# Thread: Solution to matrix with row of zeros!

1. ## Solution to matrix with row of zeros!

I have reduced a matrix to the following:

$\displaystyle \left[ \begin{array}{rrr|r} -1&1&0&0 \\ 1&-1&0&0 \\ 0&0&0&0 \end{array} \right]$

Obviously $\displaystyle x_1 = x_2$.
How do I figure out what $\displaystyle x_3$ is here?
I did this a few years ago but am lost right now... I have the answer for this problem but I don't understand how to reach it myself.

Cheers to anybody interested!

2. ## Re: Solution to matrix with row of zeros!

Originally Posted by howdigethere
Obviously $\displaystyle x_1 = x_2$ How do I figure out what $\displaystyle x_3$ is here?
Right, $\displaystyle x_1=x_2$ and there are no more conditions so, $\displaystyle x_3$ can take any real value. The solutions of the system are $\displaystyle x_1=\lambda,\;x_2=\lambda,\;x_3=\mu$ with $\displaystyle \lambda,\mu \in \mathbb{R} .$

3. ## Re: Solution to matrix with row of zeros!

Thanks!

I am actually finding eigenvectors here and I guess my question should have been: Why would I state the eigenvectors as:
$\displaystyle \lambda _1 = \lambda _2 = \begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}$
as opposed to any other vectors?

I clearly don't understand eigenvectors...

4. ## Re: Solution to matrix with row of zeros!

What FernandoRevilla told you was that the general solution to your original problem is of the form $\displaystyle \left<\lambda, \lambda, \mu\left>$. That can be written as $\displaystyle \lambda\left<1, 1, 0\right>+ \mu\left<0, 0, 1\right>$. The two eigenvectors are $\displaystyle \left<1, 1, 0\right>$ and $\displaystyle \left<0, 0, 1\right>$. A three by three matrix may have up to three independent eigenvectors- whether that is true in this case depends upon the original matrix. What did you get for the eigenvalues?

What are you given as the definition of "eigenvector"?

5. ## Re: Solution to matrix with row of zeros!

Find eigenvalues and eigenvectors of $\displaystyle \begin{bmatrix} 2&1&0\\ 1&2&0 \\ 0&0&3 \end{bmatrix}$

To which I found $\displaystyle \lambda _1 = \lambda _2 = 3, \lambda _3 = 1$.

Then, for $\displaystyle (\lambda _1 = \lambda _2 = 3), \begin{bmatrix} 2&1&0\\ 1&2&0 \\ 0&0&3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \rightarrow \left[ \begin{array}{rrr|r} -1&1&0&0 \\ 1&-1&0&0 \\ 0&0&0&0 \end{array}\right] \rightarrow x_1=x_2, \ x_3 = \ ?$

So, and I am trying to remember here (!), we describe the vector x in terms of parameters. Which I should have done as
$\displaystyle x_1=x_2=s, \ x_3 = t \rightarrow \langle s, s, t \rangle \rightarrow s \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix} + t \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ (is this right ?!)

6. ## Re: Solution to matrix with row of zeros!

That's right.

Note that the vector x represents an eigenvector (for eigenvalue 3).

So any eigenvector (with eigenvalue 3) can be expressed as a linear combination of (1,1,0) and (0,0,1) and in particular two linearly independent eigenvectors will be (1,1,0) and (0,0,1) (and any scalar multiplies of these two vectors!)