Examples

Back substitution - convert REF to RREF

Back substitution: convert REF to RREF

$$\text{REF}\left(\begin{array}{rccc}0& 2& 3& 4\\ 0& 0& 1& 0\\ 0& 0& 0& 5\\ 0& 0& 0& 0\end{array}\right)\sim \left(\begin{array}{rccc}0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{array}\right)\text{RREF}$$ $$\text{REF}\left(\begin{array}{rrrrr}2& 4& -1& 5& -2\\ 0& 3& 1& 2& -3\\ 0& 0& 0& 2& 1\\ 0& 0& 0& 0& 5\end{array}\right)\sim \left(\begin{array}{rrrrr}1& 2& -1/2& 5/2& -1\\ 0& 1& 1/3& 2/3& -1\\ 0& 0& 0& 1& 1/2\\ 0& 0& 0& 0& 1\end{array}\right)\sim \left(\begin{array}{rrrrr}1& 0& -7/6& 0& 0\\ 0& 1& 1/3& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right)\text{RREF}$$

Inverting a vector using RREF

Inverting a vector using RREF

Consider the equation $A\mathbf{x}=\mathbf{b}$, with a row-reduced $A$:

$$\left(\begin{array}{rccc}1& 0& 0& -3\\ 0& 0& 1& 2\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right)=\left(\begin{array}{c}{b}_1\\ b_2\\ b_3\\ b_4\end{array}\right).$$

This matrix translates to the system of equations:

$$\begin{array}{rl}{x}_1-3x_4& =b_1,x_1=+3x_4+b_1\\ x_3+2x_4& =b_2,x_3=-2x_4+b_2\\ 0& =b_3\\ 0& =b_4\end{array}$$

There is a solution vector $\mathbf{x}$ provided $b_3=b_4=0$. If that is true, any $x_2$ and $x_4$ can be chosen, and $x_1,x_3$ are quickly found by solving the system. Given any choice of $x_2$ and $x_4$, there is only one possibility for each of $x_1,x_3$. Notice that the leading ones (also called pivots) are in columns $1$ and $3$, and $x_1$ and $x_3$ are determined. There are no pivots in columns $2$ and $4$, and $x_2$ and $x_4$ are free to take any value.

We can go farther and write the complete set of solutions $\mathbf{x}$ as the span of two vectors plus an offset, using values of $x_2$ and $x_4$ as the coefficients:

$$\mathbf{x}=\left(\begin{array}{c}3x_4+b_1\\ x_2\\ -2x_4+b_2\\ x_4\end{array}\right)=x_2\left(\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right)+x_4\left(\begin{array}{c}3\\ 0\\ -2\\ 1\end{array}\right)+\left(\begin{array}{c}{b}_1\\ 0\\ b_2\\ 0\end{array}\right)$$

so we could write, if you like: $\mathbf{x}\in ⟨\left(\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}3\\ 0\\ -2\\ 1\end{array}\right)⟩+\left(\begin{array}{c}{b}_1\\ 0\\ b_2\\ 0\end{array}\right)$.

Inverting a vector - row reduction of augmented matrix

Inverting a vector: row reduction of augmented matrix

Suppose we are trying to solve $A\mathbf{x}=\mathbf{b}$ with $A$ and $\mathbf{b}$ given by the following data:

$$A=\left(\begin{array}{rrrrr}1& 6& 2& -5& -2\\ 0& 0& 2& -8& -1\\ 0& 0& 0& 0& 1\end{array}\right),\mathbf{b}=\left(\begin{array}{r}-4\\ 3\\ 7\end{array}\right).$$

Notice that $A$ is already in row-echelon form. So we create the augmented matrix and row-reduce it further to put $\left(\begin{array}{c}A\mid \mathbf{b}\end{array}\right)$ into RREF:

$$\begin{array}{rl}\left(\begin{array}{c}A\mid \mathbf{b}\end{array}\right)& =\left(\begin{array}{rrrrrr}1& 6& 2& -5& -2& -4\\ 0& 0& 2& -8& -1& 3\\ 0& 0& 0& 0& 1& 7\end{array}\right)\\ & \sim \left(\begin{array}{rrrrrr}1& 6& 0& 3& -1& -7\\ 0& 0& 2& -8& -1& 3\\ 0& 0& 0& 0& 1& 7\end{array}\right)\\ & \sim \left(\begin{array}{rrrrrr}1& 6& 0& 3& 0& 0\\ 0& 0& 1& -4& 0& 5\\ 0& 0& 0& 0& 1& 7\end{array}\right)\end{array}$$

and we discover that $x_2,x_4$ are free, $x_1=-6x_2-3x_4$ and $x_3=4x_4+5$ and $x_5=7$. The possible solutions for $\mathbf{x}$ are therefore:

$$\mathbf{x}=x_2\left(\begin{array}{c}-6\\ 1\\ 0\\ 0\\ 0\end{array}\right)+x_4\left(\begin{array}{c}-3\\ 0\\ 4\\ 1\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 7\end{array}\right),x_2,x_4\in \mathbb{R}.$$

Inverse of a 3x3 matrix

Inverse of a $3\times 3$ matrix

Problem: Find the inverse $A^{-1}$ of the matrix $A=\left(\begin{array}{rrr}0& 1& 2\\ 1& 0& 3\\ 4& -3& 8\end{array}\right)$.

Solution: We row reduce the augmented matrix using pivots, eliminations, and back substitutions to put $A$ into RREF:

$$\begin{array}{rl}\left(\begin{array}{rrrrrr}0& 1& 2& 1& 0& 0\\ 1& 0& 3& 0& 1& 0\\ 4& -3& 8& 0& 0& 1\end{array}\right)& \sim \left(\begin{array}{rrrrrr}1& 0& 3& 0& 1& 0\\ 0& 1& 2& 1& 0& 0\\ 4& -3& 8& 0& 0& 1\end{array}\right)\\ \sim \left(\begin{array}{rrrrrr}1& 0& 3& 0& 1& 0\\ 0& 1& 2& 1& 0& 0\\ 0& -3& -4& 0& -4& 1\end{array}\right)& \sim \left(\begin{array}{rrrrrr}1& 0& 3& 0& 1& 0\\ 0& 1& 2& 1& 0& 0\\ 0& 0& 2& 3& -4& 1\end{array}\right)\\ \sim \left(\begin{array}{rrrrrr}1& 0& 3& 0& 1& 0\\ 0& 1& 2& 1& 0& 0\\ 0& 0& 1& 3/2& -2& 1/2\end{array}\right)& \sim \left(\begin{array}{rrrrrr}1& 0& 0& -9/2& 7& -3/2\\ 0& 1& 0& -2& 4& -1\\ 0& 0& 1& 3/2& -2& 1/2\end{array}\right).\end{array}$$

Therefore we have found $A^{-1}=\left(\begin{array}{rrr}-9/2& 7& -3/2\\ -2& 4& -1\\ 3/2& -2& 1/2\end{array}\right)$.