The following matrix is the augmented matrix of a system of linear equations in the variables x, y, and z. (It is given in reduced row-echelon form.)

(a) It can be seen from the following matrix in row echelon form that the diagonal term corresponding to variable z is zero. Terms corresponding to x and y are equal to 1.
$\left[\begin{array}{cccc}1& 0& -1& 3\\ 0& 1& 2& 5\\ 0& 0& 0& 0\end{array}\right]$
Hence, the leading variables are x and y.
(b) Because the variable z is a non-leading variable. This makes the system of linear equations represented by the given augmented matrix a dependent system.
(c) After the reduced row echelon form, the corresponding system of equations needs to be written and solved using back substitution:
$\text{System}\{\begin{array}{l}x-z=3\\ y+2z=5\\ 0=0\end{array}$
Since we know that z is a non leading variable we need to use it as a parameter t. So assuming z=t. we have from first equation:
$x-z=3$
$x-t=3$
$x=t+3$
From second equation: $y+2z=5$
$y+2t=5$
$y=5-2t$
Result:
(a) The leading variables are x and y
(b) Dependent system
(c) $x=t+3,y=5-2t,z=t$