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Table of Contents
One more example
Solve the fosllowing linear system: \begin{align*} x+y+z&=2\\2x-z&=0\\x+3y+4z&=6\\3x+y&=2\end{align*}
\begin{align*} \def\go#1#2#3#4{\begin{bmatrix}#1\\#2\\#3\\#4\end{bmatrix}} \def\ar#1{\\[6pt]\xrightarrow{#1}&} &\go{3&4&-2&1}{1&1&1&4}{2&5&1&3} \ar{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} \ar{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*}
Solution
We remark that there are more equations than unknowns… but this isn't a problem! We proceed as usual:
We know that we can apply EROs to any augmented matrix into REF.
Suppose the system has $n$ equations and $m$ variables, and let $k$ be the number of non-zero rows in REF. Also suppose the system is consistent: then the REF has no row of the form $[0~0~0~\dots~1]$.
- $k\le n$, because there are only $n$ rows in the whole matrix
- $k$ is precisely the number of leading variables. So $k$ is no bigger $m$, the total number of variables; in symbols, we have $k\le m$.
- All the other variables are free variables, so $$ \text{$m-k$ is the number of free variables.} $$
What does this tell us about the set of solutions? For example, how many solutions are there?
Observation 1: free variables and the number of solutions
For consistent systems, this shows that:
- either $k=m$;
- so $m-k=0$
- there are no free variables
- the system has one solution and no more
- We say it has a unique solution.
- or $k<m$
- so $m-k>0$
- there is at least one free variable
- so the system has infinitely many solutions (one for each value of each free variable)
- The number of free variables, $m-k$, is called the dimension of the solution set.
Observation 2: systems with fewer equations than variables
For consistent systems where $n<m$ (fewer equations than variables):
- $k\le n < m$, so $k<m$.
- So there is at least one free variable.
- So in this situation we always have infinitely many solutions.