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Solve the equationsystem $$x+3y+6z=3$$$$x+y+z=-2$$$$-x+y+4z=7$$ if I use Gauss elimination it will be $$x+y+z=-2$$$$3z=9$$$$2y+5z=5$$ and then it will be$$x+y=-5$$$$y=-5$$$$z=3$$ and at the end$$x=0$$$$y=-5$$$$z=3$$ is the parameter solution$$x=3t$$$$y=-5(t+1)$$$$z=2t+3$$?

user1838781
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  • I think you're mistaken at the first step. Re-check your second equation $3z=9$. You should actually have twice the same line $2y+5z=5$. So you should only be left with two equations in your system and a free parameter ($x$, $y$ or $z$ as you wish ... $x$ would be a smarter choice though, regarding the equations you have ;)). Plus you end up with a unique solution in your computations, but if you look at the matrix of the equations: $$A=\left[\begin{array}{cc}1 && 3 && 6\ 1 && 1 && 1\ -1 && 1 && 4\end{array}\right]$$ its determinant is $0$ so you actually shouldn't have a unique solution ;) – Dolma Apr 23 '13 at 08:49
  • Oh and you could also see this from the fact that $L_3=L_1-2L_2$ if you denote $L_1$, $L_2$ and $L_3$ your 3 equations. – Dolma Apr 23 '13 at 08:52

1 Answers1

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Let's start with your equation system:

$$\left\{\begin{array}{cc}x+3y+6z=3 && L_1 \\ x+y+z=−2 && L_2 \\ −x+y+4z=7&& L_3\end{array}\right.$$

Now as I said in my comment, if you use Gauss elimination you should get:

$$\left\{\begin{array}{cc}2y+5z=5 && L_1\leftarrow L_1-L_2\\ x+y+z=−2\\ 2y+5z=5 && L_3\leftarrow L_3+L_2\end{array}\right.$$

Which simplifies in:

$$\left\{\begin{array}{cc}x+y+z=−2\\ 2y+5z=5\end{array}\right.$$

You only have two equations left and a free parameter $x$ because in the first place you only had two linearly independent equations ($L_3=L_1-2L_2$)

Once more:

$$\left\{\begin{array}{cc}2x-3z=-9 && L_1\leftarrow 2L_1-L_2\\ 2y+5z=5\end{array}\right.$$

And so:

$$\left\{\begin{array}{cc}3z=2x+9\\ 2y=5-5z\end{array}\right.\Rightarrow\left\{\begin{array}{cc}z=\frac{2}{3}x+3\\ y=\frac{5}{2}(1-z)=\frac{5}{2}\left(1-\left(\frac{2}{3}x+3\right)\right)=\frac{5}{2}\left(-2-\frac{2}{3}x\right)=-5-\frac{5}{3}x\end{array}\right.$$

You end up with:

$$\left\{\begin{array}{cc}z=\frac{2}{3}x+3\\ y=-5\left(1+\frac{x}{3}\right)\end{array}\right.,\space x\in\Bbb{R}$$

Or written in another way:

$$\left\{\begin{array}{cc}y=-5\left(\left(\frac{x}{3}\right)+1\right) \\ z=2\left(\frac{x}{3}\right)+3\end{array}\right.,\space\left(\frac{x}{3}\right)\in\Bbb{R}$$

Now you can recognize the solution you provided if you write $t=\frac{x}{3}$:

$$\left\{\begin{array}{cc}y=-5(t+1) \\ z=2t+3\end{array}\right.,\space t\in\Bbb{R}$$

Now add the fact that $t=\frac{x}{3}$ and thus $x=3t$ and here you go:

$$\left\{\begin{array}{cc}x=3t \\ y=-5(t+1) \\ z=2t+3\end{array}\right.,\space t\in\Bbb{R}$$

Dolma
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