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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{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}} \def\ar#1{\\[6pt]\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*}
So
- 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 $x=9$
The conclusion is that \[ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\-1\\-3\end{bmatrix}\] 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{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}} \def\ar#1{\\[6pt]\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} \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*}
So
- from the last row, we get $z=-3$
- from the second row, we get $y=-1$
- from the first row, we get $x=9$
The conclusion is again that \[ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\-1\\-3\end{bmatrix}\] is the only solution.
Discussion
In both of these solutions we used EROs to transform the augmented matrix into a nice form.
- In solution 1, we ended up with the matrix $\left[\begin{smallmatrix}1&0&3&0\\0&1&-2&5\\0&0&1&-3\end{smallmatrix}\right]$ which has a staircase pattern, with zeros below the staircase, and 1s just above the “steps” of the staircase. This is an example of a matrix in row echelon form (see below). We needed a bit of easy algebra, called back substitution, to finish off the solution. (Why is it called echelon form? It seems that this word has an archaic meaning which is relevant to the staircase-like pattern: “any structure or group of structures arranged in a steplike form.”)
- In solution 2, we ended up with the matrix $\left[\begin{smallmatrix}1&0&0&9\\0&1&0&-1\\0&0&1&-3\end{smallmatrix}\right]$ which has a staircase pattern with zeros below the staircase and 1s just above the “steps” of the staircase, and the additional property that we only have zeros above the 1s on the steps. This is an example of a matrix in reduced row echelon form (see below). Finding the solution from this matrix needed no extra algebra.
Row echelon form and reduced row echelon form
Row echelon form (REF)
Definition
A row of a matrix is a zero row if it contains only zeros. For example, $[0\ 0\ 0\ 0\ 0]$ is a zero row.
A row of a matrix is non-zero, or a non-zero row if contains at least one entry that is not $0$. For example $[0\ 0\ 3\ 0\ 0]$ is non-zero, and so is $[1\ 2\ 3\ 4\ -5]$.
Definition
The leading entry of a non-zero row of a matrix is the leftmost entry which is not $0$.
For example, the leading entry of the row $[0~0~0~6~2~0~3~1~0]$ is $6$.
Definition
A matrix is in row echelon form, or REF, if it has all of the following three properties:
- The zero rows of the matrix (if any) are all at the bottom of the matrix.
- In every non-zero row of the matrix, the leading entry is $1$.
- If row $i$ and row $(i+1)$ are both non-zero, then the leading entry in row $(i+1)$ is to the right of the leading entry in row $i$. <html><br /></html>In other words, as you go down the rows, the leading entries must go to the right.
For example, $\left[\begin{smallmatrix} 1&2&3&4&5\\0&1&2&3&4\\0&0&1&2&3\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 1&2&3&4&5\\0&1&2&3&4\\0&0&1&2&3\\0&0&0&0&0\end{smallmatrix}\right]$ are both in REF, but
- $\left[\begin{smallmatrix} 1&2&3&4&5\\0&0&0&0&0\\0&0&1&2&3\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 1&2&3&4&5\\0&0&0&0&0\\0&0&1&2&3\\0&0&0&0&0\end{smallmatrix}\right]$ are not in REF, since they each have a zero row which isn't at the bottom;
- $\left[\begin{smallmatrix} 1&2&3&4&5\\0&2&3&4&1\\0&0&1&2&3\end{smallmatrix}\right]$ is not in REF, since the leading entry on the second row isn't $1$;
- $\left[\begin{smallmatrix} 0&1&2&3&4\\1&2&3&4&5\\0&0&1&2&3\end{smallmatrix}\right]$ is not in REF, since the leading entry in row $2$ is not to the right of the leading entry in row $1$.
Reduced row echelon form (RREF)
Definition
A matrix is in reduced row echelon form or RREF if it is in row echelon form (REF), so that
- The zero rows of the matrix (if any) are all at the bottom of the matrix.
- In every non-zero row of the matrix, the leading entry is $1$.
- If row $i$ and row $(i+1)$ are both non-zero, then the leading entry in row $(i+1)$ is to the right of the leading entry in row $i$. <html><br /></html>In other words, as you go down the rows, the leading entries must go to the right.
and the matrix 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>
For example, \[\begin{bmatrix} {\color{blue}1}&{\color{red}2}&{\color{red}3}&4&5\\0&{\color{blue}1}&{\color{red}2}&3&4\\0&0&{\color{blue}1}&2&3\end{bmatrix}\quad\text{and}\quad \begin{bmatrix} {\color{blue}1}&0&{\color{red}3}&4&5\\0&{\color{blue}1}&0&3&4\\0&0&{\color{blue}1}&2&3\end{bmatrix}\] are both in REF, but they are not in RREF because the red entries are non-zero and are in the same column as a leading entry (in blue).
On the other hand, \[\begin{bmatrix} {\color{blue}1}&0&0&4&5\\0&{\color{blue}1}&0&3&4\\0&0&{\color{blue}1}&2&3\end{bmatrix}\] is in RREF.
Example
Use EROs to put the following matrix into RREF: \[\begin{bmatrix} 1&2&3&4&5\\0&1&2&3&4\\0&0&1&2&3\end{bmatrix}\] and solve the corresponding linear system.
Solution
\begin{align*} \def\go#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}} \def\ar#1{\\[6pt]\xrightarrow{#1}&} &\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} \end{align*} This matrix is in RREF. Write $x_i$ for the variable corresponding to the $i$th column. The solution is
- $x_4=t$, a free parameter, i.e. $t\in\mathbb{R}$. This is because the $4$th column does not contain a leading entry.
- 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: $x_1=0$
So the solution is \[ \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}0\\-2\\3\\0\end{bmatrix}+ t\begin{bmatrix}0\\1\\-2\\1\end{bmatrix},\quad t\in\mathbb{R}.\]
(Geometrically, this is a line in 4-dimensional space $\mathbb{R}^4$).