Solving for x in a 3x3 Gaussian Elimination Problem

[Saladsamurai]

Homework Statement

Examine the solutions of

$$\left[\begin{array}{c}x_1-x_2+x_3=c\\2x_1-3x_2+4x_3=0\\3x_1-4x_2+5x_3=1\end{array}\right]$$

$$\text{when }c=1\text{ and }c\ne1$$

The Attempt at a Solution

$$\left[\begin{array}{cccc}1& -1& 1& c\\2& -3& 4& 0\\3& -4& 5& 1\end{array}\right]$$

$$\left[\begin{array}{cccc}1& -1& 1& c\\0& -1& 2& -2c\\0& -1& 2& -2c\end{array}\right]$$

$$\left[\begin{array}{cccc}1& -1& 1& c\\0& 1& -2& 2c\\0& 0& 0& 0\end{array}\right]$$


What is the next step here?

 

[Pengwuino]

What this step is telling you is that x3 is arbitrary. Continue elimination if you want, but you can construct your solution from this. For example, the 2nd row shows that x2 = 2c +2x3.

 

[snipez90]

The entry at the far bottom right of the second matrix should be 1 - 3c.

 

[Pengwuino]

oops yah I didn't notice that. it is 1-3c which changes the situation...

 

[Saladsamurai]

Doh!

edit: Hey! PF changed my DOH from all Caps!

 

[Saladsamurai]

$$\left[\begin{array}{cccc}1& -1& 1& c\\0& -1& 2& -2c\\0& -1& 2& 1-2c\end{array}\right]$$

$$\left[\begin{array}{cccc}1& -1& 1& c\\0& 1& -2& 2c\\0& 0& 0& 1\end{array}\right]$$

Now... have I made another silly error?

Is the elimination finished?

 

[Pengwuino]

No, you made another mistake. You're subtracting 3 times the first row so you should have 1-3c as the bottom right coefficient, not 1-2c.

 

[Saladsamurai]

Ohh jeesh...you both already said that...last stage becomes:

$$<br /> \left[\begin{array}{cccc}1& -1& 1& c\\0& 1& -2& 2c\\0& 0& 0& (1-c)\end{array}\right]<br />$$

Now the question makes way more sense.

c=1-->infinite solutions
c$\ne$1-->inconsistent eqs

Thanks people! I should go to bed... but I think I can squeeze in one more stupid question tonight!