Linear Algebra · Book or PDF

A First Course in Linear Algebra

A.G. Hamilton · Study material generated by AI from this edition. Not affiliated with the publisher.

Chapter 2 · pp. 20–31

Solutions to Simultaneous Equations 1

Gaussian elimination (GE) gives a routine way to simplify a set of linear equations. This chapter asks what the finished elimination can tell us. A system can have no solution, exactly one solution, or infinitely many solutions, and elimination shows which of these happens before we work out the answers. We also see how a parameter in the equations can change the outcome.

Infinitely many solutions and zero rows

Sometimes elimination produces a row made only of zeros. That row says $0 = 0$ , so it gives no information. This happens when one equation is a multiple of another. Then only one real equation links the unknowns (p. 20).

With one real equation and two unknowns, you can choose a value for one unknown freely. Call it a parameter, for example $y = t$ , where $t$ is any real number. The remaining equation then fixes the other unknown. In Example 2.1 this gives $y = t$ and $x = 1 - 2t$ . Every real $t$ gives a valid solution, so there are infinitely many (p. 20).

The same idea works whenever there are fewer equations than unknowns. With two equations and three unknowns, you can put $z = t$ and solve the two equations for the other two unknowns (p. 20). A system written with three equations in three unknowns can also reduce to only two useful equations, so the original count of equations can mislead (pp. 20, 22).

Rule of thumb: if there are fewer equations than unknowns, there are infinitely many solutions, provided there are any solutions at all. Apply this rule after elimination, because the original equations may hide the true situation (p. 22).

Inconsistent equations

A set of equations is inconsistent if no values of the unknowns satisfy all of them at once (p. 21).

Sometimes this is obvious. Example 2.4 has two equations in three unknowns that contradict each other. Elimination ends with a row meaning $0x_1 + 0x_2 + 0x_3 = 1$ , which is impossible (p. 21).

Sometimes it is hidden. Example 2.5 has three equations that look harmless. Elimination produces a last row $0\ 0\ 0\ 1$ , which reads back as $0 = 1$ . Even a last row like $0\ 0\ 0\ 3$ shows inconsistency immediately, since it reads $0 = 3$ (pp. 21–22). So elimination detects inconsistency automatically. If this row appears, there are no solutions.

A common misconception

It is wrong to believe that fewer equations than unknowns always means infinitely many solutions. The rule needs the qualification "if there are any solutions." Example 2.4 has fewer equations than unknowns and no solutions at all (p. 22).

Inconsistency can happen with fewer equations than unknowns, with equally many, or with more (p. 22).

More equations than unknowns

When there are more equations than unknowns, we have more than we need. Think of using some equations to find the unknowns and then testing the values in the rest. If we are lucky, the values work. More often they fail, so such a system is likely to be inconsistent, but not always (p. 24).

Example 2.6 has three equations in two unknowns. Elimination shows they are inconsistent (p. 23).
Example 2.7 has three equations in two unknowns: $x_1 - 4x_2 = -1$ , $2x_1 + 2x_2 = 8$ and $5x_1 - x_2 = 14$ . After elimination, the third equation is redundant. The second row gives $x_2 = 1$ , and substituting into the first equation gives $x_1 - 4 = -1$ , so $x_1 = 3$ (p. 23).

The three possible outcomes

Every system of simultaneous linear equations has exactly one of these outcomes (p. 24):

no solution,
a unique solution,
infinitely many solutions.

Elimination tells us which one occurs before we find the solutions. The final array may contain a row $0\ 0\ 0\ 1$ (no solution), a full set of leading 1s (unique solution), or a zero row (infinitely many solutions) (pp. 24–26).

Classifying a system with $p$ unknowns

The array has one more column than there are unknowns, because the last column holds the right-hand sides. So $p$ unknowns means $p + 1$ columns (p. 26).

Reduce the system by GE, then look at the last non-zero row and count the non-zero rows (pp. 25–26):

No solution: the last non-zero row has a 1 at the right-hand end and zeros elsewhere (it reads $0 = 1$ ).
Unique solution: there are exactly $p$ non-zero rows, and the last has a 1 second from the right-hand end.
Infinitely many solutions: there are fewer than $p$ non-zero rows, and the no-solution case does not apply. At least one unknown is then free, and you write it as a parameter, for example $x_3 = t$ (p. 26).

Here are some small arrays of my own, with a bar before the right-hand column:

Two unknowns: rows $1\ 2 \mid 3$ and $0\ 1 \mid -1$ . There are 2 non-zero rows and the last has a 1 second from the right, so the solution is unique.
Two unknowns: rows $1\ 2 \mid 3$ and $0\ 0 \mid 1$ . The last row reads $0 = 1$ , so the system is inconsistent.
Three unknowns: rows $1\ 1\ 2 \mid 4$ , $0\ 1\ 3 \mid 2$ and $0\ 0\ 0 \mid 0$ . There are only 2 non-zero rows, and the last non-zero row is not of the form $0\ 0\ 0\ 1$ . So there are infinitely many solutions.

Equations containing a parameter

When the coefficients contain a parameter such as $c$ , both the solution values and the type of solution can depend on $c$ (p. 28).

Example 2.9. Take $x + y = c$ and $3x - cy = 2$ . Elimination gives a second row with $-c - 3$ in the $y$ position and $2 - 3c$ on the right. To get $y$ we must divide by $-c - 3$ , and that is allowed only when $c \neq -3$ . If $c = -3$ , the equations become $x + y = -3$ and $3x + 3y = 2$ . These are inconsistent, since $3x + 3y$ would have to equal both $-9$ and $2$ . So solutions exist exactly when $c \neq -3$ (pp. 27–28).

Method for any parameter problem (pp. 27–29):

Carry out elimination, keeping $c$ as a symbol.
Note every step that divides by an expression in $c$ . Find the values of $c$ that make it zero.
Treat each of those special values separately by substituting it and re-reading the resulting rows.
For all other values of $c$ , finish the elimination normally.

Example 2.10. Here the special values are $c = 1$ and $c = 2$ (p. 29):

$c = 1$ : a row $0\ 0\ 0\ 3$ appears, so there is no solution.
$c = 2$ : a zero row appears, so there are infinitely many solutions: $x_3 = t$ , $x_2 = t$ , $x_1 = -t$ .
Any other $c$ : there is a unique solution.

Check yourself

Why does a row of zeros appear in elimination, and why does it lead to infinitely many solutions rather than none?
Give an example of two equations in three unknowns that have no solution. How would elimination show this?
A system has 4 unknowns. After elimination there are 3 non-zero rows, and the last one is not of the form $0\ 0\ 0\ 0\ 1$ . What can you conclude, and why?
For $x + y = c$ and $3x - cy = 2$ , why must the case $c = -3$ be treated separately?
Can a system with more equations than unknowns have a solution? Explain using Example 2.7.

A First Course in Linear Algebra

Solutions to Simultaneous Equations 1

Infinitely many solutions and zero rows

Inconsistent equations

A common misconception

More equations than unknowns

The three possible outcomes

Classifying a system with ppp unknowns

Equations containing a parameter

Check yourself

Solutions to Simultaneous Equations 1

Infinitely many solutions and zero rows

Inconsistent equations

A common misconception

More equations than unknowns

The three possible outcomes

Classifying a system with ppp unknowns

Equations containing a parameter

Check yourself

Classifying a system with $p$ unknowns

Classifying a system with $p$ unknowns