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Table of Contents
Example
Use EROs to find the intersection of the planes \begin{align*} 3x+4y+7z&=2\\x+3z&=0\\y-2z&=5\end{align*}
Solution 1
\begin{align*} \def\go#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]} \def\ar#1{\\\xrightarrow{#1}&} &\go{3&4&7&2}{1&0&3&0}{0&1&-2&5} \ar{\text{reorder rows}}\go{1&0&3&0}{0&1&-2&5}{3&4&7&2} \ar{R3\to R3-3R1}\go{1&0&3&0}{0&1&-2&5}{0&4&-2&2} \ar{R3\to R3-4R2}\go{1&0&3&0}{0&1&-2&5}{0&0&6&-18} \ar{R3\to \tfrac16 R3}\go{1&0&3&0}{0&1&-2&5}{0&0&1&-3} \end{align*}
$\go{1&0&3&0}{0&1&-2&5}{0&0&1&-3}$
- from the last row, we get $z=-3$
- from the second row, we get $y-2z=5$
- so $y-2(-3)=5$
- so $y=-1$
- from the first row, we get $x+3z=0$
- so $x+3(-3)=0$
- so $x=9$
- Conclusion: $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\-1\\-3\end{bmatrix}$ is the only solution.
Solution 2
We start in the same way, but by performing more EROs we make the algebra at the end simpler.
\begin{align*} \def\go#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]} \def\ar#1{\\\xrightarrow{#1}&} &\go{3&4&7&2}{1&0&3&0}{0&1&-2&5} \ar{\ldots \text{same EROs as above}\ldots}\go{1&0&3&0}{0&1&-2&5}{0&0&1&-3} \ar{R2\to R2+2R3}\go{1&0&3&0}{0&1&0&-1}{0&0&1&-3} \ar{R1\to R1-3R3}\go{1&0&0&9}{0&1&0&-1}{0&0&1&-3} \end{align*}
\[ \go{1&0&0&9}{0&1&0&-1}{0&0&1&-3}\]
- from the last row, we get $z=-3$
- from the second row, we get $y=-1$
- from the first row, we get $x=9$
- So $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\-1\\-3\end{bmatrix}$ is the only solution.
Discussion
Both solutions use EROs to transform the augmented matrix.
- Solution 1: $\left[\begin{smallmatrix}1&0&3&0\\0&1&-2&5\\0&0&1&-3\end{smallmatrix}\right]$.
- “Staircase pattern”: 1s on “steps”, zeros below steps
- Called row echelon form
- Needed algebra to finish solution.
- Solution 2: $\left[\begin{smallmatrix}1&0&0&9\\0&1&0&-1\\0&0&1&-3\end{smallmatrix}\right]$
- Staircase with zeros above 1s on steps (and below).
- Called reduced row echelon form
- No extra algebra needed to finish solution.
Row echelon form and reduced row echelon form
Row echelon form (REF)
A matrix is in row echelon form, or REF, if:
- the zero rows of the matrix (if any) are all at the bottom of the matrix; and
- in every non-zero row of the matrix, the leading entry is $1$; and
- as you go down the rows, the leading entries go to the right.
- $\left[\begin{smallmatrix} 1&2&3&4&5\\0&1&2&3&4\\0&0&1&2&3\end{smallmatrix}\right]$ $\left[\begin{smallmatrix} 1&2&3&4&5\\0&1&2&3&4\\0&0&1&2&3\\0&0&0&0&0\end{smallmatrix}\right]$ $\left[\begin{smallmatrix} 1&2&3&4&5\\0&0&0&0&0\\0&0&1&2&3\end{smallmatrix}\right]$ $\left[\begin{smallmatrix} 1&2&3&4&5\\0&0&0&0&0\\0&0&1&2&3\\0&0&0&0&0\end{smallmatrix}\right]$ $\left[\begin{smallmatrix} 1&2&3&4&5\\0&2&3&4&1\\0&0&1&2&3\end{smallmatrix}\right]$ $\left[\begin{smallmatrix} 0&1&2&3&4\\1&2&3&4&5\\0&0&1&2&3\end{smallmatrix}\right]$ $\left[\begin{smallmatrix} 1&2&3&4&5\\1&5&4&3&2\\0&1&2&3&4\end{smallmatrix}\right]$
- Which are in REF?
Reduced row echelon form (RREF)
A matrix is in reduced row echelon form or RREF if it is in row echelon form (REF) and also has the property:
<html><ol start=“4”><li class=“level1”><div class=“li”></html> If a column contains the leading entry of a row, then every other entry in that column is $0$. <html></div></li></ol></html>
- e.g. $\left[\begin{smallmatrix} {\color{blue}1}&{\color{red}2}&{\color{red}3}&4&5\\0&{\color{blue}1}&{\color{red}2}&3&4\\0&0&{\color{blue}1}&2&3\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} {\color{blue}1}&0&{\color{red}3}&4&5\\0&{\color{blue}1}&0&3&4\\0&0&{\color{blue}1}&2&3\end{smallmatrix}\right]$ are in REF but not RREF
- e.g. $\left[\begin{smallmatrix} {\color{blue}1}&0&0&4&5\\0&{\color{blue}1}&0&3&4\\0&0&{\color{blue}1}&2&3\end{smallmatrix}\right]$ is in RREF.
Example
Use EROs to put $\go{1&2&3&4&5}{0&1&2&3&4}{0&0&1&2&3}$ into RREF. Solve the corresponding linear system.
Solution
\begin{align*} &\go{1&2&3&4&5}{0&1&2&3&4}{0&0&1&2&3} \ar{R2\to R2-2R3}\go{1&2&3&4&5}{0&1&0&-1&-2}{0&0&1&2&3} \ar{R1\to R1-3R3}\go{1&2&0&-2&-4}{0&1&0&-1&-2}{0&0&1&2&3} \ar{R1\to R1-2R2}\go{1&0&0&0&0}{0&1&0&-1&-2}{0&0&1&2&3},\text{ in RREF} \end{align*}
\[ \go{1&0&0&0&0}{0&1&0&-1&-2}{0&0&1&2&3} \]
- Write $x_i$ for the variable corresponding to the $i$th column.
- Column 4 has no leading variable. So we set $x_4=t$, a free parameter ($t\in\mathbb{R}$).
- From row 3: $x_3+2t=3$, so $x_3=3-2t$
- From row 2: $x_2-t=-2$, so $x_2=-2+t$
- From row 1: $x_1=0$
- Solution: $\left[\begin{smallmatrix}x_1\\x_2\\x_3\\x_4\end{smallmatrix}\right]=\left[\begin{smallmatrix}0\\-2\\3\\0\end{smallmatrix}\right]+t\left[\begin{smallmatrix}0\\1\\-2\\1\end{smallmatrix}\right],\quad t\in\mathbb{R}$.
- (Geometrically, this is a line in 4-dimensional space $\mathbb{R}^4$).
Solving a system in REF or RREF
Given an augmented matrix in REF (or RREF), each column except the last column corresponds to a variable. These come in two types:
- leading variables are variables whose column contains the leading entry of some row;
- free variables are all the other variables.
To solve the corresponding linear system:
- assign a free parameter (a letter like $r,s,t,\dots$ representing an arbitrary real number) to each free variable
- starting at the bottom of the matrix, write out each row and rearrange it to give an equation for its leading variable, substituting the other variables as needed.
Example
$\newcommand{\sm}{\left[\begin{smallmatrix}}\newcommand{\esm}{\end{smallmatrix}\right]} \sm 1&2&3&0&0&8\\0&0&1&1&1&5\\0&0&0&1&3&4\esm$ is in REF.
* Using variables $x_1,x_2,x_3,x_4,x_5$:
- $x_1$, $x_3$ and $x_4$ are leading variables, since their columns have a leading entry
- $x_2$ and $x_5$ are free variables, since their columns do not have a leading entry
- $x_2$ and $x_5$ are free, so set $x_2=s$ and $x_5=t$ where $s,t\in \mathbb{R}$
- Working from the bottom:
- $ x_4+3x_5=4\implies x_4=4-3x_5=4-3t$
- $ x_3+x_4+x_5=5\implies x_3=5-x_4-x_5=5-(4-3t)-t=1+2t$
- $ x_1+2x_2+3x_3=8\implies x_1=8-2x_2-3x_3=8-2s-3(1+2t)=5-2s-6t.$
- So $ \left[\begin{smallmatrix} x_1\\x_2\\x_3\\x_4 \\x_5\end{smallmatrix}\right]= \left[\begin{smallmatrix} 5\\0\\1\\4\\0\end{smallmatrix}\right] +s\left[\begin{smallmatrix} -2\\1\\0\\0\\0\end{smallmatrix}\right] +t\left[\begin{smallmatrix} -6\\0\\2\\-3\\1\end{smallmatrix}\right],\quad s,t\in \mathbb{R}.$
- (The solution set is a $2$-dimensional subset of $5$-dimensional space).