Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Solve the system

$$x_1+x_2+2x_3=3\\ 6x_1+7x_2-3x_3=-3$$


$$\begin{bmatrix} 1 & 1 & 2 & 3\\ 6 & 7 & -3 & -3\\ \end{bmatrix}\text{~}\begin{bmatrix} 1 & 1 & 2 & 3\\ 0 & 1 & -15 & -21\\ \end{bmatrix}\text{~}\begin{bmatrix} 1 & 0 & 17 & 24\\ 0 & 1 & -15 & -21\\ \end{bmatrix}$$

And using this I found that: $$x_1+17x_3=24\\ x_2-15x_3=-21$$

As far as I know this is correct, but my issue is how to state my solution.

$$\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} ?\\ ?\\ ? \end{bmatrix}+\begin{bmatrix} ?\\ ?\\ ? \end{bmatrix}s$$

share|improve this question
    
Just like we did here. –  JohnD Jan 27 '13 at 19:44

1 Answer 1

up vote 1 down vote accepted

Notice that all of your equations have a common factor, namely $x_3$.

Rewrite $x_1$ as a function of $x_3$ and then rewrite $x_2$ as a function of $x_3$. Now by letting $x_3$ vary over your the given field, you'll attain all solutions.

You'll get $x_1=24-17x_3$ and $x_2=-21+15x_3$, so your set of solutions should be $\{ \begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}^T : x_1, x_2, x_3\in F \wedge x_1=24-17x_3 \wedge x_2=-21+15x_3\}$ or more succinctly $\{ \begin{bmatrix} 24-17x_3 \space -21+15x_3 \space x_3 \end{bmatrix}^T : x_3\in F\}$, where $F$ is the field you're working on.

In this answer I assumed that your calculations are correct.

For similar problems check this and this.

EDIT: Your calculations are correct.

share|improve this answer
    
And then to put it in the form OP wants, $$\pmatrix{24-17x_3\cr-21+15x_3\cr x_3\cr}=\pmatrix{24\cr-21\cr0}+\pmatrix{-17\cr15\cr1\cr}x_3$$ –  Gerry Myerson Jan 28 '13 at 5:40
    
yes, thank you. –  ground.clouds1 Feb 11 '13 at 1:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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