Set of all solutions of linear equations

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

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

1 Answer

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.