16. Echelon examples

16. Echelon examples

Example 1

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: $\begin{pmatrix}1&2&1&|&5\\ 0&3&3&|&6\\ 0&-2&-2&|&-4\end{pmatrix}.$ 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: $\begin{pmatrix}1&0&-1&|&1\\ 0&1&1&|&2\\ 0&0&0&|&0\end{pmatrix}.$ 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

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: $\begin{pmatrix}1&1&0&|&1\\ 1&1&1&|&8\\ 1&1&-1&|&-5\end{pmatrix}.$ Now we clear column 1 with $R_{2}\mapsto R_{2}-R_{1}$ and $R_{3}\mapsto R_{3}-R_{1}$ , which gives: $\begin{pmatrix}1&1&0&|&1\\ 0&0&1&|&7\\ 0&0&-1&|&-6\end{pmatrix}.$ 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.

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

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 $\begin{pmatrix}1&1&0&|&1\\ 0&0&1&|&7\\ 0&0&0&|&b+6\end{pmatrix}.$ We see this only has a solution if $b=-6$ . In this case, the solution is $x=1-y$ , $z=7$ ($y$ is free).

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.