0
$\begingroup$

Having this linear equations written in matrix in $Z_{5}$

\begin{bmatrix}2&3&4&3&|&1\\1&4&0&2&|&1\\2&0&0&3&|&1\end{bmatrix}

i can GEM and get following

\begin{bmatrix}1&4&0&2&|&1\\0&2&0&4&|&4\\0&0&4&4&|&4\end{bmatrix}

we see that based on paremeter T we have this solution for this matrix

$z = 1 - t$

$y = 2 - 2t$

$ x = -7 + 6t$

We know that , to retrieve set of all solution , we have to do

$S = x^{} + S_{0}$

where x is some solution to linear equations $Ax=b$ and $S_{0}$ is set of all solutions to homogene equation e.g $Ax = 0$

So when i want to get set of homogene equations . i solve

\begin{bmatrix}1&4&0&2&|&0\\0&2&0&4&|&0\\0&0&4&4&|&0\end{bmatrix}

thus

$z = - t$

$y = - 2t$

$ x = 6t$

So to write down set of all solutions write add some solution of first matrix for example ( 3 , 2, 1, 0 ) and add it to linear span of homogenous solution so the final result would be

$S = \{ (3,2,1,0) + <(6,-2, -1 )>\}$

Is this correct or did i make mistake in process of thought solving this? Im not sure whetever the linear span of homogenous is correct, when it all depends on parameter t.

Thanks for answer

$\endgroup$
  • $\begingroup$ If you’re working in $\mathbb Z_5$ as you write at the top of your question, what are “6” and “-7?” $\endgroup$ – amd Mar 13 '17 at 18:35
0
$\begingroup$

You're missing some more steps: \begin{align} \begin{bmatrix} 1&4&0&2&|&1\\ 0&2&0&4&|&4\\ 0&0&4&4&|&4 \end{bmatrix} &\to \begin{bmatrix} 1&4&0&2&|&1\\ 0&1&0&2&|&2\\ 0&0&1&1&|&1 \end{bmatrix} &&\begin{aligned} R_2&\gets\tfrac{1}{2}R_2\\R_3&\gets\tfrac{1}{4}R_3 \end{aligned} \\[6px]&\to \begin{bmatrix} 1&0&0&-6&|&-7\\ 0&1&0&2&|&2\\ 0&0&1&1&|&1 \end{bmatrix} &&R_1\gets R_1-4R_2 \end{align} Now you see that the fourth unknown can be given arbitrary values and that the general solution is \begin{cases} x_1=-7+6t \\[4px] x_2=2-2t \\[4px] x_3=1-t \\[4px] x_4=t \end{cases} In vector form, the general solution is $$ \begin{bmatrix} -7+6t \\ 2-2t \\ 1-t \\ t \end{bmatrix} = \begin{bmatrix} -7 \\ 2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 6 \\ -2 \\ -1 \\ 1 \end{bmatrix} $$ or, with your notation, $$ S=(-7,2,1,0)+\langle(6,-2,-1,1)\rangle $$ You are forgetting the fourth component, besides doing computations wrong.

$\endgroup$

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.