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
gives that the solution is and .
Gauss' method here
gives , , and .
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
shows that and is the unique solution.
Gauss' method
gives and as the only solution.
Row reduction
shows, because the variable is not a leading variable in any
row, that there are many solutions.
Row reduction
shows that there is no solution.
Gauss' method
gives the unique solution .
Here Gauss' method gives
which shows that there are many solutions.
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
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
shows that this system is consistent if and only if both
and .
Reduction
shows that each of , , and can be any
real number— this system always has a unique solution.
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:
and get and , so the given row is
.
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).
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:
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."
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
![{\displaystyle {\xrightarrow[{}]{-(1/2)\rho _{1}+\rho _{2}}}\;{\begin{array}{*{2}{rc}r}2x&+&3y&=&13\\&-&(5/2)y&=&-15/2\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7a1cf6aec95a72b133b5a1ae9450e58a7b133c3)
![{\displaystyle {\xrightarrow[{\rho _{1}+\rho _{3}}]{-3\rho _{1}+\rho _{2}}}\;{\begin{array}{*{3}{rc}r}x&&&-&z&=&0\\&&y&+&3z&=&1\\&&y&&&=&4\end{array}}\;{\xrightarrow[{}]{-\rho _{2}+\rho _{3}}}\;{\begin{array}{*{3}{rc}r}x&&&-&z&=&0\\&&y&+&3z&=&1\\&&&&-3z&=&3\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ad004ddbc55fed7e3cc38713ce4ee8ee1a3d515)
![{\displaystyle {\begin{array}{rcl}&{\xrightarrow[{}]{-(1/2)\rho _{1}+\rho _{2}}}&{\begin{array}{*{2}{rc}r}2x&+&2y&=&5\\&&-5y&=&-5/2\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ab08b410e384f7474ada0218a67b4769177b0f3)
![{\displaystyle {\begin{array}{rcl}&{\xrightarrow[{}]{\rho _{1}+\rho _{2}}}&{\begin{array}{*{2}{rc}r}-x&+&y&=&1\\&&2y&=&3\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0b7a037a236cfad48649292791bce2647f34a4b)
![{\displaystyle {\begin{array}{rcl}&{\xrightarrow[{}]{-\rho _{1}+\rho _{2}}}&{\begin{array}{*{3}{rc}r}x&-&3y&+&z&=&1\\&&4y&+&z&=&13\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05ebb527083d24a4b67007b34baaface97216d38)
![{\displaystyle {\begin{array}{rcl}&{\xrightarrow[{}]{-3\rho _{1}+\rho _{2}}}&{\begin{array}{*{2}{rc}r}-x&-&y&=&1\\&&0&=&-1\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd501438ffd8b2073e53ae0725a937a36137d055)
![{\displaystyle {\xrightarrow[{}]{\rho _{1}\leftrightarrow \rho _{4}}}\;{\begin{array}{*{3}{rc}r}x&+&y&-&z&=&10\\2x&-&2y&+&z&=&0\\x&&&+&z&=&5\\&&4y&+&z&=&20\end{array}}\;{\xrightarrow[{-\rho _{1}+\rho _{3}}]{-2\rho _{1}+\rho _{2}}}\;{\begin{array}{*{3}{rc}r}x&+&y&-&z&=&10\\&&-4y&+&3z&=&-20\\&&-y&+&2z&=&-5\\&&4y&+&z&=&20\end{array}}\;{\xrightarrow[{\rho _{2}+\rho _{4}}]{-(1/4)\rho _{2}+\rho _{3}}}\;{\begin{array}{*{3}{rc}r}x&+&y&-&z&=&10\\&&-4y&+&3z&=&-20\\&&&&(5/4)z&=&0\\&&&&4z&=&0\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7e778d7fdcaabc5c3ce103e99c5566f9a31fd4c)
![{\displaystyle {\xrightarrow[{-2\rho _{1}+\rho _{4}}]{-(3/2)\rho _{1}+\rho _{3}}}\;{\begin{array}{*{4}{rc}r}2x&&&+&z&+&w&=&5\\&&y&&&-&w&=&-1\\&&&-&(5/2)z&-&(5/2)w&=&-15/2\\&&y&&&-&w&=&-1\end{array}}\;{\xrightarrow[{}]{-\rho _{2}+\rho _{4}}}\;{\begin{array}{*{4}{rc}r}2x&&&+&z&+&w&=&5\\&&y&&&-&w&=&-1\\&&&-&(5/2)z&-&(5/2)w&=&-15/2\\&&&&&&0&=&0\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f217b95cf41a13e70833dbbcbd67bd0fbce01f4)
![{\displaystyle {\begin{array}{rcl}&{\xrightarrow[{}]{-3\rho _{1}+\rho _{2}}}&{\begin{array}{*{2}{rc}r}x&-&y&=&1\\&&0&=&-3+k\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6544f3842b9f29b2a77958768915b2a212383796)
![{\displaystyle {\begin{array}{rcl}{\begin{array}{*{3}{rc}r}2x&-&y&+&3z&=&3\\4x&+&2y&-&2z&=&10\\6x&-&3y&+&z&=&9\end{array}}&{\xrightarrow[{-3\rho _{1}+\rho _{3}}]{-2\rho _{1}+\rho _{2}}}&{\begin{array}{*{3}{rc}r}2x&-&y&+&3z&=&3\\&&4y&-&8z&=&4\\&&&&-8z&=&0\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d79eed076af9397d9aa7ef34ecacc1eefdb53b8)
![{\displaystyle {\xrightarrow[{\begin{array}{c}\\[-19pt]\scriptstyle -\rho _{1}+\rho _{3}\\[-5pt]\scriptstyle -2\rho _{1}+\rho _{4}\end{array}}]{-3\rho _{1}+\rho _{2}}}\;{\begin{array}{*{2}{rc}r}x&-&3y&=&b_{1}\\&&10y&=&-3b_{1}+b_{2}\\&&10y&=&-b_{1}+b_{3}\\&&10y&=&-2b_{1}+b_{4}\end{array}}\;{\xrightarrow[{-\rho _{2}+\rho _{4}}]{-\rho _{2}+\rho _{3}}}\;{\begin{array}{*{2}{rc}r}x&-&3y&=&b_{1}\\&&10y&=&-3b_{1}+b_{2}\\&&0&=&2b_{1}-b_{2}+b_{3}\\&&0&=&b_{1}-b_{2}+b_{4}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/057e36a9711eba8084655b68620dfc7890a73951)
![{\displaystyle {\xrightarrow[{-\rho _{1}+\rho _{3}}]{-2\rho _{1}+\rho _{2}}}\;{\begin{array}{*{3}{rc}r}x_{1}&+&2x_{2}&+&3x_{3}&=&b_{1}\\&&x_{2}&-&3x_{3}&=&-2b_{1}+b_{2}\\&&-2x_{2}&+&5x_{3}&=&-b_{1}+b_{3}\end{array}}\;{\xrightarrow[{}]{2\rho _{2}+\rho _{3}}}\;{\begin{array}{*{3}{rc}r}x_{1}&+&2x_{2}&+&3x_{3}&=&b_{1}\\&&x_{2}&-&3x_{3}&=&-2b_{1}+b_{2}\\&&&&-x_{3}&=&-5b_{1}++2b_{2}+b_{3}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2651dc53bf2256c5643539a23a7fabb1c822c967)
![{\displaystyle {\begin{array}{rcl}{\xrightarrow[{-4\rho _{1}+\rho _{2}}]{-\rho _{1}+\rho _{2}}}\;{\begin{array}{*{3}{rc}r}a&+&b&+&c&=&2\\&&-2b&&&=&4\\&&-2b&-&3c&=&-5\end{array}}&{\xrightarrow[{}]{-\rho _{2}+\rho _{3}}}&{\begin{array}{*{3}{rc}r}a&+&b&+&c&=&2\\&&-2b&&&=&4\\&&&&-3c&=&-9\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ebdc6ec739ae5ca3d5ae8fbb9ef0538fd0eac380)
![{\displaystyle {\begin{array}{rcl}&{\xrightarrow[{}]{-(c/a)\rho _{1}+\rho _{2}}}&{\begin{array}{*{2}{rc}r}ax&+&by&=&j\\&&(-{\frac {cb}{a}}+d)y&=&-{\frac {cj}{a}}+k\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/098b5834017778f7661140fcfadd2d6e37193d63)
![{\displaystyle {\xrightarrow[{}]{\rho _{i}+\rho _{j}}}\quad {\xrightarrow[{}]{-\rho _{j}+\rho _{i}}}\quad {\xrightarrow[{}]{\rho _{i}+\rho _{j}}}\quad {\xrightarrow[{}]{-1\rho _{i}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c32c72d37700dfff279428b85c3ee6f6c0aef26)
![{\displaystyle {\begin{array}{rcl}{\begin{array}{*{3}{rc}r}a_{1,1}x_{1}&+&\cdots &+&a_{1,n}x_{n}&=&d_{1}\\&&&&&\vdots \\a_{m,1}x_{1}&+&\cdots &+&a_{m,n}x_{n}&=&d_{m}\end{array}}&{\xrightarrow[{}]{\rho _{i}\leftrightarrow \rho _{j}}}\;{\xrightarrow[{}]{\rho _{j}\leftrightarrow \rho _{i}}}&{\begin{array}{*{3}{rc}r}a_{1,1}x_{1}&+&\cdots &+&a_{1,n}x_{n}&=&d_{1}\\&&&&&\vdots \\a_{m,1}x_{1}&+&\cdots &+&a_{m,n}x_{n}&=&d_{m}\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5aab148cdb8143c09a43263f5c1f93995d7e0fe)
![{\displaystyle {\begin{array}{rcl}{\begin{array}{*{3}{rc}r}a_{1,1}x_{1}&+&\cdots &+&a_{1,n}x_{n}&=&d_{1}\\&&&&&\vdots \\a_{m,1}x_{1}&+&\cdots &+&a_{m,n}x_{n}&=&d_{m}\end{array}}&{\xrightarrow[{}]{k\rho _{i}}}\;{\xrightarrow[{}]{(1/k)\rho _{i}}}&{\begin{array}{*{3}{rc}r}a_{1,1}x_{1}&+&\cdots &+&a_{1,n}x_{n}&=&d_{1}\\&&&&&\vdots \\a_{m,1}x_{1}&+&\cdots &+&a_{m,n}x_{n}&=&d_{m}\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f46d3cf5ec6bd5d56e58c90f6ecad0caaf333265)
![{\displaystyle {\begin{array}{rcl}{\begin{array}{*{3}{rc}r}a_{1,1}x_{1}&+&\cdots &+&a_{1,n}x_{n}&=&d_{1}\\&&&&&\vdots \\a_{m,1}x_{1}&+&\cdots &+&a_{m,n}x_{n}&=&d_{m}\end{array}}&{\xrightarrow[{}]{k\rho _{i}+\rho _{j}}}\;{\xrightarrow[{}]{-k\rho _{i}+\rho _{j}}}&{\begin{array}{*{3}{rc}r}a_{1,1}x_{1}&+&\cdots &+&a_{1,n}x_{n}&=&d_{1}\\&&&&&\vdots \\a_{m,1}x_{1}&+&\cdots &+&a_{m,n}x_{n}&=&d_{m}\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b16d8279ad56bf018f19b1cdd011d0bb50a615b)
![{\displaystyle {\begin{array}{*{2}{rc}r}3x&+&2y&=&7\end{array}}\;{\xrightarrow[{}]{2\rho _{1}+\rho _{1}}}\;{\begin{array}{*{2}{rc}r}9x&+&6y&=&21\end{array}}\;{\xrightarrow[{}]{-2\rho _{1}+\rho _{1}}}\;{\begin{array}{*{2}{rc}r}-9x&-&6y&=&-21\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e06519a254a2a979f138aa934d37b9616a2227a)
![{\displaystyle {\begin{array}{rcl}{\begin{array}{*{3}{rc}r}p&+&n&+&d&=&13\\p&+&5n&+&10d&=&83\end{array}}&{\xrightarrow[{}]{-\rho _{1}+\rho _{2}}}&{\begin{array}{*{3}{rc}r}p&+&n&+&d&=&13\\&&4n&+&9d&=&70\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f836b301130374819a74ea1092e79c47fd1e29ec)
