Matrix Linear Systems Equation

1. Mar 12, 2012

zack7

1. The problem statement, all variables and given/known data
Question in the image.

2. Relevant equations
I got a but can seem to get the answer for b which is $$\vec{x} = c\begin{pmatrix} 1 \\ 0\end{pmatrix}$$

where for any scalar c
3. The attempt at a solution
I managed to get it to $$\vec{x}(A-1)=0$$

$$= \begin{pmatrix} 0 3 \\ -1 0\end{pmatrix}=\begin{pmatrix} 1 0 \\ 0 1\end{pmatrix}$$
Then I reduced the matrix to reduced row echolon form
but after that I seem not to get the answer.

Thank you

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Last edited by a moderator: Mar 12, 2012
2. Mar 12, 2012

Staff: Mentor

You made a mistake in how you calculated A - I. (It's not A - 1. IOW, you don't subtract 1 from each entry of A.)

3. Mar 12, 2012

zack7

I get (A-I),
which would then be
$$\begin{pmatrix} 0 3 \\ 0 0\end{pmatrix}$$
but this would mean x1 and x2 = 0

Last edited by a moderator: Mar 12, 2012
4. Mar 12, 2012

Staff: Mentor

No, you don't get A - 1. Here's the way it goes:

Ax = x
=> Ax - x = 0
=> (A - I)x = 0

In the last equation, that is A - I, the 2 x 2 identity matrix, NOT the matrix whose entries are all 1's.

Calculate A - I, and row reduce it to find the solution set.

5. Mar 12, 2012

zack7

Thanks, I understand that part,careless mistake but when I reduce it I get
x1 and x2 = 0

6. Mar 12, 2012

Staff: Mentor

No. The top row says 0x1 + 3x2 = 0, so x1 is arbitrary.

7. Mar 12, 2012

zack7

then why is the answer $$\vec{x} = c\begin{pmatrix} 1 \\ 0\end{pmatrix}$$

$$\vec {x} = x1\begin{pmatrix} 1 \\ 0\end{pmatrix}$$

Last edited by a moderator: Mar 12, 2012
8. Mar 12, 2012

Staff: Mentor

Fixed your LaTeX. You need to put the [noparse][/noparse] thing at the end of your expression, too.
Those are pretty much the same thing. In the first, c is an arbitrary number; in the second, x1 is an arbitrary number.