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I'm trying hard with this exercise, but it is breaking my back.

Find a basis for the solution set of the given homogeneous linear system

$3x_1+x_2+x_3=0$
$6x_1+2x_2+2x_3=0$
$-9x_1-3x_2-3x_3=0$

I do what I know I need to do. First I get the solution set of the system by reducing like this:

$\begin{pmatrix} 3 & 1 & 1 \\ 6 & 2 & 2 \\ -9 & -3 & -3 \end{pmatrix} \leadsto \begin{pmatrix} 3 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \leadsto\begin{pmatrix} 1 & 1/3 & 1/3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

So I know $\vec x = \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix} 1-\frac{1}{3}r-\frac{1}{3}s\\ r\\ s\end{bmatrix}$

That being the general solution.

Now, giving the values for $r$ and $s$ according to the standard vectors $i$, $j$

$\vec x = \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix} 1-\frac{1}{3}r-\frac{1}{3}s\\ r\\ s\end{bmatrix} = r \begin{bmatrix} \frac{2}{3}\\ 1\\ 0\end{bmatrix} + s\begin{bmatrix} \frac{2}{3}\\ 0\\ 1\end{bmatrix}$

From my results, the basis will be:

$ ( \begin{bmatrix} \frac{2}{3}\\ 1\\ 0\end{bmatrix}, \begin{bmatrix} \frac{2}{3}\\ 0\\ 1\end{bmatrix})$

But instead, the book answer (I'm self-studying )is:

$ ( \begin{bmatrix} -1\\ 3\\ 0\end{bmatrix}, \begin{bmatrix} -1\\ 0\\ 3\end{bmatrix})$

Any idea on what I'm doing wrong? Thank you :)

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  • $\begingroup$ You're basis isn't correct as you can't factor out the r that easily I think. The other thought I had was that you may be missing a sign change. $\endgroup$ – JB King Feb 6 '13 at 21:35
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Check your expression for $\vec x$. I think it should be $\vec x=\begin{bmatrix} x_1 \\x_2\\x_3\end{bmatrix}=\begin{bmatrix} -\frac{1}{3}r-\frac{1}{3}s \\r \\s \end{bmatrix}=r\begin{bmatrix} -\frac{1}{3} \\1 \\0 \end{bmatrix}+s\begin{bmatrix} -\frac{1}{3} \\0 \\1 \end{bmatrix}$

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  • $\begingroup$ Thanks! I don't know how I came to put a one right there :( $\endgroup$ – Susana Feb 6 '13 at 21:44
  • $\begingroup$ This still has a sign issue. Regards $\endgroup$ – Amzoti Feb 6 '13 at 21:46
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    $\begingroup$ I just made a one character edit: to add a negative sign before $\frac 13$ in the last vector. I hope you do not mind. $\endgroup$ – Namaste Feb 6 '13 at 21:50
  • $\begingroup$ You are absolutely correct, thanks @amWhy! $\endgroup$ – Zilliput Feb 6 '13 at 22:26
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    $\begingroup$ @user124384 I'm not sure in what sense you are using the term 'equivalent,' but, in general, one can scale the elements of any basis by any nonzero scalar to obtain a different basis. $\endgroup$ – Zilliput Oct 7 '15 at 15:44
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I don't follow this step.

$\vec x = \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix} 1-\frac{1}{3}r-\frac{1}{3}s\\ r\\ s\end{bmatrix} = r \begin{bmatrix} \frac{2}{3}\\ 1\\ 0\end{bmatrix} + s\begin{bmatrix} \frac{2}{3}\\ 0\\ 1\end{bmatrix}$

Wouldn't you want to write this as:

$\vec x = \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix} 1-\frac{1}{3}r-\frac{1}{3}s\\ r\\ s\end{bmatrix} = r \begin{bmatrix} \frac{-1}{3}\\ 1\\ 0\end{bmatrix} + s\begin{bmatrix} \frac{-1}{3}\\ 0\\ 1\end{bmatrix}$

If you write out the eigenvalues, you get $\lambda = 1, 0 , 0$

The corresponding eigenvectors are:

$(1, 0, 0), (-1/3, 0, 1), (-1/3, 1, 0)$ and these should look familiar to your solution.

Regards

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  • $\begingroup$ I understand, completely...been there...felt that way too...But, we can't resist helping, regardless ;-) $\endgroup$ – Namaste May 4 '13 at 0:35

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