## SolutionsEdit

*This exercise is recommended for all readers.*

- Problem 1

Use Gauss' method to find the unique solution for each system.

- Answer

Gauss' method can be performed in different ways, so these simply exhibit one possible way to get the answer.

- Gauss' method
- Gauss' method here

*This exercise is recommended for all readers.*

- Problem 2

Use Gauss' method to solve each system or conclude "many solutions" or "no solutions".

- Answer

- Gaussian reduction
- Gauss' method
- Row reduction
- Row reduction
- Gauss' method
- Here Gauss' method gives

*This exercise is recommended for all readers.*

- Problem 3

There are methods for solving linear systems other than Gauss' method. One often taught in high school is to solve one of the equations for a variable, then substitute the resulting expression into other equations. That step is repeated until there is an equation with only one variable. From that, the first number in the solution is derived, and then back-substitution can be done. This method takes longer than Gauss' method, since it involves more arithmetic operations, and is also more likely to lead to errors. To illustrate how it can lead to wrong conclusions, we will use the system

from Example 1.12.

- Solve the first equation for and substitute that expression into the second equation. Find the resulting .
- Again solve the first equation for , but this time substitute that expression into the third equation. Find this .

What extra step must a user of this method take to avoid erroneously concluding a system has a solution?

- Answer

- From we get that , giving .
- From we get that , leading to the conclusion that .

Users of this method must check any potential solutions by substituting back into all the equations.

*This exercise is recommended for all readers.*

- Problem 4

For which values of are there no solutions, many solutions, or a unique solution to this system?

- Answer

Do the reduction

to conclude this system has no solutions if and if then it has infinitely many solutions. It never has a unique solution.

*This exercise is recommended for all readers.*

- Problem 5

This system is not linear, in some sense,

and yet we can nonetheless apply Gauss' method. Do so. Does the system have a solution?

- Answer

Let , , and :

gives , , and . Note that no satisfies that requirement.

*This exercise is recommended for all readers.*

- Problem 6

What conditions must the constants, the 's, satisfy so that each of these systems has a solution? *Hint.* Apply Gauss' method and see what happens to the right side (Anton 1987).

- Answer

- Gauss' method
- Reduction

- Problem 7

True or false: a system with more unknowns than equations has at least one solution. (As always, to say "true" you must prove it, while to say "false" you must produce a counterexample.)

- Answer

This system with more unknowns than equations

has no solution.

- Problem 8

Must any Chemistry problem like the one that starts this subsection— a balance the reaction problem— have infinitely many solutions?

- Answer

Yes. For example, the fact that the same reaction can be performed in two different flasks shows that twice any solution is another, different, solution (if a physical reaction occurs then there must be at least one nonzero solution).

*This exercise is recommended for all readers.*

- Problem 9

Find the coefficients , , and so that the graph of passes through the points , , and .

- Answer

Because , , and we get a linear system.

Gauss' method

shows that the solution is .

- Problem 10

Gauss' method works by combining the equations in a system to make new equations.

- Can the equation be derived, by a sequence of Gaussian reduction steps, from the equations in this system?
- Can the equation be derived, by a sequence of Gaussian reduction steps, from the equations in this system?
- Can the equation be derived, by a sequence of Gaussian reduction steps, from the equations in the system?

- Answer

- Yes, by inspection the given equation results from .
- No. The given equation is satisfied by the pair . However, that pair does not satisfy the first equation in the system.
- Yes. To see if the given row is , solve the system of equations relating the coefficients of , , , and the constants:

- Problem 11

Prove that, where are real numbers and , if

has the same solution set as

then they are the same equation. What if ?

- Answer

If then the solution set of the first equation is . Taking gives the solution , and since the second equation is supposed to have the same solution set, substituting into it gives that , so . Then taking in gives that , which gives that . Hence they are the same equation.

When the equations can be different and still have the same solution set: e.g., and .

*This exercise is recommended for all readers.*

- Problem 12

Show that if then

has a unique solution.

- Answer

We take three cases: that , that and , and that both and .

For the first, we assume that . Then the reduction

shows that this system has a unique solution if and only if ; remember that so that back substitution yields a unique (observe, by the way, that and play no role in the conclusion that there is a unique solution, although if there is a unique solution then they contribute to its value). But and a fraction is not equal to if and only if its numerator is not equal to . Thus, in this first case, there is a unique solution if and only if .

In the second case, if but , then we swap

to conclude that the system has a unique solution if and only if (we use the case assumption that to get a unique in back substitution). But— where and — the condition "" is equivalent to the condition "". That finishes the second case.

Finally, for the third case, if both and are then the system

might have no solutions (if the second equation is not a multiple of the first) or it might have infinitely many solutions (if the second equation is a multiple of the first then for each satisfying both equations, any pair will do), but it never has a unique solution. Note that and gives that .

*This exercise is recommended for all readers.*

- Problem 13

In the system

each of the equations describes a line in the -plane. By geometrical reasoning, show that there are three possibilities: there is a unique solution, there is no solution, and there are infinitely many solutions.

- Answer

Recall that if a pair of lines share two distinct points then they are the same line. That's because two points determine a line, so these two points determine each of the two lines, and so they are the same line.

Thus the lines can share one point (giving a unique solution), share no points (giving no solutions), or share at least two points (which makes them the same line).

- Problem 14

Finish the proof of Theorem 1.4.

- Answer

For the reduction operation of multiplying by a nonzero real number , we have that satisfies this system

if and only if

by the definition of "satisfies". But, because , that's true if and only if

(this is straightforward cancelling on both sides of the -th equation), which says that solves

as required.

For the pivot operation , we have that satisfies

if and only if

again by the definition of "satisfies". Subtract times the -th equation from the -th equation (remark: here is where is needed; if then the two 's above are not equal) to get that the previous compound statement holds if and only if

which, after cancellation, says that solves

as required.

- Problem 15

Is there a two-unknowns linear system whose solution set is all of ?

- Answer

Yes, this one-equation system:

is satisfied by every .

*This exercise is recommended for all readers.*

- Problem 16

Are any of the operations used in Gauss' method redundant? That is, can any of the operations be synthesized from the others?

- Answer

Yes. This sequence of operations swaps rows and

so the row-swap operation is redundant in the presence of the other two.

- Problem 17

Prove that each operation of Gauss' method is reversible. That is, show that if two systems are related by a row operation then there is a row operation to go back .

- Answer

Swapping rows is reversed by swapping back.

Multiplying both sides of a row by is reversed by dividing by .

Adding times a row to another is reversed by adding times that row.

Remark: observe for the third case that if we were to allow then the result wouldn't hold.

- ? Problem 18

A box holding pennies, nickels and dimes contains thirteen coins with a total value of cents. How many coins of each type are in the box? (Anton 1987)

- Answer

Let , , and be the number of pennies, nickels, and dimes. For variables that are real numbers, this system

has infinitely many solutions. However, it has a limited number of solutions in which , , and are non-negative integers. Running through , ..., shows that is the only sensible solution.

- ? Problem 19

Four positive integers are given. Select any three of the integers, find their arithmetic average, and add this result to the fourth integer. Thus the numbers 29, 23, 21, and 17 are obtained. One of the original integers is:

- 19
- 21
- 23
- 29
- 17

(Salkind 1975, 1955 problem 38)

- Answer

Solving the system

we obtain , , , . Thus the second item, 21, is the correct answer.

*This exercise is recommended for all readers.*

- ? Problem 20

Laugh at this: . It resulted from substituting a code letter for each digit of a simple example in addition, and it is required to identify the letters and prove the solution unique (Ransom & Gupta 1935).

- Answer

*This is how the answer was given in the cited source.*

A comparison of the units and hundreds columns of this addition shows that there must be a carry from the tens column. The tens column then tells us that , so there can be no carry from the units or hundreds columns. The five columns then give the following five equations.

The five linear equations in five unknowns, if solved simultaneously, produce the unique solution: , , , and , so that the original example in addition was .

- ? Problem 21

The Wohascum County Board of Commissioners, which has 20 members, recently had to elect a President. There were three candidates (, , and ); on each ballot the three candidates were to be listed in order of preference, with no abstentions. It was found that 11 members, a majority, preferred over (thus the other 9 preferred over ). Similarly, it was found that 12 members preferred over . Given these results, it was suggested that should withdraw, to enable a runoff election between and . However, protested, and it was then found that 14 members preferred over ! The Board has not yet recovered from the resulting confusion. Given that every possible order of , , appeared on at least one ballot, how many members voted for as their first choice (Gilbert, Krusemeyer & Larson 1993, Problem number 2)?

- Answer

*This is how the answer was given in the cited source.*

Eight commissioners voted for . To see this, we will use the given information to study how many voters chose each order of , , .

The six orders of preference are , , , , , ; assume they receive , , , , , votes respectively. We know that

from the number preferring over , the number preferring over , and the number preferring over . Because 20 votes were cast, we also know that

from the preferences for over , for over , and for over .

The solution is , , , , , and . The number of commissioners voting for as their first choice is therefore .

*Comments.* The answer to this question would have been the same had we known only that *at least* 14 commissioners preferred over .

The seemingly paradoxical nature of the commissioners's preferences ( is preferred to , and is preferred to , and is preferred to ), an example of "non-transitive dominance", is not uncommon when individual choices are pooled.

- ? Problem 22

"This system of linear equations with unknowns," said the Great Mathematician, "has a curious property."

"Good heavens!" said the Poor Nut, "What is it?"

"Note," said the Great Mathematician, "that the constants are in arithmetic progression."

"It's all so clear when you explain it!" said the Poor Nut. "Do you mean like and ?"

"Quite so," said the Great Mathematician, pulling out his bassoon. "Indeed, the system has a unique solution. Can you find it?"

"Good heavens!" cried the Poor Nut, "I am baffled."

Are you? (Dudley, Lebow & Rothman 1963)

- Answer

*This is how the answer was given in the cited source.*

*We have not used "dependent" yet;* it means here that Gauss' method shows that there is not a unique solution. If the system is dependent and the solution is not unique. Hence . But the term "system" implies . Hence . If the equations are

then , .

## ReferencesEdit

- Anton, Howard (1987),
*Elementary Linear Algebra*, John Wiley & Sons. - Dudley, Underwood (proposer); Lebow, Arnold (proposer); Rothman, David (solver) (Jan. 1963), "Elemantary problem 1151",
*American Mathematical Monthly***70**(1): 93. - Gilbert, George T.; Krusemeyer, Mark; Larson, Loren C. (1993),
*The Wohascum County Problem Book*, The Mathematical Association of America. - Ransom, W. R. (proposer); Gupta, Hansraj (solver) (Jan. 1935), "Elementary problem 105",
*American Mathematical Monthly***42**(1): 47. - Salkind, Charles T. (1975),
*Contest Problem Book No 1: Annual High School Mathematics Examinations 1950-1960*.