0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $$x+3y+4z=0$$ $$2y+3z=0$$ $$3y+3z=0$$ $$3y+2z=3$$

$\endgroup$
  • $\begingroup$ What did you do? $\endgroup$ – Jonsa Jan 22 '18 at 19:06
  • $\begingroup$ i did the Gauss-algorythm $\endgroup$ – Dr. Sonnhard Graubner Jan 22 '18 at 19:06
  • $\begingroup$ But you used no matrices $\endgroup$ – Jonsa Jan 22 '18 at 19:07
  • $\begingroup$ yes this is right, you can use Matrices and compare the results $\endgroup$ – Dr. Sonnhard Graubner Jan 22 '18 at 19:08
  • $\begingroup$ I still don't understand what did you do. $\endgroup$ – Jonsa Jan 22 '18 at 19:08
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.