Find a 3x1 matrix x, with entries not all zero such that Ax = x

**Lepzed** · September 18th 2011, 03:05 AM

Given the matrix $\begin{pmatrix} 1 & 2 & -1\\ 1 & 0 & 1 \\ 4 & -4 & 5 \end{pmatrix}$ , I've to find a vector x, such that Ax = x.

I'm not quite sure what's the best way of solving a question like this, I can try some values that could be correct, doing some half educated guesses, but that just seems clumsy and slow. Any tips?

**FernandoRevilla** · September 18th 2011, 03:37 AM

If $x=(x_1,x_2,x_3)^t$ then,

$Ax=x \Leftrightarrow \ldots \Leftrightarrow\begin{Bmatrix}-2x_2-x_3=0\\ x_1-x_2+x_3=0\\ 4x_1-4x_2+4x_3=0\end{matrix}$

Solving the linear system, we obtain all the solutions. For example, $x=(3,1,-2)$ is a non zero solution.

**Lepzed** · September 18th 2011, 04:49 AM

I think I still don't understand right, if Ax=x, then if x = (3, 1, -2), then:

$3\begin{pmatrix} 1 \\ 1 \\ 4 \end{pmatrix} + \begin{pmatrix} 2 \\ 0 \\ -4 \end{pmatrix}-2\begin{pmatrix} -1 \\ 1 \\ 5 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \\ -2 \end{pmatrix} \neq \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$

What am I missing here? Why do you let the equations be 0? According to my textbook the answer (a answer?) should be $\begin{pmatrix} -\frac{1}{2}r \\ \frac{1}{2}r \\ r \end{pmatrix}$ . I see how this is a solution:

$-\frac{1}{2}\begin{pmatrix} 1 \\ 1 \\ 4 \end{pmatrix} + \frac{1}{2}\begin{pmatrix} 2 \\ 0 \\ -4 \end{pmatrix}+1\begin{pmatrix} -1 \\ 1 \\ 5 \end{pmatrix} =\begin{pmatrix} -\frac{1}{2} \\ \frac{1}{2} \\ 1 \end{pmatrix}$

I know how to solve a system of equations, but I don't get this...

**FernandoRevilla** · September 18th 2011, 04:59 AM

Originally Posted by Lepzed

What am I missing here?

$a_{12}=-2$ , and you have written $2$ .

**Lepzed** · September 18th 2011, 05:15 AM

Ah, that was a typo, it should indeed be '2', not '-2', my apologies for that. But still, how did you get to your answer? Why did you let the equations be 0?

**FernandoRevilla** · September 18th 2011, 07:08 AM

With your reformulation

and a non null solution is $x=(-1,1,2)^t$ .

**Lepzed** · September 18th 2011, 12:13 PM

Ah, so basically you subtracted an x_row from each row, to make the equation 0 and than you just fill in the blanks?

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Find a 3x1 matrix x, with entries not all zero such that Ax = x

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Re: Find a 3x1 matrix x, with entries not all zero such that Ax = x

Re: Find a 3x1 matrix x, with entries not all zero such that Ax = x

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