Problem from MIT OCW

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I found this problem on the MIT OpenCourseWare website, but a solution was not given. I tried it out, so I was wondering if my answer is correct. The problem is as follows:

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Suppose A is reduced by the usual row operations to

[tex]R=\begin{bmatrix}1 & 4 & 0 & 2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0\end{bmatrix}.[/tex]

Find the complete solution (if a solution exists) to this system involving the original A:

Ax = sum of the columns of A.

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I figured if you took the columns of A and added them together, it would be the same as multiplying a by <1,1,1>. Thus I have the following:

[tex]A\mathbf{x}=A\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}.[/tex]

So the solution is the vector <1,1,1>. Is this correct?

Thank you.
 
no,the solution is a general solution not a single solution or no solution. depending on what R is R=A or R=A|b(augmented matrix) the solution is
R=A ( id oubt its this one)
x1+4*x2+2*x4=0;
x3+2*x4 = 0

R=A|b
x1+4*x2=2
x3 =2
therefore the solution is ->
x=
|2-4t|
|..t..|
|..2..|

you arelooking at the rows.
[]if there was no solution a row would have all 0s except the last one=#(!=0)
thus 0=# which is false
[]if there was 1 solution than all the xi would be solvable...that is there must be a few rows where its just xi=# like the 2nd row. THen the remaining rows would be solved by the other linear equations in the matrix.
[]if there was many solutions than you'd have hte above.
 
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So, for R=A:

[tex]x_1+4x_2+2x_3=0[/tex]

[tex]x_3+2x_4=0[/tex]

For which the general solution is:

[tex]x_2\begin{bmatrix}-4 \\ 1 \\ 0 \\ 0\end{bmatrix}+x_4\begin{bmatrix}4 \\ 0 \\ -2 \\ 1\end{bmatrix}\quad\forall x_2,x_4\in\mathbb{R}[/tex]

I understand how to find the general solution and everything, but what I would like to know is why you set the equations equal to zero. If R=A, then the sum of the columns of A should equal the sum of the columns of R, which is non-zero. Could you just explain your steps when you set R=A and R=[A|b]?

Thanks again.
 
if the system is Ax=b; then R=[A|b] where the last column of R is b and each remaining column of R represents a xi; I believe that is the solution

however because you did not state the question the solution could be R=A where you did not state b...and so i assumed b=O
 
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Alright I understand what you are trying to say.

[tex]A\mathbf{x}=A\mathbf{u}\quad \mathbf{u}=\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}[/tex]

[tex]A(\mathbf{x}-\mathbf{u})=0[/tex]

So upon solving, I come up with the general solution which is:

[tex]x_2\begin{bmatrix}-4 \\ 1 \\ 0 \\ 0\end{bmatrix}+x_4\begin{bmatrix}-2 \\ 0 \\ -2 \\ 1\end{bmatrix}+\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}\quad\forall x_2,x_4\in\mathbb{R}[/tex]

If I didn't make any arithmetic mistakes.
 

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