Reduced Row Echelon Forms

Reducing a matrix $A$ to $rref(A)$ goes through the steps of elimination (unveiling the (in-) dependency of rows and columns from previous one’s) and makes it easy to solve $Ax=0$. Reducing a $A$ into $rref$ we first factor the matrix into LDU, leaving behind an upper triangular $U$ which we must subtract in reverse order again to clear all numbers above the pivots.

$$U=\left[\begin{array}{ccc}1& \boxed{2}& 1\\ 0& 1& 2\\ 0& 0& 0\end{array}\right]\to U=\left[\begin{array}{ccc}1& \boxed{0}& -3\\ 0& 1& 2\\ 0& 0& 0\end{array}\right]=rref(U)$$

(Subtracted 2 times row 2 from row 1)

Solving $Ax=0$

We identify the columns of the $rref$ that don’t have a pivot (in $U$ this is only col 3) as free columns. All solutions forming the Nullspace of the matrix are found through determining the missing fixed column’s multipliers inside of mock solutions, that are made of those and our free columns. We take the example of $U$ in which col 3 is free. The solution is found by finding the fixed coefficients a and b while switching our free columns to = 1, with all other fixed columns being 0 for each mock-solution, respectively.

$$x=\left[\begin{array}{c}ab1\end{array}\right]$$

In a matrix with two (non-pivot) free columns, two solutions would arise. We take a 4x4 matrix with col 2 and col 4 being free as an example, then the two solutions are

$$x\_1=\left[\begin{array}{c}a1c0\end{array}\right],x\_2=\left[\begin{array}{c}a0c1\end{array}\right]$$

with $a,c$ found.