How to find the number of solutions of an augmented matrix? Given a system of linear equations is represented as a augmented matrix:
$$M = \left[\begin{array}{cccc|c}1&0&8&0&1\\0&1&1&0&3\\0&0&0&1&1\end{array}\right]$$
How do I find the exact number of solutions in the system of equations?
If I'm not mistaken, it has unique solutions, therefore, it shouldn't be inconsistent or have an infinite amount of solutions.
Is it wrong to say that there are 3 solutions?
 A: An underdetermined system usually has infinite solutions.
For this particular problem, say we have:
$w+8y=1$, $x+y=3$ $z=1$
Note that $z$ is identically $1$. The other two equations are in $w, x, y$, which is still under represented. You may recognize two equations in 3 variables as the equation of a line (or a plane if they are not independent, or nothing if they are inconsistent). Just from inspection here we see that it is a line. We may get the equation of this line via algaebric manipulations. It comes out to be $$\frac w8 = \frac{y-\frac18}{-1} = {x-\frac{23}8}$$.
This means that the solution to this system is of the form: 
$$(8k, -k+\frac18, k+\frac{23}8, 1)$$
Hence, there are infinite solutions as we are free to choose any real $k$. 
A: In case you mean to solve the under-determined system $AX=B:$
$$\begin{pmatrix} 1 & 0 & 8 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}
\begin{pmatrix} w \\x \\ y \\ z \\ \end{pmatrix}= \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}.$$ Multiply from let with $A^T$ to get
$$\begin{pmatrix} 1 & 0 & 8 & 0\\ 0 & 1 & 1 &0\\ 8 & 1 & 65 & 0 \\ 0 & 0 & 0& 1 \end{pmatrix} \begin{pmatrix} w \\x \\ y \\ z \\ \end{pmatrix}=\begin{pmatrix} 1 \\ 3 \\ 0 \\ 1 \end{pmatrix}. $$
This simple equation can be solved to give $w=1, x=3, y=0, z=1$.
