14th September 2020 M01_LEON2218_09_SE_C01 page 15
CHAPTER
1
MC
F
1
2
1
3
1
3
1
3
1
4
1
4
1
4
1
4
1
2
Matrices and Systems of Equations
One of the most important problems in mathematics is that of solving a system of linear
equations. Well over 75 percent of all mathematical problems encountered in scientific
or industrial applications involve solving a linear system at some stage. By using the
methods of modern mathematics, it is often possible to take a sophisticated problem
and reduce it to a single system of linear equations. Linear systems arise in applications
to such areas as business, economics, sociology, ecology, demography, genetics, elec-
tronics, engineering, and physics. Therefore, it seems appropriate to begin this book
with a section on linear systems.
1.1 Systems of Linear Equations
A linear equation in n unknowns is an equation of the form
a
1
x
1
+ a
2
x
2
+···+a
n
x
n
= b
where a
1
, a
2
, ..., a
n
and b are real numbers and x
1
, x
2
, ..., x
n
are variables. A linear
system of m equations in n unknowns is then a system of the form
a
11
x
1
+ a
12
x
2
+···+ a
1n
x
n
= b
1
a
21
x
1
+ a
22
x
2
+···+ a
2n
x
n
= b
2
.
.
.
a
m1
x
1
+ a
m2
x
2
+···+a
mn
x
n
= b
m
(1)
where the a
ij
’s and the b
i
’s are all real numbers. We will refer to systems of the form (1)
as m × n linear systems. The following are examples of linear systems:
(a) x
1
+ 2x
2
= 5
2x
1
+ 3x
2
= 8
(b) x
1
− x
2
+ x
3
= 2
2x
1
+ x
2
− x
3
= 4
(c) x
1
+ x
2
= 2
x
1
− x
2
= 1
x
1
= 4
15
14th September 2020 M01_LEON2218_09_SE_C01 page 16
16 Chapter 1 Matrices and Systems of Equations
System (a) is a 2 ×2 system, (b) is a 2 × 3 system, and (c) is a 3 ×2 system.
By a solution of an m × n system, we mean an ordered n-tuple of numbers
(x
1
, x
2
, ..., x
n
) that satisfies all the equations of the system. For example, the ordered
pair (1, 2) is a solution of system (a), since
1 · (1) + 2 ·(2) = 5
2 · (1) + 3 ·(2) = 8
The ordered triple (2, 0, 0) is a solution of system (b), since
1 · (2) − 1 ·(0) + 1 ·(0) = 2
2 · (2) + 1 ·(0) − 1 ·(0) = 4
Actually, system (b) has many solutions. If α is any real number, it is easily seen that
the ordered triple (2, α, α) is a solution. However, system (c) has no solution. It follows
from the third equation that the first coordinate of any solution would have to be 4.
Using x
1
= 4 in the first two equations, we see that the second coordinate must satisfy
4 + x
2
= 2
4 − x
2
= 1
Since there is no real number that satisfies both of these equations, the system has no
solution. If a linear system has no solution, we say that the system is inconsistent.If
the system has at least one solution, we say that it is consistent. Thus, system (c) is
inconsistent, while systems (a) and (b) are both consistent.
The set of all solutions of a linear system is called the solution set of the system.
If a system is inconsistent, its solution set is empty. A consistent system will have a
nonempty solution set. To solve a consistent system, we must find its solution set.
2 × 2 Systems
Let us examine geometrically a system of the form
a
11
x
1
+ a
12
x
2
= b
1
a
21
x
1
+ a
22
x
2
= b
2
Each equation can be represented graphically as a line in the plane. The ordered pair
(x
1
, x
2
) will be a solution of the system if and only if it lies on both lines. For example,
consider the three systems
(i) x
1
+ x
2
= 2
x
1
− x
2
= 2
(ii) x
1
+ x
2
= 2
x
1
+ x
2
= 1
(iii) x
1
+ x
2
= 2
−x
1
− x
2
=−2
The two lines in system (i) intersect at the point (2, 0). Thus, {(2, 0)} is the solution
set of (i). In system (ii), the two lines are parallel. Therefore, system (ii) is inconsistent
and hence its solution set is empty. The two equations in system (iii) both represent the
same line. Any point on this line will be a solution of the system (see Figure 1.1.1).
In general, there are three possibilities: the lines intersect at a point, they are paral-
lel, or both equations represent the same line. The solution set then contains either one,
zero, or infinitely many points.
14th September 2020 M01_LEON2218_09_SE_C01 page 17
1.1 Systems of Linear Equations 17
(i) Unique Solution: Intersecting Lines
Intersecting Point (2, 0)
(iii) Infinite Solutions: Same Line
(2, 0)
(ii) No Solution: Parallel Lines
x
2
x
1
x
2
x
1
x
2
x
1
Figure 1.1.1.
The situation is the same for m × n systems. An m × n system may or may not be
consistent. If it is consistent, it must have either exactly one solution or infinitely many
solutions. These are the only possibilities. We will see why this is so in Section 1.2
when we study the row echelon form. Of more immediate concern is the problem of
finding all solutions of a given system. To tackle this problem, we introduce the notion
of equivalent systems.
Equivalent Systems
Consider the two systems
(a) 3x
1
+ 2x
2
− x
3
=−2
x
2
= 3
2x
3
= 4
(b) 3x
1
+ 2x
2
− x
3
=−2
−3x
1
− x
2
+ x
3
= 5
3x
1
+ 2x
2
+ x
3
= 2
System (a) is easy to solve because it is clear from the last two equations that x
2
= 3
and x
3
= 2. Using these values in the first equation, we get
3x
1
+ 2 · 3 − 2 =−2
x
1
=−2
14th September 2020 M01_LEON2218_09_SE_C01 page 18
18 Chapter 1 Matrices and Systems of Equations
Thus, the solution of the system is (−2, 3, 2). System (b) seems to be more dicult
to solve. Actually, system (b) has the same solution as system (a). To see this, add the
first two equations of the system:
3x
1
+ 2x
2
− x
3
=−2
−3x
1
− x
2
+ x
3
= 5
x
2
= 3
If (x
1
, x
2
, x
3
) is any solution of (b), it must satisfy all the equations of the system. Thus,
it must satisfy any new equation formed by adding two of its equations. Therefore, x
2
must equal 3. Similarly, (x
1
, x
2
, x
3
) must satisfy the new equation formed by subtracting
the first equation from the third:
3x
1
+ 2x
2
+ x
3
= 2
3x
1
+ 2x
2
− x
3
=−2
2x
3
= 4
Therefore, any solution of system (b) must also be a solution of system (a). By a similar
argument, it can be shown that any solution of (a) is also a solution of (b). This can be
done by subtracting the first equation from the second:
x
2
= 3
3x
1
+ 2x
2
− x
3
=−2
−3x
1
− x
2
+ x
3
= 5
Then add the first and third equations:
3x
1
+ 2x
2
− x
3
=−2
2x
3
= 4
3x
1
+ 2x
2
+ x
3
= 2
Thus, (x
1
, x
2
, x
3
) is a solution of system (b) if and only if it is a solution of system (a).
Therefore, both systems have the same solution set, {(−2, 3, 2)}.
Definition Two systems of equations involving the same variables are said to be equivalent if
they have the same solution set.
If we interchange the order in which two equations of a system are written, this
will have no eect on the solution set. The reordered system will be equivalent to the
original system. For example, the systems
x
1
+ 2x
2
= 4
3x
1
− x
2
= 2
4x
1
+ x
2
= 6
and
4x
1
+ x
2
= 6
3x
1
− x
2
= 2
x
1
+ 2x
2
= 4
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1.1 Systems of Linear Equations 19
both involve the same three equations and, consequently, they must have the same
solution set.
If one equation of a system is multiplied through by a nonzero real number, this
will have no eect on the solution set, and the new system will be equivalent to the
original system. For example, the systems
x
1
+ x
2
+ x
3
= 3
−2x
1
− x
2
+ 4x
3
= 1
and
2x
1
+ 2x
2
+ 2x
3
= 6
−2x
1
− x
2
+ 4x
3
= 1
are equivalent.
If a multiple of one equation is added to another equation, the new system will be
equivalent to the original system. This follows since the n-tuple (x
1
, ..., x
n
) will satisfy
the two equations
a
i1
x
1
+···+a
in
x
n
= b
i
a
j1
x
1
+···+a
jn
x
n
= b
j
if and only if it satisfies the equations
a
i1
x
1
+···+a
in
x
n
= b
i
(a
j1
+ αa
i1
)x
1
+···+(a
jn
+ αa
in
)x
n
= b
j
+ αb
i
To summarize, there are three operations that can be used on a system to obtain an
equivalent system:
I. The order in which any two equations are written may be interchanged.
II. Both sides of an equation may be multiplied by the same nonzero real number.
III. A multiple of one equation may be added to (or subtracted from) another.
Given a system of equations, we may use these operations to obtain an equivalent
system that is easier to solve.
n × n Systems
Let us restrict ourselves to n×n systems for the remainder of this section. We will show
that if an n × n system has exactly one solution, then operations I and III can be used
to obtain an equivalent “strictly triangular system.”
Definition A system is said to be in strict triangular form if, in the kth equation, the coef-
ficients of the first k − 1 variables are all zero and the coecient of x
k
is nonzero
(k = 1, ..., n).
EXAMPLE 1 The system
3x
1
+ 2x
2
+ x
3
= 1
x
2
− x
3
= 2
2x
3
= 4
14th September 2020 M01_LEON2218_09_SE_C01 page 20
20 Chapter 1 Matrices and Systems of Equations
is in strict triangular form, since in the second equation the coecients are 0, 1, −1, re-
spectively, and in the third equation the coecients are 0, 0, 2, respectively. Because of
the strict triangular form, the system is easy to solve. It follows from the third equation
that x
3
= 2. Using this value in the second equation, we obtain
x
2
− 2 = 2orx
2
= 4
Using x
2
= 4, x
3
= 2 in the first equation, we end up with
3x
1
+ 2 · 4 + 2 = 1
x
1
=−3
Thus, the solution of the system is (−3, 4, 2).
Any n × n strictly triangular system can be solved in the same manner as the last
example. First, the nth equation is solved for the value of x
n
. This value is used in the
(n − 1)st equation to solve for x
n−1
. The values x
n
and x
n−1
are used in the (n − 2)nd
equation to solve for x
n−2
, and so on. We will refer to this method of solving a strictly
triangular system as back substitution.
EXAMPLE 2 Solve the system
2x
1
− x
2
+ 3x
3
− 2x
4
= 1
x
2
− 2x
3
+ 3x
4
= 2
4x
3
+ 3x
4
= 3
4x
4
= 4
Solution
Using back substitution, we obtain
4x
4
= 4
4x
3
+ 3 · 1 = 3
x
2
− 2 · 0 + 3 ·1 = 2
2x
1
− (−1) + 3 ·0 − 2 ·1 = 1
x
4
= 1
x
3
= 0
x
2
=−1
x
1
= 1
Thus, the solution is (1, −1, 0, 1).
In general, given a system of n linear equations in n unknowns, we will use opera-
tions I and III to try to obtain an equivalent system that is strictly triangular. (We will
see in the next section of the book that it is not possible to reduce the system to strictly
triangular form in the cases where the system does not have a unique solution.)
EXAMPLE 3 Solve the system
x
1
+ 2x
2
+ x
3
= 3
3x
1
− x
2
− 3x
3
=−1
2x
1
+ 3x
2
+ x
3
= 4
14th September 2020 M01_LEON2218_09_SE_C01 page 21
1.1 Systems of Linear Equations 21
Solution
Subtracting 3 times the first row from the second row yields
−7x
2
− 6x
3
=−10
Subtracting 2 times the first row from the third row yields
−x
2
− x
3
=−2
If the second and third equations of our system, respectively, are replaced by these new
equations, we obtain the equivalent system
x
1
+ 2x
2
+ x
3
= 3
−7x
2
− 6x
3
=−10
−x
2
− x
3
=−2
If the third equation of this system is replaced by the sum of the third equation and −
1
7
times the second equation, we end up with the following strictly triangular system:
x
1
+ 2x
2
+ x
3
= 3
−7x
2
− 6x
3
=−10
−
1
7
x
3
=−
4
7
Using back substitution, we get
x
3
= 4, x
2
=−2, x
1
= 3
Let us look back at the system of equations in the last example. We can associate
with that system a 3 ×3 array of numbers whose entries are the coecients of the x
i
’s:
⎧
⎪
⎪
⎪
⎪
⎪
⎩
121
3 −1 −3
231
⎫
⎪
⎪
⎪
⎪
⎪
⎭
We will refer to this array as the coecient matrix of the system. The term matrix
means a rectangular array of numbers. A matrix having m rows and n columns is said
to be m×n. A matrix is said to be square if it has the same number of rows and columns,
that is, if m = n.
If we attach to the coecient matrix an additional column whose entries are the
numbers on the right-hand side of the system, we obtain the new matrix
⎧
⎪
⎪
⎪
⎪
⎪
⎩
121
3
3 −1 −3
−1
231
4
⎫
⎪
⎪
⎪
⎪
⎪
⎭
14th September 2020 M01_LEON2218_09_SE_C01 page 22
22 Chapter 1 Matrices and Systems of Equations
We will refer to this new matrix as the augmented matrix. In general, when an m × r
matrix B is attached to an m ×n matrix A in this way, the augmented matrix is denoted
by (A|B). Thus, if
A =
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
a
11
a
12
··· a
1n
a
21
a
22
··· a
2n
.
.
.
a
m1
a
m2
··· a
mn
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
, B =
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
b
11
b
12
··· b
1r
b
21
b
22
··· b
2r
.
.
.
b
m1
b
m2
··· b
mr
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
then
(A|B) =
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
a
11
··· a
1n
b
11
··· b
1r
.
.
.
.
.
.
a
m1
··· a
mn
b
m1
··· b
mr
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
With each system of equations, we may associate an augmented matrix of the form
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
a
11
··· a
1n
b
1
.
.
.
.
.
.
a
m1
··· a
mn
b
m
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
The system can be solved by performing operations on the augmented matrix. The x
i
’s
are placeholders that can be omitted until the end of the computation. Corresponding
to the three operations used to obtain equivalent systems, the following row operations
may be applied to the augmented matrix:
Elementary Row Operations
I. Interchange two rows.
II. Multiply a row by a nonzero real number.
III. Replace a row by the sum of that row and a multiple of another row.
Returning to the example, we find that the first row is used to eliminate the elements
in the first column of the remaining rows. We refer to the first row as the pivotal row.
For emphasis, the entries in the pivotal row are all in bold type and the entire row is
color shaded. The first nonzero entry in the pivotal row is called the pivot.
entries to be eliminated
a
21
= 3 and a
31
= 2
→
⎧
⎪
⎪
⎪
⎪
⎪
⎩
(pivot a
11
= 1) 1213 ← pivotal row
3 −1 −3
−1
231
4
⎫
⎪
⎪
⎪
⎪
⎪
⎭
By using row operation III, 3 times the first row is subtracted from the second row and
2 times the first row is subtracted from the third. When this is done, we end up with the
matrix
⎧
⎪
⎪
⎪
⎪
⎪
⎩
121
3
0 −7 −6 −10 ← pivotal row
0 −1 −1
−2
⎫
⎪
⎪
⎪
⎪
⎪
⎭
14th September 2020 M01_LEON2218_09_SE_C01 page 23
1.1 Systems of Linear Equations 23
At this step, we choose the second row as our new pivotal row and apply row opera-
tion III to eliminate the last element in the second column. This time the pivot is −7
and the quotient
−1
−7
=
1
7
is the multiple of the pivotal row that is subtracted from the
third row. We end up with the matrix
⎧
⎪
⎪
⎪
⎪
⎪
⎩
121
3
0 −7 −6
−10
00−
1
7
−
4
7
⎫
⎪
⎪
⎪
⎪
⎪
⎭
This is the augmented matrix for the strictly triangular system, which is equivalent to
the original system. The solution of the system is easily obtained by back substitution.
EXAMPLE 4 Solve the system
− x
2
− x
3
+ x
4
= 0
x
1
+ x
2
+ x
3
+ x
4
= 6
2x
1
+ 4x
2
+ x
3
− 2x
4
=−1
3x
1
+ x
2
− 2x
3
+ 2x
4
= 3
Solution
The augmented matrix for this system is
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
0 −1 −11
0
1111
6
241−2
−1
31−22
3
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
Since it is not possible to eliminate any entries by using 0 as a pivot element, we will
use row operation I to interchange the first two rows of the augmented matrix. The new
first row will be the pivotal row and the pivot element will be 1:
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
(pivot a
11
= 1) 11116 ← pivotal row
0 −1 −11
0
241−2
−1
31−22
3
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
Row operation III is then used twice to eliminate the two nonzero entries in the first
column:
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
1111
6
0 −1 −11 0
02−1 −4
−13
0 −2 −5 −1
−15
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
Next, the second row is used as the pivotal row to eliminate the entries in the second
column below the pivot element −1:
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
1111
6
0 −1 −11
0
00−3 −2 −13
00−3 −3
−15
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
14th September 2020 M01_LEON2218_09_SE_C01 page 24
24 Chapter 1 Matrices and Systems of Equations
Finally, the third row is used as the pivotal row to eliminate the last element in the third
column:
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
1111
6
0 −1 −11
0
00−3 −2
−13
000−1
−2
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
This augmented matrix represents a strictly triangular system. Solving by back substi-
tution, we obtain the solution (2, −1, 3, 2).
In general, if an n ×n linear system can be reduced to strictly triangular form, then
it will have a unique solution that can be obtained by performing back substitution on
the triangular system. We can think of the reduction process as an algorithm involving
n −1 steps. At the first step, a pivot element is chosen from among the nonzero entries
in the first column of the matrix. The row containing the pivot element is called the
pivotal row. We interchange rows (if necessary) so that the pivotal row is the new first
row. Multiples of the pivotal row are then subtracted from each of the remaining n − 1
rows so as to obtain 0’s in the first entries of rows 2 through n. At the second step, a
pivot element is chosen from the nonzero entries in column 2, rows 2 through n,of
the matrix. The row containing the pivot is then interchanged with the second row of
the matrix and is used as the new pivotal row. Multiples of the pivotal row are then
subtracted from the remaining n −2 rows so as to eliminate all entries below the pivot
in the second column. The same procedure is repeated for columns 3 through n − 1.
Note that at the second step row 1 and column 1 remain unchanged, at the third step
the first two rows and first two columns remain unchanged, and so on. At each step, the
overall dimensions of the system are eectively reduced by 1 (see Figure 1.1.2).
If the elimination process can be carried out as described, we will arrive at an
equivalent strictly triangular system after n −1 steps. However, the procedure will break
down if, at any step, all possible choices for a pivot element are equal to 0. When this
happens, the alternative is to reduce the system to certain special echelon, or staircase-
shaped, forms. These echelon forms will be studied in the next section. They will also
be used for m × n systems, where m = n.
x
x
x
x
x
x
x
x
x
x
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Step 1
Step 2
Step 3
Figure 1.1.2.
14th September 2020 M01_LEON2218_09_SE_C01 page 25
1.1 Systems of Linear Equations 25
SECTION 1.1 EXERCISES
1. Use back substitution to solve each of the following
systems of equations:
(a)
x
1
+ x
2
= 7
2x
2
= 6
(b) x
1
+ x
2
+ x
3
= 10
2x
2
+ x
3
= 11
2x
3
= 14
(c)
x
1
+ 2x
2
+ 3x
3
+ 4x
4
= 6
7x
2
− x
3
+ 2x
4
= 5
x
3
− 4x
4
=−9
4x
4
= 8
(d)
x
1
+ x
2
+ 16x
3
+ 3x
4
+ x
5
= 5
4x
2
+ 4x
3
+ 6x
4
+ 3x
5
= 1
−8x
3
+ 27x
4
− 7x
5
= 7
3x
4
+ 11x
5
= 1
x
5
= 0
2. Write out the coecient matrix for each of the systems
in Exercise 1.
3. In each of the following systems, interpret each equa-
tion as a line in the plane. For each system, graph
the lines and determine geometrically the number of
solutions.
(a)
x
1
+ x
2
= 4
x
1
− x
2
= 2
(b) x
1
+ 2x
2
= 4
−2x
1
− 4x
2
= 4
(c)
2x
1
− x
2
= 3
−4x
1
+ 2x
2
=−6
(d) x
1
+ x
2
= 1
x
1
− x
2
= 1
−x
1
+ 3x
2
= 3
4. Write an augmented matrix for each of the systems in
Exercise 3.
5. Write out the system of equations that corresponds to
each of the following augmented matrices:
(a)
⎧
⎪
⎩
30
6
02
4
⎫
⎪
⎭
(b)
⎧
⎪
⎩
1 −15
8
302
0
⎫
⎪
⎭
(c)
⎧
⎪
⎪
⎪
⎪
⎪
⎩
1 −21
4
705
2
−320
0
⎫
⎪
⎪
⎪
⎪
⎪
⎭
(d)
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
1 −20−8
5
2134
6
0 −31−1
7
8411
9
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
6. Solve each of the following systems:
(a)
x
1
− x
2
= 11
x
1
+ x
2
=−1
(b) 3x
1
− 2x
2
=−5
2x
1
+ 3x
2
= 27
(c)
4x
1
+
1
2
x
2
= 2
7
3
x
1
+ 14x
2
= 9
(d) x
1
+ 2x
2
− x
3
=−6
2x
1
− x
2
+ x
3
= 7
−x
1
+ x
2
+ 2x
3
= 3
(e)
x
1
+ 3x
2
+ 5x
3
= 27
2x
1
+ 4x
2
+ 6x
3
= 30
2x
1
+ 2x
2
+ 3x
3
= 11
(f)
2x
1
− x
2
+ 4x
3
=−4
x
1
+ 3x
2
− x
3
= 8
3x
1
− x
2
− x
3
= 2
(g)
3
5
x
1
+
1
3
x
2
+
2
3
x
3
= 1
5
7
x
1
−
2
5
x
2
−
3
5
x
3
=−1
1
10
x
1
+
2
10
x
2
+
3
10
x
3
=
1
2
(h)
x
1
+ 2x
2
+ 2x
3
+ x
4
= 7
x
1
− 3x
2
+ x
3
− x
4
= 2
3x
1
− x
2
+ x
3
+ x
4
= 0
2x
1
+ 2x
3
= 8
7. The two systems
x
1
+ 2x
2
= 8
4x
1
− 3x
2
=−1
and
x
1
+ 2x
2
= 7
4x
1
− 3x
2
= 6
have the same coecient matrix but dierent right-hand
sides. Solve both systems simultaneously by eliminating
the first entry in the second row of the augmented matrix:
⎧
⎪
⎪
⎩
12
87
4 −3
−16
⎫
⎪
⎪
⎭
and then performing back substitutions for each of the
columns corresponding to the right-hand sides.
8. Solve the two systems
x
1
+ 2x
2
− x
3
= 6
2x
1
− x
2
+ 3x
3
=−3
x
1
+ x
2
− 4x
3
= 7
and
x
1
+ 2x
2
− x
3
= 9
2x
1
− x
2
+ 3x
3
=−2
x
1
+ x
2
− 4x
3
= 9
by doing elimination on a 3 × 5 augmented matrix and
then performing two back substitutions.
9. Given a system of the form
−m
1
x
1
+ x
2
= b
1
−m
2
x
1
+ x
2
= b
2
where m
1
, m
2
, b
1
,andb
2
are constants:
(a)
Show that the system will have a unique solution if
m
1
= m
2
.