TITLE: Ax=b problem solving
QUESTION [5 upvotes]: Can you please tell how to solve this:
Ax=b
$$A=\begin{pmatrix}
 2 & 4   & -8 &  4 & 1\\
 4 & 10  &-16 &  8 &-8\\
-4 & -12 & 17 & -8 & 20\\
-2 & -10 & 10 & -3 & 34  \end{pmatrix}$$
$$b=\begin{pmatrix}-33\\-36\\4\\-62\end{pmatrix}$$
I have to give a fundamental and
particular solutions.
I have tried so:

But I do not know whether it is right and what to do next ...
Please, help me! Thanks.

REPLY [3 votes]: $$\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\ 4&10&-16&8&-8&-36\\ -4&-12&17&-8&20&4\\-2&-10&10&-3&34&-62\end{array}\right)\to\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\0&2&0&0&-10&30\\ 0&-2&1&0&12&-32\\0&-6&2&1&35&-95\end{array}\right)\to$$
$$\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\0&1&0&0&-5&15\\ 0&0&1&0&2&-2\\0&0&2&1&5&-5\end{array}\right)\to\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\0&1&0&0&-5&15\\ 0&0&1&0&2&-2\\0&0&0&1&1&-9\end{array}\right)\to$$
$$\left(\begin{array}{ccccc|c}2&0&0&0&33&-73\\0&1&0&0&-5&15\\ 0&0&1&0&2&-2\\0&0&0&1&1&-9\end{array}\right)$$ So your general solution is $(-\frac{73}{2},15,-2,-9,0)+x(-\frac{33}{2},5,-2,-1,1)$. Choosing a particular value of $x$ (e.g., $x=0$) you will get a particular solution.