Download Chapter 1 Systems of Linear Equations and Matrices

Chapter 1
Systems of Linear Equations and
Matrices
System of linear algebraic equations and their solution constitute one of the major topics
studied in the course known as “linear algebra.” In the first section we shall introduce
some basic terminology and discuss a method for solving such system.
1.1
Systems of Linear Equations
Linear Equations
Any straight line in the xy-plane can be represented algebraically by an equation of the
form
a1 x + a2 y = b
where a1 , a2 , and b are real constants and a1 and a2 are not both zero. An equation of
this form is called a linear equation in the variables x and y. More generally, we define
a linear equation in the n veriables x1 , x2 , . . . , xn to be one that can be expressed in
the form
a1 x1 + a2 x2 + · · · + an xn = b
where a1 , a2 , . . . , an , and b are real constants. The variables in a linear equation are
sometimes called unknowns. For example, the equations
3x + y = 7, y = 51 x + 2z + 4, and x1 + 3x2 − 2x3 + 5x4 = 7
are linear. The equations
√
x + 3y = 2,
3x − 2y − 5z + yz = 4,
and y = cos x
are not linear.
Observe that a linear equation does not involve any products or roots of variables. All
variables occur only to the first power and do not appear as arguments for trigonometric,
logarithmic, or exponential functions.
A solution of a linear equation a1 x1 + a2 x2 + · · · + an xn = b is a sequence of n
numbers s1 , s2 , . . ., sn such that the equation satisfied when we substitute x1 = s1 ,
x2 = s2 , . . ., xn = sn . The set of all solutions of the equation is called its solution set
or sometimes the general solution of the equation.
1
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2
Example 1.1 Find the solution set of
(a) 5x − 2y = 3 and
(b) x1 − 3x2 + 4x3 = 2.
Solution . . . . . . . . .
Linear Systems
A finite set of linear equations in the variables x1 , x2 , . . ., xn is called system of linear
equations or a linear system. A sequence of numbers s1 , s2 , . . ., sn is called a
solution of the system if x1 = s1 , x2 = s2 , . . ., xn = sn is a solution of every equation
in the system. For example, the system
4x1 − x2 + 3x3 = −1
3x1 + x2 + 9x3 = −4
the the solution x1 = 1, x2 = 2, and x3 = −1 since these values satisfy both equations.
However, x1 = 1, x2 = 8, x3 = 1 is not a solution since these values satisfy only the first
equation in the system. Thus, not all system of linear equations have solutions.
A system of equations that has no solutions is said to be inconsistent; if there is
at least one solution of the system, it is called consistent. To illustrate the possibilities
that can occur in solving systems of linear equations, consider a general system of two
linear equations in the unknowns x and y:
a1 x + b1 y = c1
a2 x + b2 y = c2
(a1 , b1 not both zero)
(a2 , b2 not both zero)
The graphs of these equations are lines; call them `1 and `2 . Since a point (x, y) lies on
a line if and only if the number x and y satisfy the equation of the line, the solutions of
the system of equations correspond to points of the intersection of `1 and `2 . There are
three possibilities, illustrated in Figure 1.1
y
`1
y
`2
y
`1
x
(a) No solution
`1 and `2
`2
x
(b) One solution
x
(c) Infinitely many solutions
Figure 1.1:
• The lines `1 and `2 may be parallel, in which case there is no intersection and
consequently no solution to the system.
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3
• The lines `1 and `2 may intersect at only one point, in which case the system has
exactly one solution.
• The lines `1 and `2 may coincide, in which case there are infinitely many points of
intersection and consequently infinitely many solutions to the system.
Although we have considered only two equations with two unknowns here, we will
show later that the same three possibilities hold for arbitrary linear systems:
Every system of linear equations has no solutions, or has exactly one solution,
or has infinitely many solutions.
An arbitrary system of m linear equations in n unknowns can be written as
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
..
..
..
..
.
.
.
.
am1 x1 + am2 x2 + · · · + amn xn = bm
(1.1)
where x1 , x2 , . . ., xn are the unknowns and the subscripted a’s and b’s denote constants.
For example, a general system of three linear equations in four unknowns can be written
as
a11 x1 + a12 x2 + a13 x3 + a14 x4 = b1
a21 x1 + a22 x2 + a23 x3 + a24 x4 = b2
a31 x1 + a32 x2 + a33 x3 + a34 x4 = b3
The double subscripting on the coefficients of the unknowns is a useful device that
is used to specify the location of the coefficient in the system. The first subscript on
the coefficient aij indicates the equation in which the coefficient occurs, and the second
subscript indicates which unknown it multiplies. Thus, a12 is in the first equation and
multiplies unknown x2 .
Augmented Matrices
A system of m linear equations in
rectangular array of numbers

a11
 a21

 ..
 .
am1
n unknowns can be abbreviated by writing only the
a12
a22
..
.
···
···
a1n
a2n
..
.
b1
b2
..
.
am2
···
amn
bm



.

This is called the augmented matrix for the system. For example, the augmented
matrix for the system of equations
x1 + x2 + 2x3 = 9
2x1 + 4x2 − 3x3 = 1
3x1 + 6x2 − 5x3 = 0
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is

1
 2
3
4

9
1 .
0
1 2
4 −3
6 −5
Remark When constructing an augmented matrix, we must write the unknowns in the
same order in each equation, and the constants must be on the right.
Exercise 1.1
1. Which of the following are linear equations in x1 , x2 and x3 ?
√
(b) x1 + 3x2 + x1 x3 = 2
(a) x1 + 5x2 − 2 x3 = 1
(c) x1 = −7x2 + 3x3
3/5
(e) x1 − 2x2 + x3 = 4
(d) x−2
1 + x2 + 8x3 = 5
√
(f) πx1 − 2 x2 + 13 x3 = 71/3
2. Find the solution set of each of the following linear systems.
(a) 7x − 5y = 3
(b) 3x1 − 5x2 + 4x3 = 7
(c) −8x1 + 2x2 − 5x3 + 6x4 = 1
(d) 3v − 8w + 2x − y + 4z = 0
3. Which of the following are linear systems?
(a) x1 − 3x2 = x3 − 4
x4 = 1 − x1
x1 + x4 + x3 − 2 = 0
√
(b) 2x − y + 3z = −1
x + 2y − z = 2
4x − y
= −1
(c) 3x − xy
= 1
x + 2xy − y = 0
(d) y = 2x − 1
y = −x
(e) 2x1 − sin x2 = 3
x2 = x1 + x3
−x1 + x2 − 3x3 = 0
(f)
x1 + x2 + x3 = 1
−2x21
− 2x3 = −1
3x2 − x3 = 2
4. Find the augmented matrix for each of the following systems of linear equations.
(a) 3x1 − 2x2 = −1
4x1 + 5x2 = 3
7x1 + 3x2 = 2
(b)
(c) x1
(d) x1 + 2x2
− x4 + x5 = 1
3x2 + x3
− x5 = 2
x3 + 7x4
= 1
x2
x3
= 1
= 2
= 3
2x1
+ 2x3 = 1
3x1 − x2 + 4x3 = 7
6x1 + x2 − x3 = 0
5. Find a system of linear equations corresponding to the augmented matrix.

2 0
(a)  3 −4
0 1

0
0 
1

3
(b)  7
0
0
1
−2

−2 5
4 −3 
1
7
MA131 (Section 750001): Prepared by Dr. Archara Pacheenburawana
(c)
7 2
1 2
1 −3
4 0
5
1

1
 0
(d) 
 0
0
0
1
0
0
0
0
1
0
0
0
0
1
5

7
−2 

3 
4
6. (a) Find a linear equation in the variables x and y that has the general solution
x = 5 + 2t, y = t.
(b) Show that
x = t, y =
1
2
t−
5
2
is also the general solution of the equation in part (a).
7. Consider the system of equations
ax + by = k
cx + dy = `
ex + f y = m
Indicate what we can say about the relative positions of the line ax + by = k,
cx + dy = `, and ex + f y = m when
(a) the system has no solutions.
(b) the system has exactly one solution.
(c) the system has infinitely many solutions.
8. Show that the linear system
ax + by = e
cx + dy = f
where a, b, c, d, e, f are constants, has a unique solution if ad − bc 6= 0. Express this
solution in terms of a, b, c, d, e, and f .
Answer to Exercise 1.1
1. (a), (c), (f)
3
+ 75 t, y = t (b) x1 = 35 s − 34 t − 18 , x2 =
7
x1 = 41 r − 58 s + 34 t − 81 , x2 = r, x3 = s, x4 = t
v = 38 q − 23 r + 31 s − 34 t, w = q, x = r, y = s, z
2. (a) x =
(c)
(d)
3. (a), (d)
s, x3 = t
=t
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

3 −2 −1
3 
4. (a)  4 5
7 3
2

1 2 0 −1

(d) 0 3 1 0
0 0 1 7
6



2 0
2 1
1 0 0 1
(b)  3 −1 4 7  (c)  0 1 0 2 
6 1 −1 0
0 0 1 3

1 1
−1 2 
0 1

5. (a) 2x1
=0
3x1 − 4x2 = 0
x2 = 1
(b) 3x1
− 2x3 = 5
7x1 + x2 + 4x3 = −3
−2x2 + x3 = 7
(c) 7x1 + 2x2 + x3 − 3x4 = 5
x1 + 2x2 + 4x3
=1
(d) x1
x2
=7
= −2
x3
=3
x4 = 4
6. (a) x − 2y = 5 (b) Let x = t; then t − 2y = 5. Solving for y yields y = 21 t − 52 .
7. (a) The line have no common point of intersection.
(b) The line intersect in exactly one point.
(c) The three lines coincide.
1.2
Elementary Operations
Elementary Row Operations
The basic method for solving a system of linear equations is to replace the given system
by a new system that has the same solution set but is easier to solve. This new system is
generally obtained in a series of steps by applying the following three types of operations
to eliminate unknowns systematically:
1. Multiply any equation, say the i th, by a nonzero constant α. This is denoted by
α ri .
2. Interchange of two equations, say the i th and j th. This is denoted by ri ↔ rj .
3. Add α times the i th equation to the j th equation. This is denoted by rj + α ri .
(This keeps the i th equation unchanged.)
Since the rows of an augmented matrix correspond to the equations in the associated
system, these three operations correspond to the following operations on the rows of the
augmented matrix:
1. Multiply any row by a nonzero constant: α ri .
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2. Interchange of two rows: ri ↔ rj .
3. Add a multiple of one row to another: rj + α ri .
These are called elementary row operations. The following illustrates how these
operations can be used to solve system of linear equations.
Example 1.2 Solve the following system by using elementary row operations.
x + y + 2z = 9
2x + 4y − 3z = 1
3x + 6y − 5z = 0
Solution In the left column we solve a system of linear equations by operating on the
equations in the system, and in the right column we solve the same system by operating
on the rows of the augmented matrix.


x + y + 2z = 9
1 1 2 9
 2 4 −3 1 
2x + 4y − 3z = 1
3x + 6y − 5z = 0
3 6 −5 0
Add −2 times the first equation
to the second to obtain
x + y + 2z
2y − 7z
3x + 6y − 5z
= 9
= −17
= 0
Add −3 times the first equation
to the third to obtain
x + y + 2z
2y − 7z
3y − 11z
= 9
= −17
= −27
Multiply the second equation
by 21 to obtain
x + y + 2z
y − 27 z
3y − 11z
= 9
= − 17
2
= −27
Add −3 times the second equation
to the third to obtain
x + y + 2z = 9
y − 72 z
− 21 z
= − 17
2
= − 23
Add −2 times the first row
to the second to obtain


1 1 2
9
 0 2 −7 −17 
3 6 −5
0
Add −3 times the first row
to the third to obtain


1 1
2
9
 0 2 −7 −17 
0 3 −11 −27
Multiply the second row
by 21 to obtain


1 1
2
9
 0 1 − 7 − 17 
2
2
0 3 −11 −27
Add −3 times the second row
to the third to obtain


9
1 1 2


 0 1 − 27 − 17

2
0 0 − 12
− 23
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Multiply the third equation
by −2 to obtain
x + y + 2z
y − 72 z
z
Multiply the third row
by −2 to obtain


1 1 2
9
 0 1 − 7 − 17 
2
2
3
0 0 1
= 9
= − 17
2
= 3
Add −1 times the second equation
to the first to obtain
x
+ 11
z
2
7
y − 2z
z
= 35
2
= − 17
2
= 3
Add − 11
times the third equation
2
to the first and 27 times the third
equation to the second to obtain
x
y
z
= 1
= 2
= 3
8
Add −1 times the second row
to the first to obtain


35
1 0 11
2
2
 0 1 − 7 − 17 
2
2
0 0 1
3
times the third row
Add − 11
2
to the first and 27 times the third
row to the second to obtain


1 0 0 1
 0 1 0 2 
0 0 1 3
The solution x = 1, y = 2, z = 3 is now evident.
♣
Echelon Forms
In previous example, we solved a linear
the augmented matrix to the form

1
 0
0
system in the unknowns x, y, and z by reducing

0 0 1
1 0 2 .
0 1 3
This is an example of a matrix that is in reduced row-echelon form. To be of this
form, a matrix must have the following properties:
1. If the row does not consist entirely of zeros, then the first nonzero number in the
row is a 1. We call this a leading 1.
2. If there are any rows that consist entirely of zeros, then they are grouped together
at the bottom of the matrix.
3. In any two successive rows that do not consist entirely of zeros, the leading 1 in
the lower row occurs farther to the right than the leading 1 in the higher row.
4. Each column that contains a leading 1 has zeros everywhere else in that column.
A matrix that has the first three properties is said to be in row-echelon matrix.
(Thus, a matrix in reduced row-echelon form is of necessity in row-echelon form, but not
conversely.)
MA131 (Section 750001): Prepared by Dr. Archara Pacheenburawana
The following matrices are in reduced row-echelon form.





0 1 −2
1 0 0 4
1 0 0
0 0 0
0 1 0 7  , 0 1 0 , 
0 0 0
0 0 1 −1
0 0 1
0 0 0
The following matrices are in row-echelon



1 4 −3 7
1
0 1 6


2 ,
0
0 0 1 −1
0
form.

0 0
1 0 ,
0 0
0
1
0
0

0
3
,
0
0
0 0
0 0
9


0 1 2 6 0
0 0 1 −1 0
0 0 0 0 1
Example 1.3 Suppose that the augmented matrix for a linear equations has been reduced
by row operations to the given reduced row-echelon form. Solve the system.




1 0 0 3 −5
1 0 0 2
(b)  0 1 0 4 −1 
(a)  0 1 0 4 
0 0 1 −3
0 0 1 2 6




1 0 2 −3
1 0 4 0
(c)  0 1 1 4 
(d)  0 1 −5 0 
0 0 0 0
0 0 0 1
Solution . . . . . . . . .
Elimination Methods
We have just seen how easy it is to solve a system of linear equations once its augmented
matrix is in reduced row-echelon form.
• The process of using row operations to transform a linear system into one whose
augmented matrix is in row-echelon form is called Gaussian elimination.
• The process of using row operations to transform a linear system into one whose
augmented matrix is in reduced row-echelon form is called Gauss-Jordan elimination.
Remark It can be show that every matrix has a unique reduced row-echelon form.
Example 1.4 Solve by Gauss-Jordan elimination.
x1 + 3x2 − 2x3
+ 2x5
2x1 + 6x2 − 5x3 − 2x4 + 4x5 − 3x6
5x3 + 10x4
+ 15x6
2x1 + 6x2
+ 8x4 + 4x5 + 18x6
Solution . . . . . . . . .
=
=
=
=
0
−1
5
6
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10
Back-Substitution
It is sometimes preferable to solve a system of linear equations by using Gaussian elimination to bring the augmented matrix into row-echelon form without continuing all the
way to the reduced row-echelon form. When this is done, the corresponding system of
equations can be solved by a technique called back-substitution. The next example
illustrates the idea.
Example 1.5 From Example 1.4, solve by Gaussian elimination and back-substitution.
Solution . . . . . . . . .
Example 1.6 Solve
x + y + 2z = 9
2x + 4y − 3z = 1
3x + 6y − 5z = 0
by Gaussian elimination and back-substitution.
Solution . . . . . . . . .
Homogeneous Linear Systems
A system of linear equations is said to be homogeneous if the constant terms are all
zero; that is the system has the form
a11 x1 + a12 x2 + · · · + a1n xn = 0
a21 x1 + a22 x2 + · · · + a2n xn = 0
..
..
..
..
.
.
.
.
am1 x1 + am2 x2 + · · · + amn xn = 0
Every homogeneous system of linear equations is consistent, since all such systems
have x1 = 0, x2 = 0, . . ., xn = 0 as a solution. This solution is called the trivial
solution; if there are other solutions, they are called nontrivial solution.
Because a homogeneous linear system always has the trivial solution, there are only
two possibilities for its solutions:
• The system has only the trivial solution.
• The system has infinitely many solutions in addition to the trivial solution.
In the special case of a homogeneous linear system of two equations in two unknowns,
say
a1 x + b1 y = 0 (a1 , b1 not both zero)
a2 x + b2 y = 0 (a2 , b2 not both zero)
the graphs of the equations are lines through the origin, and the trivial solution corresponds to the point of intersection at the origin.
MA131 (Section 750001): Prepared by Dr. Archara Pacheenburawana
y
11
y
a1 x + b1 y = 0
x
a2 x + b2 y = 0
x
a1 x + b1 y = 0
and
a2 x + b2 y = 0
(b) Infinitely many solutions
(a) Only the trivial solution
There is one case in which a homogeneous system is assured of having nontrivial
solution — namely, whenever the system involves more unknowns than equations. To
see why, consider the following example of four equations in five unknowns.
Example 1.7 Solve the following homogeneous system of linear equations by using
Gauss-Jordan elimination.
2x1 + 2x2 − x3
+ x5
−x1 − x2 + 2x3 − 3x4 + x5
x1 + x2 − 2x3
− x5
x3 + x4 + x5
=
=
=
=
0
0
0
0
Solution . . . . . . . . .
Theorem 1.1.
A homogeneous system of linear equations with more unknowns than equations
has infinitely many solutions.
Remark Note that Theorem 1.1 applied only to homogeneous systems. A nonhomogeneous system with more unknowns than equations need not be consistent; however, if
the system is consistent, it will have infinitely many solutions.
Exercise 1.2
1. Which of the following 3 × 3 matrices are in reduced row-echelon form?


1 0 0
(a) 0 1 0
0 0 1


1 0 0
(e) 0 0 0
0 0 1


0 1 0
(b) 1 0 0
0 0 0


1 1 0
(f) 0 1 0
0 0 0


0 1 0
(c) 0 0 1
0 0 0


1 0 0
(d) 0 0 1
0 0 0

1 0 2
(g) 0 1 3
0 0 0


0 0 0
(h) 0 0 0
0 0 0

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12
2. Which of the following 3 × 3 matrices are in row-echelon form?


1 0 0
(a) 0 1 0
0 0 1


1 4 6
(e) 0 0 1
0 1 3


1 3 0
(b) 0 0 1
0 0 0


1 1 0
(f) 0 1 0
0 0 0
3. In each part determine whether
echelon form, both, or neither.



0 1 3 5 7
1



(a) 0 0 1 2 3
(b) 0
0 0 0 0 0
0

1
1 −7 5 5

(e) 0
(d)
0 1 3 2
0


1 1 1
(c) 0 1 2
0 0 3


1 3 4
(g) 0 0 1
0 0 0


1 0 0
(d) 0 1 0
0 2 0


1 2 3
(h) 0 0 0
0 0 1
the matrix is in row-echelon form, reduced row
0 0 5
0 1 3
1 0 4
1 0 3 1
(c)
0 1 2 4

0 0 1 2
1 0 2 4
0 1 3 6


0 1
(f) 0 0
0 0
4. Find the reduced row-echelon form of A when


2
2 −1 0
1 0
−1 −1 2 −3 1 0
.
A=
1
1 −2 0 −1 0
0
0
1
1
1 0
5. Find the reduced row-echelon form of B when


0 0 −2 0 7 12
B = 2 4 −10 6 12 28  .
2 4 −5 6 −5 −1
6. In each part suppose that the augmented matrix for a system of linear equations
has been reduced by row operations to the given reduced row-echelon form. Solve
the system.




1 0 0 −7 8
1 0 0 −3
2 
(b)  0 1 0 3
(a)  0 1 0 0 
0 0 1 7
0 0 1 1 −5


1 −3 0 2
5
1
2
0
1
(c)  0 0 1 −2 
(d)
0 0 1 3 4
0 0 0 0




1 −6 0 0 3 −2
1 −3 0 0
 0 0 1 0 4 7 

(f)  0 0 1 0 
(e) 
 0 0 0 1 5 8 
0 0 0 1
0 0 0 0 0 0
MA131 (Section 750001): Prepared by Dr. Archara Pacheenburawana
13
7. Solve each of the following systems by Gauss-Jordan elimination.
(a)
x1 + x2 + 2x3 = 8
−x1 − 2x2 + 3x3 = 1
3x1 − 7x2 + 4x3 = 10
(b)
2x1 + 2x2 + 2x3 = 0
−2x1 + 5x2 + 2x3 = 1
8x1 + x2 + 4x3 = −1
(c) 2x1 − 3x2 = −2
2x1 + x2 = 1
3x1 + 2x2 = 1
(d)
−2b + 3c = 1
3a + 6b − 3c = −2
6a + 6b + 3c = 5
(e)
4x1 − 8x2 = 12
3x1 − 6x2 = 9
−2x1 + 4x2 = −6
(f)
x1 + 3x2 + x3 + x4 = 3
2x1 − 2x2 + x3 + 2x4 = 8
3x1 + x2 + 2x3 − x4 = −1
(g)
x − y + 2z − w = −1
2x + y − 2z − 2w = −2
−x + 2y − 4z + w = 1
3x
− 3w = −3
(h)
3x1 + 2x2 − x3 = −15
5x1 + 3x2 + 2x3 =
0
3x1 + x2 + 3x3 = 11
−6x1 − 4x2 + 2x3 = 30
(i)
x1 + x2 + x3 + x4 = 0
2x1 + x2 − x3 + 3x4 = 0
x1 − 2x2 + x3 + x4 = 0
(j)
10y − 4z + w = 1
x + 4y − z + w = 2
3x + 2y + z + 2w = 5
−2x − 8y + 2z − 2w = −4
x − 6y + 3z
= 1
8. Solve each of the systems in Exercise 2 by Gaussian elimination.
9. Determine which of the following homogeneous systems have nontrivial solutions.
(a) 2x1 − 3x2 + 4x3 − x4 = 0
7x1 + x2 − 8x3 + 9x4 = 0
2x1 + 8x2 + x3 − x4 = 0
(b) x1 + 3x2 − x3 = 0
x2 − 8x3 = 0
4x3 = 0
(c) a11 x1 + a12 x2 + a13 x3 = 0
a21 x1 + a22 x2 + a23 x3 = 0
(d) 3x1 − 2x2 = 0
6x1 − 4x2 = 0
10. Solve the following homogeneous systems of linear equations by any method.
(a) 2x1 + x2 + 3x3 = 0
x1 + 2x2
=0
x2 + x3 = 0
(b) 3x1 + x2 + x3 + x4 = 0
5x1 − x2 + x3 − x4 = 0
(c) −x1 + x2 − x3 + 3x4 = 0
3x1 + x2 − x3 − x4 = 0
2x1 − x2 − 2x3 − x4 = 0
(d) − 21 x1 + 13 x2 + 21 x3 = 0
1
x − 23 x2 + 41 x3 = 0
4 1
1
x + 13 x2 − 34 x3 = 0
4 1
(e)
(f)
2x + 2y + 4z
w
− y − 3z
2w + 3x + y + z
−2w + x + 3y − 2z
=0
=0
=0
=0
2x − y − 3z = 0
−x + 2y − 3z = 0
x + y + 4z = 0
MA131 (Section 750001): Prepared by Dr. Archara Pacheenburawana
14
11. Consider a linear system whose augmented matrix is of the form


1
2 1 1
 −1 4 3 2 
2 −2 α 3
For what values of α will the system have a unique solution?
12. Consider a linear system whose augmented matrix is of the form


1 2 1 0
 2 5 3 0 
−1 1 β 0
(a) Is it possible for the system to be consistent? Explain.
(b) For what values of β will the system have infinitely many solutions?
13. Consider a linear system whose augmented

1 1 3
 1 2 4
1 3 α
matrix is of the form

2
3 
β
(a) For what values of α and β will the system have infinitely many solutions?
(b) For what values of α and β will the system be inconsistent?
14. Solve the system
2x1 − x2
= λ x1
2x1 − x2 + x3 = λ x2
−2x1 + 2x2 + x3 = λ x3
for x1 , x2 , and x3 in the two cases λ = 1, λ = 2.
Answer to Exercise 1.2
1. (a), (b), (d), (g), (h) 2. (a), (b), (c), (f), (g)
3. (a)

1
0
4. 
0
0
Row-echelon
1
0
0
0
0
1
0
0
0
0
1
0
1
1
0
0
(b) Neither (c) Both (d) Row-echelon



0
1 2 0 3 0 7

0
5. 0 0 1 0 0 1
0
0 0 0 0 1 2
0
(e) Both
(f) Both
6. (a) x1 = 3, x2 = 0, x3 = 7 (b) x1 = 7t + 8, x2 = −3t + 2, x3 = −t − 5, x4 = t
(c) x1 = 2 + 3t, x2 = t, x3 = −2 (d) x1 = 5 − 2s − t, x2 = s, x3 = 4 − 3t, x4 = t
(e) x1 = 6s − 3t − 2, x2 = s, x3 = −4t + 7, x4 = −5t + 8, x5 = t
(f) Inconsistent
MA131 (Section 750001): Prepared by Dr. Archara Pacheenburawana
7. (a) x1 = 3, x2 = 1, x3 = 2 (b) x1 = − 17 − 73 t, x2 =
1
7
15
− 74 t, x3 = t
(c) Inconsistent (d) Inconsistent (e) x1 = 3 + 2t, x2 = t
(f) x1 =
3
4
− 85 t, x2 = − 14 − 18 t, x3 = t, x4 = 3
(g) x = t − 1, y = 2s, z = s, w = t (h) x1 = −4, x2 = 2, x3 = 7
(i) x1 = − 34 t, x2 = 0, x3 = 13 t, x4 = t
(j) x =
14
5
7
− 65 t, y = − 10
+
3
10
t−
1
s,
10
z = −2 + t, w = t
9. (a), (c), (d)
10. (a) x1 = 0, x2 = 0, x3 = 0
(b) x1 = −s, x2 = −t − s, x3 = 4s, x4 = t
(c) x1 = t, x2 = −t, x3 = t (d) x1 = 53 t, x2 = t, x3 = t
(e) w = t, x = −t, y = t, z = 0 (f) x = 0, y = 0, z = 0
11. α 6= −2 12. β = 2 13. (a) α = 5, β = 4, (b) α = 5, β 6= 4
14. If λ = 1, then x1 = x2 = s, x3 = 0. If λ = 2, then x1 = − 21 s, x2 = 0, x3 = s
1.3
Matrices and Matrix Operations
In this section we begin our study of matrix theory by giving some of the fundamental
definitions of the subject. We shall see how matrices can be combined through the
arithmetic operations of addition, subtraction, and multiplication.
Matrix Notation and Terminology
Definition 1.1.
A matrix is a rectangular array of numbers, each of which is called an entry
of the matrix.
For example,


2 1
−1 2
0 5
1 −1 0
0 5 0
and
√
2 π
1
(1.2)
(1.3)
(1.4)
Each horizontal line of number is called a row of the matrix; each vertical one is
called a column. For example, in matrix (1.2) the third row consist of the entries 0 and
5, whereas the entries in the first column are 2, −1, and 0.