MHB Find an A and b such that Ax=b has the solution set......

[TomSavage]

the solution set for the problem is

x1=s+1

x2=t-2

x3=2s+2t

x4=s-t+1I was thinking that I would have to isolate all the variables to one side and create a matrix and then get all the integers to another side and multiply them in order to get b but that doesn't seem correct to me. Can anyone show me/tell me how to complete this and more importantly why it is completed the way that they are showing, thanks.

 

[Olinguito]

Hi TomSavage.

By inspecting the equations, the easiest thing to do first would be to express $x_3$ and $x_4$ in terms of $x_1$ and $x_2$:
$$x_3=2s+2t=2(s+1)+2(t-2)+2=2x_1+2x_2+2 \\ x_4=s-t+1=(s+1)-(t-2)-2=x_1-x_2-2.$$
So
$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} x_1 \\ x_2 \\ 2x_1+2x_2+2 \\ x_1-x_2-2 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 2 & 2 & 0 & 0 \\ 1 & -1 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}+\begin{pmatrix} 0 \\ 0 \\ 2 \\ -2 \end{pmatrix}$$
Hence you can take
$$\mathbf A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}-\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 2 & 2 & 0 & 0 \\ 1 & -1 & 0 & 0 \end{pmatrix}=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -2 & -2 & 1 & 0 \\ -1 & 1 & 0 & 1 \end{pmatrix}$$
and
$$\mathbf b=\begin{pmatrix} 0 \\ 0 \\ 2 \\ -2 \end{pmatrix}.$$

 

[topsquark]

$$\mathbf A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}-\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 2 & 2 & 0 & 0 \\ 1 & -1 & 0 & 0 \end{pmatrix}=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -2 & -2 & 1 & 0 \\ -1 & 1 & 0 & 1 \end{pmatrix}$$

There's something here I'm not following. Why are you doing the line above? I don't understand how that's supposed to work.

Thanks!

-Dan

 

[Olinguito]

In

$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} x_1 \\ x_2 \\ 2x_1+2x_2+2 \\ x_1-x_2-2 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 2 & 2 & 0 & 0 \\ 1 & -1 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}+\begin{pmatrix} 0 \\ 0 \\ 2 \\ -2 \end{pmatrix}$$

the LHS is
$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}.$$

 

[topsquark]

In

the LHS is
$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}.$$

So obvious! Thanks for explaining that. :)

-Dan