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Table of Contents
Example 1
Solve the linear system \begin{align*}3x+4y-2z&=1\\x+y+z&=4\\2x+5y+z&=3\end{align*} by transforming the augmented matrix into reduced row echelon form.
\begin{align*} \def\go#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]} \def\ar#1{\\\xrightarrow{#1}&} \go{3&4&-2&1}{1&1&1&4}{2&5&1&3} \xrightarrow{R1\leftrightarrow R2}& \go{1&1&1&4}{3&4&-2&1}{2&5&1&3} \ar{R2\to R2-3R1\text{ and }R3\to R3-2R1} \go{1&1&1&4}{0&1&-5&-11}{0&3&-1&-5} \ar{R3\to R3-3R1} \go{1&1&1&4}{0&1&-5&-11}{0&0&14&28} \ar{R3\to \tfrac1{14}R3} \go{1&1&1&4}{0&1&-5&-11}{0&0&1&2} \end{align*}
- in REF, keep going for RREF…
\begin{align*} \go{1&1&1&4}{0&1&-5&-11}{0&0&1&2} \xrightarrow{R1\to R1-R3\text{ and }R2\to R2+5R3}& \go{1&1&0&2}{0&1&0&-1}{0&0&1&2} \ar{R1\to R1-R2} \go{1&0&0&3}{0&1&0&-1}{0&0&1&2} \end{align*}
- $x=3$, $y=-1$, $z=2$
- There is a unique solution: $(3,-1,2)$.
- No free variables.
Example 2
Solve the linear system \begin{align*}3x+y-2z+4w&=5\\x+z+w&=2\\4x+2y-6z+6w&=0\end{align*}
\begin{align*} \go{3&1&-2&4&5}{1&0&1&1&2}{4&2&-6&6&0} \xrightarrow{R1\leftrightarrow R2}& \go{1&0&1&1&2}{3&1&-2&4&5}{4&2&-6&6&0} \ar{R2\to R2-3R1\text{ and }R3\to R3-4R1} \go{1&0&1&1&2}{0&1&-5&1&-1}{0&2&-10&2&-8} \ar{R3\to R3-2R2} \go{1&0&1&1&2}{0&1&-5&1&-1}{0&0&0&0&-6} \ar{R3\to -\tfrac16 R3} \go{1&0&1&1&2}{0&1&-5&1&-1}{0&0&0&0&1} \end{align*}
\[ \go{1&0&1&1&2}{0&1&-5&1&-1}{0&0&0&0&1}\]
- This is in REF.
- The last row corresponds to the equation $0=1$
- This has no solution!
- So the original linear system has no solutions.
- Call the system inconsistent (no solutions).
- To detect this: put in REF and find a row $[0~0~\dots~0~1]$.
Example 3
For which value(s) of $k$ does the following linear system have infinitely many solutions?
\begin{gather*}x+y+z=1\\x-z=5\\2x+3y+kz=-2\end{gather*}
\begin{align*} \go{1&1&1&1}{1&0&-1&5}{2&3&k&-2} \xrightarrow{R2\to R2-R1\text{ and }R3\to R3-2R1}& \go{1&1&1&1}{0&-1&-2&4}{0&1&k-2&-4} \ar{R2\to -R2} \go{1&1&1&1}{0&1&2&-4}{0&1&k-2&-4} \ar{R3\to R3-R2} \go{1&1&1&1}{0&1&2&-4}{0&0&k-4&0} \end{align*}
\[\go{1&1&1&1}{0&1&2&-4}{0&0&k-4&0}\]
- If $k=4$:
- get $\go{1&1&1&1}{0&1&2&-4}{0&0&0&0}$, in REF
- $z$ free, so infinitely many solutions
- If $k\ne4$ then $k-4\ne0$.
- $R3\to \tfrac1{k-4}R3$: $\go{1&1&1&1}{0&1&2&-4}{0&0&1&0}$, in REF
- no free vars, so not infinitely many solutions
- So infinitely many solutions $\iff k=4$.
Observations
For a system of linear equations with<html><br /></html> #vars variables, #eqs equations:
- If #vars > #eqs, at least one var is free (see REF!)
- either system is inconsistent….
- ….or it has infinitely many solutions, one for each value of the free vars.
- The dimension of the set of solutions is the number of free variables.
- If there's a unique solution:<html><br /></html>always have #vars ≤ #eqs