
First, you construct the extended coefficient matrix.
$(\begin{array}{rrr} 1 & - 2 & 3 \\ - 2 & 4 & - 6 \\ - 1 & 2 & - 3 \end{array})$
Then divide the second line by -2 and the third by -1.
$\overset{\overset{}{I I : (- 2), I I I : (- 1)}}{\to} (\begin{array}{rrr} 1 & - 2 & 3 \\ 1 & - 2 & 3 \\ 1 & - 2 & 3 \end{array})$
Now subtract the first from the second line and from the third.
$\overset{\overset{}{I I - I, I I I - I}}{\to} (\begin{array}{rrr} 1 & - 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array})$
According to the criterion for Solvability of linear systems, it already follows that the system of equations is not uniquely solvable, because:
$rank (\begin{array}{rrr} 1 & - 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}) = 1 < 3$
Nevertheless, you can write down a solution set:
$L = {(x, y) \in ℝ^{2} | x - 2 y = 3}$
Remark: The (infinitely many) solutions are on a straight line with the equation $x - 2 y = 3 \Leftrightarrow y = \frac{1}{2} x - \frac{3}{2}$ .

$\begin{array}{rcrcrcr} 6 x & - & y & + & 2 z & = & 1 \\ 5 x & - & 3 y & + & 3 z & = & 4 \\ 3 x & - & 2 y & + & z & = & 14 \end{array}$

$\begin{array}{cccc} \frac{3}{4} x & - & \frac{7}{6} y & = & \frac{1}{8} \\ - 9 x & + & 14 y & = & - \frac{3}{2} \\ \frac{1}{3} x & + & \frac{1}{9} y & = & 0 \end{array}$

For this task you need the following basic knowledge: Gaussian elimination
$\begin{array}{cccc} \frac{3}{4} x & - & \frac{7}{6} y & = & \frac{1}{8} \\ - 9 x & + & 14 y & = & - \frac{3}{2} \\ \frac{1}{3} x & + & \frac{1}{9} y & = & 0 \end{array}$
First, you construct the extended coefficient matrix.
$(\begin{array}{ccc} \frac{3}{4} & - \frac{7}{6} & \frac{1}{8} \\ - 9 & 14 & - \frac{3}{2} \\ \frac{1}{3} & \frac{1}{9} & 0 \end{array})$
Then divide the first line by $\frac{3}{4}$ , the second by $- 9$ and the third by $\frac{1}{3}$ .
$\underset{III : \frac{1}{3}}{\overset{\overset{}{I : \frac{3}{4}, I I : (- 9)}}{\to}} (\begin{array}{ccc} 1 & - \frac{14}{9} & \frac{1}{6} \\ 1 & - \frac{14}{9} & \frac{1}{6} \\ 1 & \frac{1}{3} & 0 \end{array})$
Now subtract the first line from the second and third lines.
$\underset{I I I - I}{\overset{\overset{}{I I - I}}{\to}} (\begin{array}{ccc} 1 & - \frac{14}{9} & \frac{1}{6} \\ 0 & 0 & 0 \\ 0 & \frac{17}{9} & - \frac{1}{6} \end{array})$
Sicne $rank (A) = 2 = rank (A | b)$ and we have two variables, the system of equations is uniquely solvable according to the rank criterion for the solvability of linear systems.
Now divide the third line by $\frac{17}{9}$ .
$\overset{\overset{}{I I I : \frac{17}{9}}}{\to} (\begin{array}{ccc} 1 & - \frac{14}{9} & \frac{1}{6} \\ 0 & 0 & 0 \\ 0 & 1 & - \frac{3}{34} \end{array})$
Then subtract from the first $- \frac{14}{9}$ times the third line.
$\overset{\overset{}{I - (- \frac{14}{9}) \cdot I I I}}{\to} (\begin{array}{ccc} 1 & 0 & \frac{1}{34} \\ 0 & 0 & 0 \\ 0 & 1 & - \frac{3}{34} \end{array})$
Now you can read the solution in the right column:
$L = {(\frac{1}{34}; - \frac{3}{34})}$
$\begin{array}{rcrcrcr} 6 x & - & z & + & 2 y & = & 48 \\ 5 y & - & 3 x & + & 3 z & = & 49 \\ 3 z & - & 2 x & + & y & = & 24 \end{array}$

$\begin{array}{rcrcrcr} 6 x & - & z & + & 2 y & = & 48 \\ 5 y & - & 3 x & + & 3 z & = & 49 \\ 3 z & - & 2 x & + & y & = & 24 \end{array}$
Sort the variables.
$\begin{array}{rcrcrcr} 6 x & + & 2 y & - & z & = & 48 \\ - 3 x & + & 5 y & + & 3 z & = & 49 \\ - 2 x & + & y & + & 3 z & = & 24 \end{array}$
Construct the extended coefficient matrix and transform it according to the Gaussian method.
$(\begin{array}{crcc} 6 & 2 & - 1 & 48 \\ - 3 & 5 & 3 & 49 \\ - 2 & 1 & 3 & 24 \end{array}) \overset{1 \cdot I + 2 \cdot I I}{\underset{1 \cdot I + 3 \cdot I I I}{⟶}} (\begin{array}{crcc} 6 & 2 & - 1 & 48 \\ 0 & 12 & 5 & 146 \\ 0 & 5 & 8 & 120 \end{array}) \underset{5 \cdot I I - 12 \cdot I I I}{⟶} (\begin{array}{crcc} 6 & 2 & - 1 & 48 \\ 0 & 12 & 5 & 146 \\ 0 & 0 & - 71 & - 710 \end{array})$
The third line implies
$- 71 z$ $=$ $- 710$
$z$ $=$ $10$
Plug this z into the second line.
$12 y + 5 \cdot 10$ $=$ $146$
$12 y + 50$ $=$ $146$
$2 y$ $=$ $96$
$y$ $=$ $8$
Plug the known y- and z-value into the first line.
$6 x + 2 \cdot 8 - 10$ $=$ $48$
$6 x \cdot 16 - 10$ $=$ $48$
$6 x + 6$ $=$ $48$
$6 x$ $=$ $42$
$x$ $=$ $7$
$\Rightarrow x = 7; y = 8; z = 10$
$\begin{array}{rcrcrcr} x & - & 2 y & = & 4 \\ - & y & - & z & = & - 1 \\ - x & + & y & + & 3 z & = & - 1 \end{array}$

For this task you need the following basic knowledge: Gaussian elimination
$\begin{array}{rcrcrcr} x & - & 2 y & = & 4 \\ - & y & - & z & = & - 1 \\ - x & + & y & + & 3 z & = & - 1 \end{array}$
First, construct the extended coefficient matrix and transform it according to the Gaussian method.
$(\begin{array}{cccc} 1 & - 2 & 0 & 4 \\ 0 & - 1 & - 1 & - 1 \\ - 1 & 1 & 3 & - 1 \end{array}) \overset{1 \cdot I + 1 \cdot I I I}{⟶} (\begin{array}{cccc} 1 & - 2 & 0 & 4 \\ 0 & - 1 & - 1 & - 1 \\ 0 & - 1 & 3 & 3 \end{array}) \overset{1 \cdot I I - 1 \cdot I I I}{⟶} (\begin{array}{cccc} 1 & - 2 & 0 & 4 \\ 0 & - 1 & - 1 & - 1 \\ 0 & 0 & - 4 & - 4 \end{array})$
The third line implies
$- 4 z$ $=$ $- 4$ $: (- 4)$
$z$ $=$ $1$
↓
Plug into the second line.
$- y - 1$ $=$ $- 1$ $+ 1$
$- y$ $=$ $0$ $: (- 1)$
$y$ $=$ $0$
↓
Plug into the first line.
$x - 2 \cdot 0$ $=$ $4$
$x$ $=$ $4$
$\Rightarrow x = 4; y = 0; z = 1$

$\begin{array}{rcrcrcr} 2 x & + & 3 y & - & z & = & 3 \\ x & + & 2 z & = & 9 \\ x & - & y & = & 2 \end{array}$

$\begin{array}{rcrcrcr} 5 x & + & 2 y & - & 2 z & = & - 1 \\ 3 x & + & y & - & 3 z & = & - 4 \\ 2 x & + & z & = & 4 \end{array}$
For this task you need the following basic knowledge: Gaussian elimination
$\begin{array}{ccrcrcr} 5 x & + & 2 y & - & 2 z & = & - 1 \\ 3 x & + & y & - & 3 z & = & - 4 \\ 2 x & + & z & = & 4 \end{array}$
First, construct the extended coefficient matrix and transform it according to the Gaussian method.
$(\begin{array}{ccrr} 5 & 2 & - 2 & - 1 \\ 3 & 1 & - 3 & - 4 \\ 2 & 0 & 1 & 4 \end{array}) \overset{3 \cdot I - 5 \cdot I I}{\underset{2 \cdot I - 5 \cdot I I I}{⟶}} (\begin{array}{ccrr} 5 & 2 & - 2 & - 1 \\ 0 & 1 & 9 & 17 \\ 0 & 4 & - 9 & - 22 \end{array}) \overset{4 \cdot I I - 1 \cdot I I I}{⟶} (\begin{array}{ccrr} 5 & 2 & - 2 & - 1 \\ 0 & 1 & 9 & 17 \\ 0 & 0 & 45 & 90 \end{array})$
The third line implies
$45 z$ $=$ $90_{}$ $: 45$
$z$ $=$ $2$
↓
Plug into the second line.
$y + 9 \cdot 2$ $=$ $17$
↓
Multiply out.
$y + 18$ $=$ $17$ $- 18$
$y$ $=$ $- 1$
↓
Plug into the first line.
$5 x + 2 \cdot (- 1) - 2 \cdot 2$ $=$ $- 1$
↓
Multiply out.
$5 x - 2 - 4$ $=$ $- 1$
$5 x - 6$ $=$ $- 1$ $+ 6$
$5 x$ $=$ $5$ $: 5$
$x$ $=$ $1$
$\Rightarrow x = 1; y = - 1; z = 2$
$\begin{array}{ccrcrcr} 4 x & + & 3 y & + & z & = & 13 \\ 2 x & - & 5 y & + & 3 z & = & 1 \\ 7 x & - & y & - & 2 z & = & - 1 \end{array}$

For this task you need the following basic knowledge: Gaussian elimination
$\begin{array}{ccrcrcr} 4 x & + & 3 y & + & z & = & 13 \\ 2 x & - & 5 y & + & 3 z & = & 1 \\ 7 x & - & y & - & 2 z & = & - 1 \end{array}$
First, construct the extended coefficient matrix and transform it according to the Gaussian method.
$(\begin{array}{crrr} 4 & 3 & 1 & 13 \\ 2 & - 5 & 3 & 1 \\ 7 & - 1 & - 2 & - 1 \end{array}) \overset{1 \cdot I - 2 \cdot I I}{\underset{7 \cdot I - 4 \cdot I I I}{⟶}} (\begin{array}{crrr} 4 & 3 & 1 & 13 \\ 0 & 13 & - 5 & 11 \\ 0 & 25 & 15 & 95 \end{array}) \overset{25 \cdot I I - 13 \cdot I I I}{⟶} (\begin{array}{crrr} 4 & 3 & 1 & 13 \\ 0 & 13 & - 5 & 11 \\ 0 & 0 & - 320 & - 960 \end{array})$
The third line implies
$- 320 z$ $=$ $- 960$ $: (- 320)$
$z$ $=$ $3$
↓
Plug into the second line.
$13 y - 5 \cdot 3$ $=$ $11$
$13 y - 15$ $=$ $11$ $+ 15$
$13 y$ $=$ $26$ $: 13$
$y$ $=$ $2$
↓
Plug into the second line.
$4 x + 3 \cdot 2 + 3$ $=$ $13$
↓
Conclude.
$4 x + 9$ $=$ $13$ $- 9$
$4 x$ $=$ $4$ $: 4$
$x$ $=$ $1$
$\Rightarrow x = 1; y = 2; z = 3$

$\begin{array}{rcrcrcr} 2 x & + & 9 y & - & 14 z & = & 39 \\ 3 x & + & 6 y & + & 2 z & = & 36 \\ \frac{x}{2} & + & \frac{y}{3} & + & 7 z & = & 2 \end{array}$

$\begin{array}{rcrcrcr} x & + & y & - & z & = & 4 \\ 4 x & - & 2 y & - & 2 z & = & 3 \\ - 5 x & + & 4 y & + & 2 z & = & 0 \end{array}$

For this task you need the following basic knowledge: Gaussian elimination
$\begin{array}{rcrcrcr} x & + & y & - & z & = & 4 \\ 4 x & - & 2 y & - & 2 z & = & 3 \\ - 5 x & + & 4 y & + & 2 z & = & 0 \end{array}$
First, construct the extended coefficient matrix and transform it according to the Gaussian method.
$(\begin{array}{rrrc} 1 & 1 & - 1 & 4 \\ 4 & - 2 & - 2 & 3 \\ - 5 & 4 & 2 & 0 \end{array}) \overset{4 \cdot I - 1 \cdot I I}{\underset{5 \cdot I + 1 \cdot I I I}{⟶}} (\begin{array}{rrrc} 1 & 1 & - 1 & 4 \\ 0 & 6 & - 2 & 13 \\ 0 & 9 & - 3 & 20 \end{array}) \overset{9 \cdot I - 1 \cdot I I I}{⟶} (\begin{array}{rrrc} 1 & 1 & - 1 & 4 \\ 0 & 6 & - 2 & 13 \\ 0 & 0 & 0 & - 1 \end{array})$
The third line implies
$0 \cdot z = - 1$
$\Rightarrow$ So there is no solution!

2
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.
$x$ $=$ $\frac{8}{2} + \frac{3 - 3 r}{2} - \frac{2 r}{2}$
$x$ $=$ $\frac{- 5 r + 11}{2}$
$\Rightarrow x = \frac{- 5 r + 11}{2}; y = \frac{1 - r}{2}; z = r$
3
Solve the systems of equations using the Gauss-Jordan method.
$\begin{array}{ccccc} 3 x & + & 4 y & = & - 1 \\ 2 x & + & 5 y & = & - 3 \end{array}$

$\begin{array}{ccccc} 3 x & + & 4 y & = & - 1 \\ 2 x & + & 5 y & = & - 3 \end{array}$
First, write down the extended coefficient matrix.
$(\begin{array}{ccc} 3 & 4 & - 1 \\ 2 & 5 & - 3 \end{array})$
Then divide the first line by 3, the second by 2.
$\overset{\overset{}{I : 3, I I : 2}}{\to} (\begin{array}{ccc} 1 & \frac{4}{3} & - \frac{1}{3} \\ 1 & \frac{5}{2} & - \frac{3}{2} \end{array})$
Now subtract the first line from the second line.
$\overset{\overset{}{I I - I}}{\to} (\begin{array}{ccc} 1 & \frac{4}{3} & - \frac{1}{3} \\ 0 & \frac{7}{6} & - \frac{7}{6} \end{array})$
Then divide the second line by $\frac{7}{6}$ .
$\overset{\overset{}{I I : \frac{7}{6}}}{\to} (\begin{array}{ccc} 1 & \frac{4}{3} & - \frac{1}{3} \\ 0 & 1 & - 1 \end{array})$
Then subtract from the first line $\frac{4}{3}$ times the second line.
$\overset{\overset{}{I - \frac{4}{3} \cdot I I}}{\to} (\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & - 1 \end{array})$
Now you can read the solution in the right column:
$\Rightarrow x = 1; y = - 1$

$\begin{array}{ccrcc} 3 x & - & 4 y & = & - 26 \\ 2 x & + & 3 y & = & 28 \end{array}$

$\begin{array}{ccrcc} 3 x & - & 4 y & = & - 26 \\ 2 x & + & 3 y & = & 28 \end{array}$
First, write down the extended coefficient matrix.
$(\begin{array}{ccc} 3 & - 4 & - 26 \\ 2 & 3 & 28 \end{array})$
Then divide the first line by 3 and the second by 2.
$\overset{\overset{}{I : 3, I I : 2}}{\to} (\begin{array}{ccc} 1 & - \frac{4}{3} & - \frac{26}{3} \\ 1 & \frac{3}{2} & 14 \end{array})$
Now subtract the first line from the second line.
$\overset{\overset{}{I I - I}}{\to} (\begin{array}{ccc} 1 & - \frac{4}{3} & - \frac{26}{3} \\ 0 & \frac{17}{6} & \frac{68}{3} \end{array})$
Then divide the second line by $\frac{17}{6}$ .
$\overset{\overset{}{I I : \frac{17}{6}}}{\to} (\begin{array}{ccc} 1 & - \frac{4}{3} & - \frac{26}{3} \\ 0 & 1 & 8 \end{array})$
Then subtract from the first line $- \frac{4}{3}$ times the second line.
$\overset{\overset{}{I - (- \frac{4}{3}) \cdot I I}}{\to} (\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & 8 \end{array})$
Now you can read the solution in the right column:
$\Rightarrow x = 2; y = 8$

$\begin{array}{rcrcrcr} x & + & 2 y & - & z & = & 2 \\ x & + & y & + & 2 z & = & 9 \\ 2 x & + & 3 y & - & 3 z & = & - 1 \end{array}$

$\begin{array}{rcrcrcr} x & + & 2 y & - & z & = & 2 \\ x & + & y & + & 2 z & = & 9 \\ 2 x & + & 3 y & - & 3 z & = & - 1 \end{array}$
First, write down the extended coefficient matrix.
$(\begin{array}{ccr} 1 & 2 & - 1 \\ 1 & 1 & 2 \\ 2 & 3 & - 3 \end{array} | \begin{array}{r} 2 \\ 9 \\ - 1 \end{array}) \overset{I I - I}{\underset{I I I - 2 \cdot I}{⟶}} (\begin{array}{crr} 1 & 2 & - 1 \\ 0 & - 1 & 3 \\ 0 & - 1 & - 1 \end{array} | \begin{array}{r} 2 \\ 7 \\ - 5 \end{array}) \overset{I I I - I I}{⟶} (\begin{array}{crr} 1 & 2 & - 1 \\ 0 & - 1 & 3 \\ 0 & 0 & - 4 \end{array} | \begin{array}{r} 2 \\ 7 \\ - 12 \end{array})$
$\overset{I I : (- 1)}{\underset{I I I : (- 4)}{⟶}} (\begin{array}{ccc} 1 & 2 & - 1 \\ 0 & 1 & - 3 \\ 0 & 0 & 1 \end{array} | \begin{array}{c} 2 \\ - 7 \\ 3 \end{array})$
The third line implies:
$z = 3$
Plug this into the second line.
$y - 9 = - 7$
$\Rightarrow y = 2$
Plug $y$ and $z$ into the first line.
$x + 4 - 3 = 2$
$x = 1$
$\Rightarrow x = 1; y = 2; z = 3$

$\frac{832}{3} z$	$=$	$- 120$	$: \frac{832}{3}$
$z$	$=$	$- \frac{45}{104}$
	↓	Plug into the second line.
$15 \cdot y - 46 \cdot (- \frac{45}{104})$	$=$	$45$
	↓	Multiply out.
$15 \cdot y + \frac{1035}{52}$	$=$	$45$	$- \frac{1035}{52}$
$15 \cdot y$	$=$	$\frac{1025}{52}$	$: 15$
$y$	$=$	$\frac{87}{52}$
	↓	Plug into the first line.
$2 \cdot x + 9 \cdot \frac{87}{52} - 14 \cdot (- \frac{45}{104})$	$=$	$39$
	↓	Multiply out.
$2 \cdot x + \frac{783}{52} + \frac{315}{52}$	$=$	$39$
$2 \cdot x + \frac{1098}{52}$	$=$	$39$	$- \frac{1098}{52}$
$2 \cdot x$	$=$	$\frac{465}{26}$	$: 2$
$x$	$=$	$\frac{465}{52}$

$13 y - 8 \cdot (- 12.25)$	$=$	$- 19$
	↓	Multiply out.
$13 y + 98$	$=$	$- 19$
$13 y$	$=$	$- 117$
$y$	$=$	$- 9$
	↓	Insert the found y- and z-value into the first line.
$6 x - (- 9) + 2 \cdot (- 12.25)$	$=$	$1$
	↓	Multiply out.
$6 x + 9 - 24.5$	$=$	$1$
$6 x - 15.5$	$=$	$1$
$6 x$	$=$	$16.5$
$x$	$=$	$2.75$

$12 y + 5 \cdot 10$	$=$	$146$
$12 y + 50$	$=$	$146$
$2 y$	$=$	$96$
$y$	$=$	$8$

$6 x + 2 \cdot 8 - 10$	$=$	$48$
$6 x \cdot 16 - 10$	$=$	$48$
$6 x + 6$	$=$	$48$
$6 x$	$=$	$42$
$x$	$=$	$7$

$- 4 z$	$=$	$- 4$	$: (- 4)$
$z$	$=$	$1$
	↓	Plug into the second line.
$- y - 1$	$=$	$- 1$	$+ 1$
$- y$	$=$	$0$	$: (- 1)$
$y$	$=$	$0$
	↓	Plug into the first line.
$x - 2 \cdot 0$	$=$	$4$
$x$	$=$	$4$

$- 22 z$	$=$	$- 72$	$: (- 22)$
$z$	$=$	$\frac{36}{11}$
	↓	Plug into the second line.
$3 \cdot y - 5 \cdot \frac{36}{11}$	$=$	$- 15$
	↓	Multiply out.
$3 \cdot y - \frac{180}{11}$	$=$	$- 15$	$+ \frac{180}{11}$
$3 \cdot y$	$=$	$\frac{15}{11}$	$: 3$
$y$	$=$	$\frac{5}{11}$
	↓	Plug into the first line
$2 \cdot x + 3 \cdot \frac{5}{11} - \frac{36}{11}$	$=$	$3$
	↓	Multiply.
$2 \cdot x + \frac{15}{11} - \frac{36}{11}$	$=$	$3$
	↓	Subtract.
$2 \cdot x + (- \frac{21}{11})$	$=$	$3$	$+ \frac{21}{11}$
$2 \cdot x$	$=$	$\frac{54}{11}$	$: 2$
$x$	$=$	$\frac{27}{11}$

$45 z$	$=$	$90_{}$	$: 45$
$z$	$=$	$2$
	↓	Plug into the second line.
$y + 9 \cdot 2$	$=$	$17$
	↓	Multiply out.
$y + 18$	$=$	$17$	$- 18$
$y$	$=$	$- 1$
	↓	Plug into the first line.
$5 x + 2 \cdot (- 1) - 2 \cdot 2$	$=$	$- 1$
	↓	Multiply out.
$5 x - 2 - 4$	$=$	$- 1$
$5 x - 6$	$=$	$- 1$	$+ 6$
$5 x$	$=$	$5$	$: 5$
$x$	$=$	$1$

$- 320 z$	$=$	$- 960$	$: (- 320)$
$z$	$=$	$3$
	↓	Plug into the second line.
$13 y - 5 \cdot 3$	$=$	$11$
$13 y - 15$	$=$	$11$	$+ 15$
$13 y$	$=$	$26$	$: 13$
$y$	$=$	$2$
	↓	Plug into the second line.
$4 x + 3 \cdot 2 + 3$	$=$	$13$
	↓	Conclude.
$4 x + 9$	$=$	$13$	$- 9$
$4 x$	$=$	$4$	$: 4$
$x$	$=$	$1$

$- 3 y + 4 \cdot (- 1)$	$=$	$- 7$
	↓	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.
$x$	$=$	$\frac{8}{2} + \frac{3 - 3 r}{2} - \frac{2 r}{2}$
$x$	$=$	$\frac{- 5 r + 11}{2}$

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