Matrix Linear Systems Equation

[zack7]

Homework Statement

Question in the image.

Homework 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

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

 

[Mark44]

Homework Statement

Question in the image.

Homework 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

The Attempt at a Solution

I managed to get it to $$\vec{x}(A-1)=0

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.)

= \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

 

[zack7]

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.)

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

 

[Mark44]

but if I factorize the x out I get (A-1), how would I simplfy that since it is not possible to subtract?

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.

 

[zack7]

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.

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

 

[Mark44]

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

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

 

[zack7]

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

then why is the answer $$\vec{x} = c\begin{pmatrix} 1 \\ 0\end{pmatrix}$$
$$\vec {x} = x1\begin{pmatrix} 1 \\ 0\end{pmatrix}$$

 

[Mark44]

Fixed your LaTeX. You need to put the [noparse]$$[/noparse] thing at the end of your expression, too.

then why is the answer $$\vec{x} = c\begin{pmatrix} 1 \\ 0\end{pmatrix}$$isn't it then $$\vec{x} = x1\begin{pmatrix} 1 \\ 0\end{pmatrix}$$

Those are pretty much the same thing. In the first, c is an arbitrary number; in the second, x₁ is an arbitrary number.