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

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?

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  • $\begingroup$ Why do you use the array environment in MathJax if you're not going to use alignment tabs? $\endgroup$ – Michael Hardy Jun 19 '17 at 14:10
  • $\begingroup$ 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 $\endgroup$ – Bucephalus Jun 19 '17 at 14:38
  • $\begingroup$ Are you referring to the {rcrcrcl} that you have put in? I will have to go and investigate this. @MichaelHardy $\endgroup$ – Bucephalus Jun 19 '17 at 14:40
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    $\begingroup$ 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 $\endgroup$ – Bucephalus Jun 19 '17 at 14:44
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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 $s=0$.

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  • $\begingroup$ Do you mean "If you plug in your solution to the original equation, you get" , not "the books solution", @5xum $\endgroup$ – Bucephalus Jun 19 '17 at 13:32
  • $\begingroup$ @Bucephalus Yeah, sorry. I edited my answer $\endgroup$ – 5xum Jun 19 '17 at 13:33
  • $\begingroup$ oh nice, thanks. $\endgroup$ – Bucephalus Jun 19 '17 at 13:34
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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 $2x_1-x_2+x_3=1\neq2=x_3$.

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  • $\begingroup$ Where did the very last $x_3$ come from @JoseCarlosSantos? $\endgroup$ – Bucephalus Jun 19 '17 at 13:34
  • $\begingroup$ @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 $2$. $\endgroup$ – José Carlos Santos Jun 19 '17 at 13:36
  • $\begingroup$ Ok thanks @Jose $\endgroup$ – Bucephalus Jun 19 '17 at 13:38

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