# Gauss-Jordan for a system of four equations

Given the system in $Z_5$

$$\left\{ \begin{array}{c} x+3y+4z=0 \\ 3x+2y+4z=0 \\ x+z=0 \\ 3x+y=2 \end{array} \right.$$

Using Gauss-Jordan ($5R_2, R_2<->R_4, 5R_3, 5R_2$) I get to this point:

$$\begin{matrix} 1 & 3 & 4 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}$$

How can I do?

## 2 Answers

i have got this here $$x+3y+4z=0$$ multiplying the first equation by $-3$ and adding to the second we get: $$-7y-8z=0$$ multiplying the first equation by $-1$ and adding to the third we get: $$-3y-3z=0$$ and multiplying the first equation by $-3$ and adding to the fourth we get $$-8y-12z=2$$ in $\mathbb{Z_5}$ we get $$x+3y+4z=0$$ $$-2y-3z=0$$ $$-3y-3z=0$$ $$-3y-2z=2$$ the new System $$x+3y+4z=0$$ $$2y+3z=0$$ $$3y+3z=0$$ $$3y+2z=3$$

• What did you do? – Jonsa Jan 22 '18 at 19:06
• i did the Gauss-algorythm – Dr. Sonnhard Graubner Jan 22 '18 at 19:06
• But you used no matrices – Jonsa Jan 22 '18 at 19:07
• yes this is right, you can use Matrices and compare the results – Dr. Sonnhard Graubner Jan 22 '18 at 19:08
• I still don't understand what did you do. – Jonsa Jan 22 '18 at 19:08

It will be simpler to take as set of representatives $\;\{0,\pm 1,\pm 2 \}$.

Let's proceed with row reduction of augmented matrix:

\begin{align} &\begin{bmatrix} 1&-2&-1&{}\mid 0\\-2&2&-1&{}\mid 0\\ 1&0&1&{}\mid 0\\-2&1&0&{}\mid 2 \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1&-2&-1&{}\mid 0\\0&-2&2&{}\mid 0\\ 0&2&2&{}\mid 0\\0&2&-2&{}\mid 2 \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1&-2&-1&{}\mid 0\\0&-2&2&{}\mid 0\\ 0&0&-1&{}\mid 0\\0&0&0&{}\mid 2 \end{bmatrix} \end{align}

So the system is inconsistent and there's no solution.