I'm trying to find the general solution to this matrix \begin{bmatrix}1&-2&1&3&0\\2&-4&4&6&4\\ -2&4&-1&-6&2\\1&-2&-3&3&-8\end{bmatrix}
Ax=$\begin{bmatrix}1&6&0&-7&\end{bmatrix}^T$
I think I'm supposed to get it in x=x*+z format, I'm still not sure if this is the correct way to do it. But I ended up getting this matrix in row echelon form. -2(r1)+(r2)
2(r1)+(r3)
-(r1)+(r4) \begin{bmatrix}1&-2&1&3&0\\0&0&2&0&4\\ 0&0&1&0&2\\0&0&-4&0&-8\end{bmatrix}
And then
2(r2)+(r4)
-1/2(r2)+(r3)
\begin{bmatrix}1&-2&1&3&0\\0&0&2&0&4\\ 0&0&0&0&0\\0&0&0&0&0\end{bmatrix}
then
1/2(r2)
\begin{bmatrix}1&-2&1&3&0\\0&0&1&0&1\\ 0&0&0&0&0\\0&0&0&0&0\end{bmatrix}
and lastly -r2 + r2
\begin{bmatrix}1&-2&0&3&-2\\0&0&1&0&1\\ 0&0&0&0&0\\0&0&0&0&0\end{bmatrix}
after doing some algebra i ended up getting
x1 = 2(x2)-3(x4)-2(x5)
x3 = -2(x5)
and set x5 = x4 = x2 = 1
and got z = $\begin{bmatrix}1&1&-2&1&1\end{bmatrix}^T$
But I'm when i try to solve for Ax = $\begin{bmatrix}1&6&0&-7\end{bmatrix}^T$
The last two rows are full of zeros
so I can't have 0=-7
How would i solve this?
You have to work with the augmented matrix, and do the same operations on the rows of this matrix (including the supplementary column). Here is how it goes:
\begin{align} &\begin{bmatrix}\begin{array}{@{}rrrrr|r@{}} 1&-2&1&3&0&1\\2&-4&4&6&4&6\\-2&4&-1&-6&2&0\\1&-2&-3&3&-8&-7 \end{array}\end{bmatrix}\rightsquigarrow \begin{bmatrix}\begin{array}{@{}rrrrr|r@{}} 1&-2&1&3&0&1\\0&0&2&0&4&4\\0&0&1&0&2&2\\0&0&-4&0&-8&-8 \end{array}\end{bmatrix}\rightsquigarrow\\ &\begin{bmatrix}\begin{array}{@{}rrrrr|r@{}} 1&-2&1&3&0&1\\0&0&1&0&2&2\\0&0&0&0&0&0\\0&0&0&0&0&0 \end{array}\end{bmatrix}\rightsquigarrow\begin{bmatrix}\begin{array}{@{}rrrrr|r@{}} 1&-2&0&3&-2&-1\\0&0&1&0&2&2\\0&0&0&0&0&0\\0&0&0&0&0&0 \end{array}\end{bmatrix} \end{align}
so the solutions are given by the equations \begin{cases} x_1=2x_2-3x_4-2x_5-1,\\ x_3=-2x_5+2. \end{cases}