Ax=b problem solving 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.
 A: The first thing to notice is that you have four equations in five variables -- this means you will have an infinite number of solutions or zero solutions. In the case of an infinite number, you can solve for the set of solutions in terms of one  parameter.
The easiest way to do this by hand is to reduce the augmented matrix $\left( A  |  b\right)$ to reduced row-echelon form. This means proceed with Gaussian elimination until it reaches the form
$$
\left( \begin{array}{cccccc} 
  1 && 0 && 0 && 0 && \star && \star\\
  0 && 1 && 0 && 0 && \star && \star \\
  0 && 0 && 1 && 0 && \star && \star\\
  0 && 0 && 0 && 1 && \star && \star
\end{array}
\right)
$$
where $\star$ is some unknown value. 
For your matrix, this results in 
$$
\left( \begin{array}{cccccc} 
  1 && 0 && 0 && 0 && 16.5 && -52.5\\
  0 && 1 && 0 && 0 && -5 && 15\\
  0 && 0 && 1 && 0 && 2 && -2\\
  0 && 0 && 0 && 1 && 1 && -1
\end{array}
\right)
$$
Thus, we see that 
$$ 
\begin{align*}
x_1 + 16.5x_5 &= -52.5
\\ x_2 - 5 x_5 &= 15 
\\ x_3 + 2 x_5 &= -2
\\ x_4 + x_5 &= -1
\end{align*}
$$
which I assume is the "general" solution. For a particular solution you can just pick any $x_5$ and solve for the remaining variables.
A: $$\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\ 4&10&-16&8&-8&-36\\ -4&-12&17&-8&20&4\\-2&-10&10&-3&34&-62\end{array}\right)\to\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\0&2&0&0&-10&30\\ 0&-2&1&0&12&-32\\0&-6&2&1&35&-95\end{array}\right)\to$$
$$\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\0&1&0&0&-5&15\\ 0&0&1&0&2&-2\\0&0&2&1&5&-5\end{array}\right)\to\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\0&1&0&0&-5&15\\ 0&0&1&0&2&-2\\0&0&0&1&1&-9\end{array}\right)\to$$
$$\left(\begin{array}{ccccc|c}2&0&0&0&33&-73\\0&1&0&0&-5&15\\ 0&0&1&0&2&-2\\0&0&0&1&1&-9\end{array}\right)$$ 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.
