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FUNDAMENTALS OF ALGEBRA
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8. Linear Systems (Last updated 3/24/11)
8.1. Dimension. In this section and the next, we will be touching on some topics in the
field of linear algebra. We will be working in Euclidean spaces of various dimensions, so a
quick summary of dimension is in order.
(1) R0 is 0-dimensional Euclidean space. It can be thought of as a single point.
(2) R1 = R is 1-dimensional Euclidean space. It can be thought of as the usual real
line; every point in R1 is determined by one coordinate.
(3) R2 = {(x, y) | x and y are real numbers} is 2-dimensional Euclidean space. It
can be thought of as the usual Cartesian plane; every point in R2 is determined by
two coordinates.
(4) R3 = {(x, y, z) | x, y, and z are real numbers} is 3-dimensional Euclidean space.
It can be thought of as the space in which we live; every point in R3 is determined
by three coordinates.
(5) Rn = {(x1 , x2 , . . . , xn ) | each xi is a real number} is n-dimensional Euclidean
space. Every point in Rn is determined by three coordinates.
8.2. General Linear Equations. Recall the standard form for a line in the plane is Ax +
By = C. This form generalizes from two variables (x and y) to arbitrarily many variables.
Definition 8.1. A linear equation in the variables x1 , x2 , . . . xn is an equation of the form
a1 x1 + a2 x2 + · · · + an xn = b
where each of the ai are real numbers. Each ai is called the coefficient of xi .
The graph of such an equation in Rn is a hyperplane of dimension n − 1, just like a line
Ax + By = C is really a copy of R1 in R2 and a plane Ax + By + Cz = D is really a copy
of R2 in R3 .
Example 8.1. The equation 3x1 − 27 x3 + x4 =
√
2 is a linear equation in the variables
x1 , x2 , x3 , x4 . Notice the coefficient of x2 is zero.
Definition 8.2. A system of linear equations in n variables is a collection of linear equations in n variables. A solution to such a system is the collection of all points
(x1 , x2 , . . . , xn ) that satisfies each equation in the system.
We will be considering systems with the same number of equations as variables.
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8.3. Systems of Two Linear Equations in 2 Variables. A system of two linear equations
in two variables is of the form
a1 x + b1 y = c1
a2 x + b2 y = c2
Since the graph of each of these equations is a line in R2 , (x, y) is a solution to the system
if and only if it lies on both lines. There are three possible relative configurations of lines in
the plane:
!
!1!
!!
!
!!
2
!!
!!
!!
!
!
!!
!!
Parallel
"
"
!
"
#1#
## "
#
"##
##
"
"
"
"!2 Intersecting
$$
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$$ !1 = !2
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Coincident
When the corresponding lines are parallel37, the system has no solution. When the lines are
intersecting, the system has exactly one solution, corresponding to the point of intersection.
When the lines are coincident, the system has infinitely many solutions, corresponding to
the set of points on the line.
Example 8.2.
(1) The system
3x − 5y = 7
−6x + 10y = −14
consists of two identical lines; notice the second equation is can be obtained from
the first by multiplying by −2. So every point on this common line is a solution to
the system. Solving the first (or second) equation for y yields y = 35 x − 75 . Thus the
solution to this system consists of all points of the form (x, 35 x − 75 ), where x is any
real number.
(2) The system
3x − 5y = 7
−6x + 10y = −15
consists of two parallel lines. Comparing with the previous example, the coefficients
of x and y in the second equations are the same; changing the right hand side from
14
= − 57 to − 15
= − 23 .
−14 to −15 results in changing the y-intercept from − 10
10
37Following
the literature, we will assume parallel lines have no points in common. Notice the slight
distinction between this and defining parallel lines as “lines having the same slope”: the latter definition
implies identical lines are parallel, which is not the standard definition of parallel.
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We can also verify algebraically that no point lies on both lines. For if (x0 , y0) is a
point on the first line, then 3x0 − 5y0 = 7, so −6x0 + 10y0 = −14 #= −15, so (x0 , y0 )
does not lie on the second line.
(3) The system
x − 2y = 4
3x + y = −1
consists of two lines with different slopes, so there must be a unique solution. In fact,
).
that solution is ( 72 , − 13
7
One way to find this solution is to solve the second equation for y, yielding y =
−3x − 1, substitute that into the first equation, yielding x − 2(−3x − 1) = 4, solve the
result for x, yielding x = 27 , and then substitute that into either equation, yielding
y = − 13
. While this particular method works reasonably well for systems with 2
7
variables, it can be awkward for systems with 3 or more variables.
Another way to find the solution is to combine the equations in a certain way:
multiply the first equation by −3 and add it to the second equation:
−3x + 6y = −12
3x + y = −1
7y = −13
− 13
.
7
So y =
We can then substitute that value into either equation to get x = 72 . This
second method of combining the linear equations in certain ways is what generalizes
nicely to higher dimensions.
8.4. Elementary Row Operations. Given a system of n linear equations in n unknowns,
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
a31 x1 + a32 x2 + · · · + a3n xn = b3
..
.
an1 x1 + an2 x2 + · · · + ann xn = bn ,
we would like to replace this system with a system that has the same solution set, but is
easier to solve. We can accomplish this by performing a sequence of the following operations
on the given system.
Elementary Row Operations. There are three elementary operations that can be performed on a system of equations without changing the solution set:
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VAN WYK’S MATH 304
Type 1. Interchange any two equations.
Type 2. Multiply any equation by a nonzero real number.
Type 3. Replace any equation by the sum of itself and a multiple of another equation.
Types (1) and (2) obviously don’t change the solution set, but what about Type (3)?
Given that a Type (2) operation doesn’t change the solution set, we need only show that
the solution set of two linear equations E1 and E2 is the same as the solution set of the
two equations E1 + E2 and E2. Clearly if "x satisfies38 both equation E1 and equation E2,
it satisfies E1 + E2, so one direction is clear. Conversely, suppose "x satisfies both equation
E1+E2 and equation E2. But since "x also satisfies −E2, it must satisfy (E1+E2)−E2 = E1
as well. Thus Type 3 row operations don’t change the solution set of the system either.
Notice we used a Type (3) operation in the second solution of Example 8.2 (3) when
we essentially replaced the second equation with the sum of itself and −3 times the first
equation.
8.5. Systems of Linear Equations in 3 Variables. A system of three linear equations
in three variables is of the form
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d2
a3 x + b3 y + c3 z = d3
Since the graph of each of these equations is a plane in R3 , (x, y, z) is a solution to the
system if and only if it lies on all three planes. There are eight possible relative configurations
of three planes in R3 :
Configurations corresponding to no solution.
(1) Three parallel planes.
(2) Two parallel planes, the third intersecting them.
(3) Two coincident planes, the third parallel to them.
(4) Two intersecting planes, the third parallel to their line of intersection.
Configurations corresponding to a 0-dimensional solution (a point).
(5) Three planes intersecting in a single point.
Configurations corresponding to a 1-dimensional solution (a line).
(6) Three planes intersecting in a line.
(7) Two coincident planes, the third intersecting them.
Configurations corresponding to a 2-dimensional solution (a plane).
(8) Three coincident planes.
38We
are using !x here to denote some point in Rn , (x1 , x2 , . . . , xn ).
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Figure 1. Cases (4) and (6) above.
Most of these are easily visualized. An example of case (5) can be seen in a corner of a
room, with the planes corresponding to the two walls and the floor/ceiling. Cases (4) and
(6) are show in the figure.
Unlike linear systems consisting of two lines in the plane, which can easily be categorized
into one of the three possible cases mentioned above, linear systems consisting of three planes
in space are more difficult to categorize geometrically. Usually performing elementary row
operations on the system is the best way to find the solution set. The general procedure is
to use the first equation to eliminate x from the other two equations, then use the second
equation to eliminate y from the third equation, resulting in a system with the same solution
set, but in an “upper triangular form:”
a1 x + b1 y + c1 z = d1
b2 y + c2 z = d2
c3 z = d 3
Working from the third equation upward, it is then easy to determine z, y, and x.
Example 8.3.
(1) Consider the system
3x − 7y + 7z = −8
x − 3y + 4z = −4
−4x + 6y − z = 7
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Since the coefficient of x in the second equation is 1, interchange the first and
second equations:
x − 3y + 4z = −4
3x − 7y + 7z = −8
−4x + 6y − z = 7
Replace the second equation with itself minus 3 times the first equation, and replace
the third equation with itself plus 4 times the first equation:
x − 3y + 4z = −4
2y − 5z = 4
−6y + 15z = −9
Replace the third equation with itself plus 3 times the second equation:
x − 3y + 4z = −4
2y − 5z = 4
0 = 3
Since 0 #= 3, this system has no solution39. Since no two of the corresponding
planes are parallel – none of the left sides of the original equations is a multiple of
another – the configuration of the three planes must be as in Case (4) above.
(2) Consider the system
x − 3y + 4z = −4
−4x + 6y − z = 7
−2x + 6y − 8z = 8
Replace the second equation with itself plus 4 times the first equation, and replace
the third equation with itself plus 2 times the first equation:
x − 3y + 4z = −4
−6y + 15z = −9
0 = 0
9
6
y − 15
= 52 y − 53 ,
15
.
3y − 4( 52 y − 35 ) + 4 = 57 y + 32
5
The second equation yields z =
and substituting that into the
Since there is no restriction on
first equation yields x =
y, the solution set of this system consists of all points of the form ( 75 y + 32
, y, 52 y − 35 ).
5
The value of y can range over the entire set of real numbers, and the values of both
39Such
a system is called inconsistent.
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x and z are dependent upon the value of y; thus this set of points forms a line. Since
the original third equation is a multiple of the original first equation, those two planes
are coincident, and the configuration of the three planes is as in Case (7) above.
(3) Consider the system
x+z = 3
x + 2y + z = −1
y−z = 7
Replace the second equation with itself minus the first equation:
x+z = 3
2y = −4
y−z = 7
We could reduce this further, but it isn’t too much work solving for the system as
is. The second equation yields y = −2, so the third equation yields z = −9, and the
first equation then yields x = 12. So the solution set of this system consists of the
single point (12, −2, −9). The configuration of the three planes is thus as in Case (5)
above.
In fact, a few more elementary row operations would yield the system
x = 12
y = −2
z = −9
(4) Consider the system
x+y+z = 1
2x + 2y + 2z = 2
3x + 3y + 3z = 3
This system clearly corresponds to Case (8) above, since each of these planes
is identical. To see what happens to the system when performing elementary row
operations, replace the second equation with itself minus 2 times the first, and replace
the third equation with itself minus 3 times the first:
x+y+z = 1
0 = 0
0 = 0
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Thus the solution set consists of the set of points on the plane with equation
x + y + z = 1. Once two of these variables are given, the third is determined. Solving
for x, for example, gives x = 1 − y − z. So the solution set consists of the set of points
of the form (1 − y − z, y, z). The values of y and z have no constraint, so are free
to run over all reals. Since they do not depend on each other, their combined values
run over the entire Cartesian plane.
Problems.
1. Suppose (3, 1) lies on a line with rational slope m = ab . Show that there are infinitely
many points with integer coordinates that lie on that line.
2. Find the slope-intercept form of the line that contains all points of the form (n, an ), where
(an ) is the arithmetic sequence 5, 2, −1, −4, −7, . . . .
3. Find the solution set of the system
2x − y = 5
x + 3y = 7
4. Find the solution set of the system
2x − y = 5
−6x + 3y = −15
5. Determine all possible equations ax + by = c so that the solution set of the system
2x − y = 5
ax + by = c
is the single point (3, 1). Hint. Think geometrically.
6. Determine all possible equations ax + by = c so that the system
2x − y = 5
ax + by = c
has infinitely many solutions.
7. Eight thousand people attended a basketball game. Some tickets sold for $15 and some
sold for $25. The total revenue generated from ticket sales was $141, 200. How many
people bought each type of ticket?
8. The sum of the digits of a two-digit number is 11. When the two digits are reversed,
the resulting number is 27 more than the original. What is the original number? Hint.
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If the original number has base ten representation a1 a0 , then the value of the number is
10a1 + a0 .
9. Find the solution set of the system
x+y+z = 1
2x + y + 3z = 6
4x + 3y + 5z = 7
10. Find the solution set of the system
x − y + 2z = 2
x + 2y − 3z = −1
2x + y − z = 1
11. Find the solution set of the system
x − y + 2z = 2
−x + y − 2z = −1
2x − 2y + 4z = 3
12. The local movie theater sold 3500 tickets one Saturday and made $27, 200 in revenue.
There were three types of tickets sold: adult tickets for $10 each, children’s tickets for $7
each, and senior’s tickets for $8 each. There were three times as many children’s tickets
sold as adult tickets. How many of each type of tickets were sold that day?