How do you find free variables in a matrix?

Hint: A variable is a basic variable if it corresponds to a pivot column. Otherwise, the variable is called a free variable. To identify free variables, you have to convert the augmented matrix to echelon form by applying the method of row reduction.

Complete step by step solution:
Take such a system of linear equations in which the number of variables is more than the number of equations.
$x + y + 3w = 4\\x + y + z + 5w = 9$
The corresponding augmented matrix is $A|b = \begin{bmatrix}1&1&0&3&|&4\\0&0&1&2&|&5\end{bmatrix}$
Let us apply the row operation on $A$ to find its row echelon form.
Subtract the elements of the second row from the first row.
The matrix becomes $\begin{bmatrix}1&1&0&3&|&4\\0&0&1&2&|&5\end{bmatrix}$
It is the row echelon form of the matrix $A$.
So, $B = \begin{bmatrix}1&1&0&3&|&4\\0&0&1&2&|&5\end{bmatrix}$
The elements ${b_{1,1}}$ (at the point of intersection of the first row and the first column) and ${b_{2,3}}$ (at the point of intersection of the second row and the third column) are the pivots. So, the variables $x$ and $z$ are the basic variables.
The total number of variables in the given system is $4$ and the number of pivots is $2$.
So, the number of free variables is $4 - 2 = 2$ and they are $y$ and $w$.
This is the procedure of finding the free variables.

Note: If $A$ be a matrix of order $m \times n$. So, the order of its augmented matrix is $m \times \left( {n + 1} \right)$ and let $B$ be the row reduced echelon form of $A$. Then there are $m$ linear equations in $n$ variables and if $r$ be the number of pivots in $B$, then