16. Echelon examples

Example 1

0.00 Take the system

$x+2y+z=5,$

$-x+y+2z=1,$

$x-z=1.$ The augmented matrix is

$\begin{pmatrix}1&2&1&|&5\\ -1&1&2&|&1\\ 1&0&-1&|&1\end{pmatrix}.$ To clear the first column, we do

$R_{2}\mapsto R_{2}+R_{1}$ and

$R_{3}\mapsto R_{3}-R_{1}$ , which gives:

$\begin{pmatrix}1&2&1&|&5\\ 0&3&3&|&6\\ 0&-2&-2&|&-4\end{pmatrix}.$ 2.50 Now we do

$R_{2}\mapsto\frac{1}{3}R_{2}$ and

$R_{3}\mapsto-\frac{1}{2}R_{3}$ to get:

$\begin{pmatrix}1&2&1&|&5\\ 0&1&1&|&2\\ 0&1&1&|&2\end{pmatrix}.$ Finally, clear column 2 with

$R_{3}\mapsto R_{3}-R_{2}$ and

$R_{1}\mapsto R_{1}-2R_{2}$ , giving:

$\begin{pmatrix}1&0&-1&|&1\\ 0&1&1&|&2\\ 0&0&0&|&0\end{pmatrix}.$ 5.33 This is now in reduced echelon form. It corresponds to the equations

$x-z=1,$

$y+z=2,$

$0=0.$ The third equation is trivially satisfied: it is just a check that our equation really has a solution. The other two can be used to express the dependent variables

$x, y$ in terms of the free variable

$z$ . Note that

$x, y$ are chosen as dependent variables because they are the variables whose coefficients are leading entries.

Example 2

6.50 Take the system

$2x+2y=2,$

$x+y+z=8,$

$x+y-z=-5.$ The augmented matrix is

$\begin{pmatrix}2&2&0&|&2\\ 1&1&1&|&8\\ 1&1&-1&|&-5\end{pmatrix}.$ We do

$R_{1}\mapsto\frac{1}{2}R_{1}$ , yielding:

$\begin{pmatrix}1&1&0&|&1\\ 1&1&1&|&8\\ 1&1&-1&|&-5\end{pmatrix}.$ 8.04 Now we clear column 1 with

$R_{2}\mapsto R_{2}-R_{1}$ and

$R_{3}\mapsto R_{3}-R_{1}$ , which gives:

$\begin{pmatrix}1&1&0&|&1\\ 0&0&1&|&7\\ 0&0&-1&|&-6\end{pmatrix}.$ 9.06 Finally,

$R_{3}\mapsto R_{3}+R_{2}$ yields:

$\begin{pmatrix}1&1&0&|&1\\ 0&0&1&|&7\\ 0&0&0&|&1\end{pmatrix}.$ The left-hand block is now in reduced echelon form.

9.44 Translating this back into equations gives

$x+y=1,\qquad z=7,\qquad 0=1.$ This means there is no solution, because

$0\neq 1$ .

Example 3

11.07 Take the very similar system

$2x+2y=2,$

$x+y+z=8,$

$x+y-z=b.$ Here,

$b$ is a parameter; we'll perform the same analysis and figure out what value

$b$ needs to take in order for the system to have a solution. The augmented matrix is

$\begin{pmatrix}2&2&0&|&2\\ 1&1&1&|&8\\ 1&1&-1&|&b\end{pmatrix}.$ The same sequence of row operations yield

$\begin{pmatrix}1&1&0&|&1\\ 0&0&1&|&7\\ 0&0&0&|&b+6\end{pmatrix}.$ 13.11 We see this only has a solution if

$b=-6$ . In this case, the solution is

$x=1-y$ ,

$z=7$ (

$y$ is free).

13.57 The moral of this is: for an augmented matrix $(M|v)$ with $M$ in reduced echelon form,

any row of zeros in $M$ gives a constraint on the vector $v$ which has to be satisfied if a solution is to exist,
any other row gives you a way of expressing a single dependent variable in terms of free variables.