# Solve the system for $x_1, x_2, x_3$ when $\lambda = 1$

$$\begin{array}{rcrcrcl} 2x_1 & - & x_2 & & & = & \lambda x_1\\ 2x_1 & - & x_2 & + & x_3 & = & \lambda x_2\\ -2x_1 & + & 2x_2 & + & x_3 & = & \lambda x_3\end{array}$$

So when $\lambda = 1$ we have

$$\begin{array}{1} \,\,\,\,2x_1 - \,\,x_2 \quad\quad\,\, = x_1\\ \,\,\,\, 2x_1 - \,\,x_2 + x_3 \,= x_2\\ -2x_1 + 2x_2 + x_3 = x_3 \end{array}$$

So then I brought over the right-hand side to the left-handside.

$$\begin{array}{1} \,\,\,\,\,\,x_1 - \,\,x_2 \quad\quad\,\,\,\, = 0\\ \,\,\,\, 2x_1 - 2x_2 + x_3 \,= 0\\ -2x_1 + 2x_2 + \quad\,\,\, = 0 \end{array}$$

So now I reduced it go get the following augmented matrix:

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

From here I'm stuck because the answer in the back of the book says that
$x_1 = x_2 = -\frac12 s$, $x_3 = s$

I thought it would have been more like:
$x_1 = x_2 = s$, $x_3 = 0$

What am I doing wrong?

• Why do you use the array environment in MathJax if you're not going to use alignment tabs? – Michael Hardy Jun 19 '17 at 14:10
• Well I have only 111 points, I'm sure there is a lot that I'm doing wrong here, apart from my maths. So are you saying I should be using the format that you corrected me with? @MichaelHardy – Bucephalus Jun 19 '17 at 14:38
• Are you referring to the {rcrcrcl} that you have put in? I will have to go and investigate this. @MichaelHardy – Bucephalus Jun 19 '17 at 14:40
• Ok, thanks @Michael. I have looked these arrays now, so right, centre and left justification. That would have come in handy instead of me putting all those \,\, in the array. Cheers – Bucephalus Jun 19 '17 at 14:44

## 2 Answers

Your solution is correct, the books is not.

If you plug in your solution to the original equation, you get

$2s-s=s\\ 2s-s+0=s\\ -2s+2s+0=0$

and all three equations are correct.

On the other hand, if you plug in the book's solution, the first equation becomes

$$2\cdot(-\frac12)s -(-\frac12 s) = s$$ which, already, is clearly not true since it is equivalent to $$\frac32 s = s$$ which is only ever true for $s=0$.

• Do you mean "If you plug in your solution to the original equation, you get" , not "the books solution", @5xum – Bucephalus Jun 19 '17 at 13:32
• @Bucephalus Yeah, sorry. I edited my answer – 5xum Jun 19 '17 at 13:33
• oh nice, thanks. – Bucephalus Jun 19 '17 at 13:34

Obviously, the solution from the back of your book is not correct. Take $s=2$. From the solution from the back of your book, you get that a solution of the system (when $\lambda=1$) is $x_1=x_2=-1$ and $x_3=2$. But then $2x_1-x_2+x_3=1\neq2=x_3$.

• Where did the very last $x_3$ come from @JoseCarlosSantos? – Bucephalus Jun 19 '17 at 13:34
• @Bucephalus I don't understand your question. I got that $2x_1-x_2+x_3=1$. But if the solution from your book was correct, then I should have obtained $x_3$, which happens to be equal to $2$. – José Carlos Santos Jun 19 '17 at 13:36
• Ok thanks @Jose – Bucephalus Jun 19 '17 at 13:38