Understanding Gauss-Jordan elimination

Question

I have a following system:

$$x_1 + x_2 - x_3 = 5$$ $$2x_1 + 2x_2 - 4x_3 = 6$$ $$x_1 + x_2 - 2x_3 = 3$$

I dont understand how to solve this system using Gauss-Jordan elimination. I was told it I had to have a final result of something like this:

let $x$ represents any other number

$$\begin{bmatrix}1&x&x&x&x\\0&1&x&x&x\\0&0&1&x&x\end{bmatrix}$$

but I had seen some solutions like this:

$$\begin{bmatrix}1&x&x&x&x\\0&0&1&x&x\\0&0&0&1&x\end{bmatrix}$$

If someone can explain clearly how I can go about solving this, that would be great!

I have done some steps at the moment but had stopped because I was not sure. This is what I have before me:

$$\begin{bmatrix}1&1&-1&5\\0&0&1&2\\0&0&-1&-2\end{bmatrix}$$

Thanks!

Answer 1

Hints:

• $R_3 = 2R_3 - R_2$, that is, it is an all zeros row
• Swap $R_1$ with $R_2$
• $R_2 = R_2 - (1/2)R_1$
• $R_2 = R_2/6$

Can you see the final form and tell your solution set from this Gaussian Elimination?

Answer 2

Yes, very good. You can now add the second row to the third giving $$\pmatrix{1&1&-1&{\it 5} \\ 0&0&1&{\it 2} \\ 0&0&0&{\it 0} }$$ Now rewrite it to equations: $$\begin{align} x_1+x_2-x_3&=5 \\ x_3 &=2 \end{align}$$ This will give you all the solutions:

$x_3$ must be $2$, and then by the other equation,
$x_1+x_2=7$ in arbitrary way.

So, say, $x_1\in\Bbb R$ can be arbitrary and then $x_2=7-x_1$.