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Table of Contents
Examples
Example 1
Solve the linear system \begin{align*}2x+4y-2z&=1\\x+y+z&=4\\2x+5y+z&=3\end{align*} by transforming the augmented matrix into reduced row echelon form.
Solution
\begin{align*} \def\go#1#2#3{\begin{bmatrix}#1\\#2\\#3\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*} The solution is $x=3$, $y=-1$, $z=2$. So there is a unique solution: just one point in $\mathbb{R}^3$, namely $(3,-1,2)$.
Example 2
Solve the linear system \begin{align*}3x+y-2z+4w&=5\\x+z+w&=2\\4x+2y-6z+6w&=0\end{align*}
Solution
\begin{align*} &\go{3&1&-2&4&5}{1&0&1&1&2}{4&2&-6&6&0} \ar{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*}
This is in REF. The last row corresponds to the equation \[ 0=1\] which clearly has no solution! We conclude that this system has no solutions, and hence the original linear system has no solutions either.
A linear system with no solutions is called inconsistent.
We can detect an inconsistent linear system, since whenever we apply EROs to put the augmented matrix into REF, we will get a row of the form $[0~0~0~\dots~0~*]$ where $*$ is non-zero.
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*}
Solution
\begin{align*} &\go{1&1&1&1}{1&0&-1&5}{2&3&k&-2} \ar{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*}
If $k-4=0$, then this matrix is in REF: \[\go{1&1&1&1}{0&1&2&-4}{0&0&0&0}\] In this situation, $z$ is a free variable (since there's no leading entry in the third column). For each value of $z$ we get a different solution, so if $k-4=0$, or equivalently, if $k=4$, then there are infinitely many different solutions.
If $k-4\ne0$, then we can divide the third row by $k-4$ to get the REF: \[\go{1&1&1&1}{0&1&2&-4}{0&0&1&0}\] In this situtation, there are no free variables since $x$, $y$ and $z$ are all leading variables. So if $k-4\ne 0$, or equivalently if $k\ne 4$, then there are no free variables so there is not an infinite number of solutions. (The only possibilities are that there is is a unique solution or that the system is inconsistent; and in this case you can check that there is a unique solution, although we don't need to know this to answer the question).
In conclusion, the system has infinitely many solutions if and only if $k=4$.
One more example
Solve the following linear system: \begin{align*} x+y+z&=2\\2x-z&=0\\x+3y+4z&=6\\3x+y&=2\end{align*}
Solution
We remark that there are more equations than unknowns… but this isn't a problem! We proceed as usual:
\begin{align*} \def\go#1#2#3#4{\begin{bmatrix}#1\\#2\\#3\\#4\end{bmatrix}} \def\ar#1{\\[6pt]\xrightarrow{#1}&} &\go{1&1&1&2}{2&0&-1&0}{1&3&4&6}{3&1&0&2} \ar{R2\to R2-2R1,\ R3\to R3-R1\text{ and }R4\to R4-3R1} \go{1&1&1&2}{0&-2&-3&-4}{0&2&3&4}{0&-2&-3&-4} \ar{R3\to R3+R2\text{ and }R4\to R4-R2} \go{1&1&1&2}{0&1&1.5&2}{0&0&0&0}{0&0&0&0} \ar{R1\to R1-R2} \go{1&0&-0.5&0}{0&1&1.5&2}{0&0&0&0}{0&0&0&0} \end{align*}
Now $z$ is a free variable, say $z=t$ where $t\in \mathbb{R}$, and from row 2 we get $y=2-1.5t$ and row 1 gives $x=0.5t$, so the solution is \[ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\2\\0\end{bmatrix}+t\begin{bmatrix}0.5\\-1.5\\1\end{bmatrix},\quad t\in\mathbb{R}.\]
Observations about Gaussian elimination
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.