Chapter Three

Matrices, Determinant and Systems of Linear Equation

Matrices, which are also known as rectangular arrays of numbers or functions, are the main
tools of linear algebra. Matrices are very important to express large amounts of data in
an organized and concise form. Furthermore, since matrices are single objects, we denote
them by single letters and calculate with them directly. All these features have made matri-
ces very popular for expressing scientific and mathematical ideas. Moreover, application of
matrices are found in most scientific fields; such as economics, finance, probability theory
and statistics, computer science, engineering, physics, geometry, and other areas.

Main Objectives of this Chapter

At the end of this chapter, students will be able to:-

• Understand the notion of matrices and determinants

• Use matrices and determinants to solve system of linear equations

• Apply matrices and determinants to solve real life problems

3.1 Definition of Matrix

Consider an automobile company that manufactures two types of vehicles, Trucks and
Passenger cars in two different colors, red and blue. The company’s sales for the month of
January are 15 Trucks and 20 Passenger cars in red color, and 10 Trucks and 16 Passenger
cars in blue color. This data is presented in Table 1.

Table 1

Trucks Passenger Cars

Red 15 20
Blue 10 16

The information in the table can be given in the form of rectangular arrays of numbers as

C1 C2
 
R1 15 20
.
R2 10 16

In this arrangement, the horizontal and vertical lines of numbers are called rows (R1 , R2 )
and columns (C1 , C2 ), respectively. The columns C1 and C2 represent the Trucks and
Passenger cars, respectively, which are sold in January. And the rows R1 and R2 represent

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the red and blue colored vehicles, respectively. An arrangement of this type is called a
matrix. Note that the above matrix has two rows and two columns. This shows us the
usefulness of matrix to organize information.

Definition 3.1 (Definition of Matrix). If m and n are positive integers, then by a

matrix of size m by n, or an m × n matrix, we shall mean a rectangular array
consisting of mn numbers, or symbols, or expressions in a boxed display consisting
of m rows and n columns. This can be denoted by

C1 C2 C3 Cn
 
R1 a11 a12 a13 ... a1n
R2 
 a21 a22 a23 ... a2n 

R3 
 a31 a32 a33 ... a3n 

..  .. .. .. .. 
.  . . . . 
Rm am1 am2 am3 . . . amn

where (R1 , R2 , R3 , ..., Rm ) and (C1 , C2 , C3 , ..., Cn ) represent the m rows and n
columns, respectively.

Remark.

1. Note that the first suffix denotes the number of a row (or position) and the second
suffix that of a column, so that aij appears at the intersection of the i-th row and the
j-th column.

2. Matrix A of size m × n may also be expressed by

A = [aij ]m×n ,

where aij represents the (i, j)-th entry of the matrix [aij ].

Example 3.1. The following are matrices of different

 size.

  a b c
a b
A= is a 2 × 2 matrix B =  b c d is 3 × 3 matrix
c d
c d e
   
1 2 1
2 3 2
C= 3 4 is 4 × 2 matrix D= 3 is 4 × 1 matrix
 

4 5 4
 
a b c d  
E= is 2 × 4 matrix, F = b c d e is 1 × 4 matrix
b c d e

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Definition 3.2. Matrices which are n×1 or 1×n are called vectors. Thus, the n×1 matrix
 
a11
 a21 
A= . 
 
 .. 
a2n
is called a column vector, and the 1 × n matrix
 
B = b11 b12 . . . b1n

is called a row vector.

Definition 3.3 (Submatrix). Let A be an m × n matrix. A submatrix of matrix A is

any matrix of size r × s with r ≤ m and s ≤ n, which is obtained by deleting any
collection of rows and/or columns of matrix A.
 
1 2 3
Example 3.2. For the given matrix A = 2 3 4 ,
3 4 5
 
1 2 3
(i) is a submatrix of A, which is obtained by deleting the third row of A.
2 3 4
 
1 3
(ii) 2 4 is a submatrix of A, which is obtained by deleting the second column of A.
3 5
 
3 4
(iii) is a submatrix of A, which is obtained by deleting the first column and first
4 5
row of A.

Definition 3.4 (Equality of Matrices). Two matrices of the same size, A = [aij ]m×n
and B = [bij ]m×n , are said to be equal (and write A = B) if and only if

aij = bij , for all ij.

Example 3.3.
(a) Determine the values of a, b, c and d for which the matrices A and B are equal:
   
5 4 a b
A= , B= .
0 2 c d

Solution: By Definition 3.4, we have a11 = b11 implies a = 5, a12 = b12 implies
b = 4, a21 = b21 implies c = 0 and a22 = b22 implies d = 2.

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 (b) Find the values of α and β for which the given matrices A and B are equal.
   
1 2 α−β 2
A= , B=
3 −1 α −1

Solution: Similarly, we have a11 = b11 implies α − β = 1, a21 = b21 implies α = 3,

and hence β = 2.

Definition 3.5 (Zero Matrix). An m × n matrix A = [aij ] is said to be the zero

matrix if aij = 0 for all ij.

Example 3.4. The following are zero matrices.

 
  0 0
  0 0 0  
0 0 0 0 0 0 0 0
, 0 0 0 ,  ,
0 0 0 0 0 0 0 0
0 0 0
0 0

Exercise 3.1.

1. Write out the matrix of size 3 × 3 whose entries are given by xij = i + j.

2. Write out the matrix of size 4 × 4 whose entries are given by


 1 if i > j
xij = 0 if i = j
−1 if i < j.


 
1 2 3
3. For the matrix A = 2 3 4, give all the submatrices of size 2 × 2.
3 4 5

3.2 Matrix Algebra

In this section, we discuss addition of matrices, scalar multiplication, and matrix multipli-
cation.

3.2.1 Addition and Scalar Multiplication

Addition and scalar multiplication are the basic matrix operations. To see the usefulness of
these operations, let us observe the following simple application.

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Consider again an automobile company that manufactures two types of vehicles, Trucks
and Passenger cars in two different colors, red and blue. If the sales for the months of
January and February, respectively, are given by
   
15 20 12 28
J= and F = ,
10 16 20 14

then the total sales for two months can be given as follows. The total number of red Trucks
sold in two months is 15 + 12 = 27. Similarly, the total number of blue Trucks, red Pas-
senger cars and blue Passenger cars sold in the two months are given by 10 + 20 = 30,
20 + 28 = 48 and 16 + 14 = 30, respectively.

The preceding computations are examples of matrix addition. We can write the sum of two
2 × 2 matrices indicating the sales of January and February as
       
15 20 12 28 15 + 12 20 + 28 27 48
J +F = + = = .
10 16 20 14 10 + 20 14 + 16 30 30

Definition 3.6. Let A = [aij ]m×n and B = [bij ]m×n be two matrices of the same
size. Then the sum of A and B, denoted by A + B, is the m × n matrix defined by
the formula
A + B = [aij + bij ].
The sum of two matrices of different sizes is undefined.

Example 3.5. For the given matrices A, B, C, D compute A + B and C + D.

       
a b w x 1 0 4 1 1 0
A= , B= , C= , D=
c d y z −1 1 1 0 −2 3

Solution: Using Definition 3.6, we have

     
a b w x a+w b+x
A+B = + =
c d y z c+y d+z

and
     
2 0 4 1 1 0 3 1 4
C +D = + = .
−1 1 1 0 −2 3 −1 −1 4

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 Theorem 3.1 (Laws of Matrix Addition ). Let A, B, C be matrices of the same size
m × n, 0 the m × n zero matrix. Then

1. Closure Law of Addition: A + B is an m × n matrix.

2. Associative Law: (A + B) + C = A + (B + C).

3. Commutative Law : A + B = B + A.

4. Identity Law : A + 0 = A.

5. Inverse Law : A + (−A) = 0.

Definition 3.7 (Scalar Multiplication). Let A = [aij ] be an m × n matrix, and α a

scalar. Then the product of the scalar α with matrix A, denoted by αA, is defined by

αA = [αaij ]m×n .

Example 3.6. Consider the automobile manufacturing company once again. Suppose the
company’s sales for the months of January and March, respectively, are given by
   
15 20 18 22
J= , and M = .
10 16 14 20
(a) If the sales of January is to be doubled in February, then the sales of February should
be    
2(15) 2(20) 30 40
2J = = .
2(10) 2(16) 20 32
(b) If the sales of March is to be declined by 50% in April, then the sales of April should
be 1 1
  
1 (18) (22) 9 11
( )J = 12 2 = .
2 2
(14) 12 (20) 7 10
Example 3.7. Given the matrices A and B, compute 4A and A + (−1)B.
   
1 2 2 4
A= , B=
3 4 1 3

Solution: Using Definition 3.7, we have

     
1 2 4(1) 4(2) 4 8
4A = 4 = = .
3 4 4(3) 4(4) 12 16
And, from the definitions 3.6 and 3.7, we have
         
1 2 2 4 1 2 −2 −4 −1 −2
A + (−1)B = + (−1) = + = .
3 4 1 3 3 4 −1 −3 2 1

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From this example, we observe that the difference of two matrices A and B, which is
denoted by A − B, can be defined by the formula

A − B = A + (−1)B = [aij − bij ]m×n .

Theorem 3.2 (Laws of Scalar Multiplication). Let A, B be matrices of the same

size m × n, and α and β scalars. Then

1. Closure Law of Scalar Multiplication: αA is an m × n matrix.

2. Associative Law: α(βA) = (αβ)A.

3. Distributive Law: α(A + B) = αA + αB.

4. Distributive Law: (α + β)A = αA + βA.

5. Monoidal Law: 1A = A.

Example 3.8. Let      

1 2 2 0 a b
A= , B= , C=
0 1 1 1 c d
be the given matrices. Then,
       
1+2 2+0 3 2 (2)3 (2)2 6 4
2(A + B) = 2 =2 = =
0+1 1+1 1 2 (2)1 (2)2 2 4

and
         
(2)1 (2)2 (2)2 (2)0 2 4 4 0 6 4
2A + 2B = + = + = .
(2)0 (2)1 (2)1 (2)1 0 2 2 2 2 4

Thus, we have 2(A + B) = 2A + 2B.

Example 3.9. Solve for X in the matrix equation 2X + A = B, where
   
4 0 6 −4
A= , and B = .
−2 2 8 0

Solution: We begin by solving the equation for X to obtain

1
2X = B − A implies X = ( )(B − A).
2
Thus, we have the solution
     
1 6−4 −4 − 0 1 2 −4 1 −2
X= = = .
2 8 − (−2) 0 − 2 2 10 −2 5 −1

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3.2.2 Matrix Multiplication
An other important matrix operation is matrix multiplication. To see the usefulness of this
operation, consider the application below, in which matrices are helpful for organizing in-
formation.

A football stadium has three concession areas, located in South, North and West stands.
The top-selling items are, peanuts, hot dogs and soda. Sales for one day are given in the
first matrix below, and the prices (in dollar) of the three items are given in the second matrix
(note that the price per Peanuts, Hot dogs and Soda are given by $2.00, $3.00 and $2.75,
respectively).

Peanuts Hot dogs Sodas

   
South Stand 120 250 305 2.00
North Stand  207 140 419   3.00 .
West Stand 29 120 190 2.75

To calculate the total sales of the three top-selling items at the South stand, multiply each
entry in the first row of the matrix on the left by the corresponding entry in the price column
matrix on the right and add the results. Thus, we have

120(2.00) + 250(3.00) + 305(2.75) = 1828.75$ (South stand sales).

Similarly, the sales for the other two stands are given below:

207(2.00) + 140(3.00) + 419(2.75) = 1986.25$ (North stand sales).

29(2.00) + 120(3.00) + 190(2.75) = 940.5$ (West stand sales).

The preceding computations are examples of matrix multiplication. We can write the prod-
uct of the 3 × 3 matrix indicating the number of items sold and the 3 × 1 matrix indicating
the selling prices as shown below.

     
120 250 305 2.00 1828.75
 207 140 419   3.00  =  1986.25 
29 120 190 2.75 940.5

The product of these matrices is the 3 × 1 matrix giving the total sales for each of the three
stands.

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 Definition 3.8 (Matrix Multiplication). Let A = [aij ]m×n and B = [bij ]n×p be two
matrices. Then the product of A and B, denoted by AB, is an m × p matrix whose
(i, j)-th entry is defined by the formula
n
X
[AB]ij = aik bkj = ai1 b1j + ai2 b2j + ai3 b3j + ... + ain bnj.
k=1

In the other words, the (i, j)-th entry of the product AB is obtained by summing the
products of the elements in the i-th row of A with corresponding elements in the j-th
column of B.

The above definition can be understood as follows. If

 
A = a11 a12 . . . a1n

has only one row (R1 ), and  

b11
 b21 
B= . 
 
 .. 
bn1
has only one column (C1 ), then product AB is given by
 
b11
   b21 

AB = [R1 C1 ] = a11 a12 . . . a1n  .  = a11 b11 + a12 b21 + ... + a1n bn1 .

.
 . 
bn1

If A has m rows R1 , R2 , ..., Rm , and B has n columns C1 , C2 , ..., Cp , then the product AB
can be given by the formula
 
R1 C1 R1 C2 . . . R1 Cp
 R2 C1 R2 C2 . . . R2 Cp 
AB =  . ..  .
 
..
 .. . ... . 
Rm C1 Rm C2 . . . Rm Cp

That is, the (i, j)-th entry of AB is Ri Cj .

Remark. The product AB of two matrices A and B is defined only if the number of
columns in A and the number of rows in B are equal.

113
  
  b11 b12
a a a
Example 3.10. Let A = 11 12 13 and B = b21 b22  be two matrices. Clearly,
a21 a22 a23
b31 b32
the product AB is defined in this case, since the number of column of A and the number of
rows of B are equal. Thus, we have
   
R1 C1 R1 C2 a11 b11 + a12 b21 + a13 b31 a11 b12 + a12 b22 + a13 b32
AB = = .
R2 C1 R2 C2 a21 b11 + a22 b21 + a23 b31 a21 b12 + a22 b22 + a23 b32
In this example, the matrices A and B, respectively, are 2 × 3 and 3 × 2 matrices, whereas
the product AB is a 2 × 2 matrix.
Example 3.11. Compute the product AB of the given matrices
 
  1 1
A = 1 2 3 and B = 1 −1 .
1 2

Solution: The product AB is defined since the number of columns in matrix A and the
number of rows in matrix B are equal. Thus, we have AB is given by
 
  1 1    
1 2 3 1 −1 = (1)(1) + (2)(1) + (3)(1) (1)(1) + (2)(−1) + (3)(2) = 6 5 .
1 2

Note that the product BA is not defined in this case.

   
0 1 1 0
Example 3.12. Let A = and B = be the given matrices. Then, we have
0 0 0 0
         
0 1 1 0 0 0 1 0 0 1 0 1
AB = = , and BA = = .
0 0 0 0 0 0 0 0 0 0 0 0
In this example, we observe that both the products AB and BA are defined. This is true in
general i.e., the products AB and BA are defined for any two square matrices A and B of
the same size. For the matrices A and B given above, we have AB 6= BA. Hence, matrix
multiplication is not commutative.
Example 3.13. Consider the following diagonal matrices.
   
a11 0 0 b11 0 0
A =  0 a22 0  , and B =  0 b22 0 
0 0 a33 0 0 b33

The product AB is given by

    
a11 0 0 b11 0 0 a11 b11 0 0
AB =  0 a22 0   0 b22 0  =  0 a22 b22 0 
0 0 a33 0 0 b33 0 0 a33 b33

114
Similarly, we have  
b11 a11 0 0
BA =  0 b22 a22 0 .
0 0 b33 a33
In this case, we have AB = BA , and hence the given matrices A and B commute. More
generally, if A and B are any two diagonal matrices of the same size, then AB = BA.

Theorem 3.3. Matrix multiplication is associative, i.e., whenever the products are
defined, we have A(BC) = (AB)C.

From Theorem 3.3, we shall write ABC for either A(BC) or (AB)C. Also, for every
positive integer n, we shall write An for the product AAA...A (n terms).

Theorem 3.4. If all multiplications and additions make sense, the following hold for
matrices, A, B, C and α, β scalars.

1. A(αB + βC) = α(AB) + β(AC).

2. (αB + βC)A = α(BA) + β(CA) .

Exercise 3.2.

1. Find your own examples:

(i) 2 × 2 matrices A and B such that A 6= 0, B 6= 0 with AB 6= BA.

(ii) 2 × 2 matrices A and B such that A 6= 0, B 6= 0 but AB = 0.
(iii) 2 × 2 matrix A such that A2 = I2 and yet A 6= I2 and A 6= −I2 .
 
−1 −1
2. Let A = . Find all 2 × 2 matrices, B such that AB = 0.
3 3
   
1 2 1 2
3. Let A = and B = . Is it possible to choose c so that AB = BA? If
3 4 1 c
so, what should be the value of c?
     
1 3 −1 2 2 1
4. Given the matrices A = ,B = , and C = and α a scalar
2 4 0 1 4 0

i. Compute the products A(BC), (AB)C, and verify that A(BC) = (AB)C.
ii. Compute the products α(AB), (αA)B, A(αB)), and verify that

α(AB) = (αA)B = A(αB).

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 5. Consider the automobile producer whose agency’s sales for the month of January
were given by  
15 20
J= .
10 16
Suppose that the price of a Truck is $200 and that of a Passenger car is $100. Use
matrix multiplication to find the total values of the red and blue vehicles for the
month of January.

3.3 Types of Matrices

There are certain types of matrices that are so important that they have acquired names
of their own. In this section we are going to discuss some of these matrices and their
properties.

Definition 3.9 (Square Matrix). A matrix A is said to be square if it has the same
number of rows and columns. If A has n-rows and n-columns, we call it a square
matrix of size n.

Example 3.14. The following are square matrices.

 
a b
A= (Square matrix of size 2)
c d
 
1 2 −1
B= 0 1
 3  (Square matrix of size 3)
4 2 −2
 
c11 c12 . . . c1n
 c21 c22 . . . c2n 
C= . ..  (Square matrix of size n)
 
.. . .
 .. . . . 
cn1 cn2 . . . cnn

Definition 3.10 (Identity Matrix). A square matrix A = [aij ]n×n is called an iden-
tity matrix if

1, if i = j
aij =
0, otherwise

and it is denoted by In .

Example 3.15. The following are identity matrices.

 
1 0
I2 = (Identity matrix of size 2)
0 1

116
  
1 0 0
I3 = 0
 1 0 (Identity matrix of size 3)
0 0 1
 
1 0 ... 0
0 1 . . . 0
In =  . .. . . ..  (Identity matrix of size n)
 
 .. . . .
0 0 ... 1

Definition 3.11 (Diagonal Matrix). A square matrix D = [dij ]n×n is said to be

diagonal if dij = 0 whenever i 6= j. Less formally, D is said to be diagonal when all
the entries off the main diagonal are 0.

Example 3.16. The following are diagonal matrices.

 
1 0
D= (Diagonal matrix of size 2)
0 1
 
2 0 0
D = 0 4 0 (Diagonal matrix of size 3)
0 0 5
 
0 0 0
D = 0 3 0  (Diagonal matrix of size 3)
0 0 −2
 
0 0 0
D = 0 0 0 (Diagonal matrix of size 3)
0 0 0
Note that the identity matrix is the special case of diagonal matrix where all the entries in
the main diagonal are 1.

Definition 3.12 (Scalar Matrix). A diagonal matrix in which all diagonal entries
are equal is called a scalar matrix.

Example 3.17. The following are scalar matrices.

   
  2 0 0 1 0 0
3 0
(a) (b) 0 2 0 (c) 0 1 0
0 3
0 0 2 0 0 1

Definition 3.13 (Triangular Matrix). A square matrix A = [aij ]n×n is said to be

lower triangular if and only if aij = 0 whenever i < j. A is said to be upper
triangular if and only if aij = 0 whenever i > j.

117
Example 3.18.
     
3 2 1 0 −1 1 0 0 0
(i) 0 2 7 , 0 0 0 , 0 4 0 (Upper triangular matrices).
0 0 3 0 0 0 0 0 0
     
3 0 0 0 0 0 1 0 0
(ii)  1 2 0 ,
 1 0 0 , 0 2 0 (Lower triangular matrices).
−2 4 3 0 2 0 0 0 3
Remark.

(a) In the lower triangular matrix all the entries above the main diagonal are zero, whereas
in the upper triangular matrix all the entries below the main diagonal are zero.

(b) Any diagonal matrix is both upper and lower triangular.

Definition 3.14 (Transpose of Matrix). Let A = [aij ] be an m × n matrix . Then by

the transpose of A we mean the n × m matrix, denoted by At , whose (i, j)-th entry
is the (j, i)-th entry of A. More precisely, if A = [aij ]m×n , then At = [aji ]n×m . That
is,    
a11 a12 . . . a1n a11 a21 . . . am1
 a21 a22 . . . a2n   a12 a22 . . . am2 
t
A= . ..  , then A =  .. ..  .
   
. .. ..
 . . ... .   . . ... . 
am1 am2 . . . amn a1n a2n . . . anm

Note that the k-th row of matrix A becomes k-th column of At , and the k-th column of A
becomes k-th row of At .
Example 3.19. Compute the transposes of the following matrices.
 
  2 1 3
1 −1 −1
A= , B = 1 5 −3
1 2 3
3 −3 7

Solution: First let us consider matrix A. Now, row 1 of matrix A becomes column 1 of At ,
and row 2 of A becomes column 2 of At . Thus, we have
 
1 1
At = −1 2 .
−1 3

Similarly,  
2 1 3
B t = 1 5 −3 .
3 −3 7

118
 Definition 3.15 (Symmetric Matrix). A square matrix A is said to be Symmetric if
A = At .

Example 3.20. Distinguish whether the given matrix is symmetric or not.

   
2 1 3 1 1 3
(a) A = 1 5 −3 (b) B = 1 2 2
3 −3 7 3 2 3

Solution:
   
0 1 3 0 −1 −3
(a) For the matrix A = −1 0 2, At = 1 0 −2. Thus, we have A 6= At ,
−3 −2 0 3 2 0
and hence A is not symmetric.
   
1 1 3 1 1 3
t
(b) For the matrix B = 1 2 2 , B = 1 2
   2. Thus, we have B = B t , and
3 2 3 3 2 3
hence B is symmetric.

Theorem 3.5 (Properties of Matrix Transpose). When the relevant sums and prod-
ucts are defined, and α is a scalar. Then

1. (At )t = A.

2. (A + B)t = At + B t .

3. (αA)t = α(At ).

3. (AB)t = B t At .

Exercise 3.3.
   
1 −1 3 −2
For the given matrices A = , and B = :
3 2 0 1

(a) Show that (At )t = A.

(b) Show that (A + B)t = At + B t .
(c) Show that (4A)t = 4(At ).
(d) Show that(AB)t = B t At .

119
3.4 Elementary row operations
Elementary row operations are useful to find the rank of a matrix (see Section 3.6), to com-
pute the determinants of matrices (see Section 3.7), and to find the inverse of a matrix (see
Section 3.8). Furthermore, elementary row operations are widely used in solving systems
of linear equations (see Section 3.9).

In this section, we introduce the elementary row operations and apply these operations to
transform the given matrix into different form.

Definition 3.16 (Elementary Row Operations).

Let A be an m × n matrix. The following are known as elementary row operations.

1. Interchanging two rows: Ri ↔ Rj .(Rule of Interchanging)

2. Multiplying a row by a nonzero scalar: Ri → αRi (α is a nonzero scalar).

(Rule of Scaling)

3. Adding a multiple of one row to another: Ri → Ri + αRj (α is a nonzero

scalar). (Rule of Replacement)

Example 3.21. Use elementary row operations to transform the given matrix A into, (a) an
upper triangular matrix, (b) an identity matrix.
 
3 12 6
A = 1 1 −1
1 2 3

Solution: Consider the given matrix A:

(a) First let us transform the matrix A into an upper triangular. This can be done as fol-
lows:   
3 12 6 1 4 2
A = 1 1 −1 R1 → ( 13 )R1 1 1 −1(Scaling R1 )
1 2 3 1 2 3
 
1 4 2
R2 → R2 + (−1)R1 , R3 → R3 + (−1)R1 0 −3 −3 (Replacing R2 and R3 )
0 −2 1
 
1 4 2
1
R2 → (− 3 )R2 0 1 1 (Scaling R2 )
0 −2 1
 
1 4 2
R3 → R3 + 2R2 0 1 1(Replacing R3 )
0 0 3

120
  
1 4 2
Hence, the matrix 0 1 1 is an upper triangular, which is obtained from A by
0 0 3
elementary row operations.

(b) To transform the matrix A into a diagonal matrix, we simply change all the entries
above the main diagonal into zeros and the entries in the main diagonal into 1. Let
us denote
 the above
 upper triangular
  by B. Then we have
matrix
1 4 2 1 4 2
1
B = 0 1 1 R3 → ( 3 )R3 0 1 1 (Scaling R3 )
  
0 0 3 0 0 1
 
1 4 0
R2 → R2 + (−1)R3 , R1 → R1 + (−2)R3 0 1 0 (Replacing R1 and R2 )
0 0 1
   
1 0 0 1 0 0
R1 → R1 + (−4)R2 0 1 0 (Replacing R1 ). Thus, I3 = 0 1 0 is the
0 0 1 0 0 1
identity matrix obtained from A.

Definition 3.17. Two matrices are said to be raw equivalent if one can be obtained
from the other by a sequence of elementary row operations.

Example 3.22. Let A, B, I3 be the matrices in Example 3.21. Then, A is row equivalent
to both B and the identity matrix I3 . Also the matrix B is row equivalent to the identity
matrix I3 .
Exercise 3.4.
 
4 3
1. Given the matrix A = , use elementary row operations to find the lower trian-
2 1
gular matrix which are row equivalent to A.
 
0 1 1
2. Given the matrix B = 1 0 1, use elementary row operations to find an identity
1 1 0
matrix which is row equivalent to B.

3.5 Row Echelon Form and Reduced Row Echelon Form of a Matrix
In order to find the rank, or to compute the inverse of a matrix, or to solve a linear system,
we usually write the matrix either in its row echelon form or reduced row echelon form.

121
 Definition 3.18. An m × n matrix is said to be in echelon form (or row echelon
form) if the following conditions are satisfied:

1. All nonzero rows are above any rows of all zeros.

2. Each leading entry of a row is in a column to the right of the leading entry of
the row above it. (A leading entry refers to the left most nonzero entry in a
nonzero row)

3. All entries in a column below a leading entry are zeros.

If a matrix in row echelon form satisfies the following additional conditions, then it
is in reduced echelon form (or reduced row echelon form)

4. The leading entry in each nonzero row is 1.

5. Each leading 1 is the only nonzero entry in its column.

A matrix in row echelon form is said to be in reduced row echelon when every
column that has a leading 1 has zeros in every position above and below the leading
entry.

Example 3.23. The given matrices A, B, C, D are in row echelon form

     
1 −1 0 4 0 0   1 2 0
1 0 5 2
A = 0 5 0 , B = 0 0 0 , C = , D = 0 5 1
0 0 1 3
0 0 1 0 0 0 0 0 0

and the following are in reduced row echelon form.

     
1 0 0 1 0 0   1 1 0
1 0 0 2
P = 0 1 0 , Q= 0 1 4 , R=
    , S = 0 0 1
0 0 1 2
0 0 1 0 0 0 0 0 0

Theorem 3.6 (Uniqueness of the Reduced Echelon Form). Each matrix is row equivalent
to one and only one reduced echelon matrix.

Definition 3.19. A pivot position in a matrix A is a location in A that corresponds

to a leading 1 in the reduced row echelon form of A. A pivot column is a column
of A that contains a pivot position. A pivot element is a nonzero number in a pivot
position that is used as needed to create zeros via row operations.

122
 To write a matrix in reduced echelon form:

1. Begin with the leftmost nonzero column. This is a pivot column. The pivot
position is at the top.

2. Select a nonzero entry in the pivot column as a pivot. If necessary, interchange

rows to move this entry into the pivot position.

3. Use row replacement operations to create zeros in all positions below the pivot.

4. Cover (or ignore) the row containing the pivot position and cover all rows,
if any, above it. Apply steps 1-3 to the submatrix that remains. Repeat the
process until there are no more nonzero rows to modify.

5. Beginning with the rightmost pivot column and working upward and to the
left, create zeros above each pivot. If a pivot is not 1, make it 1 by a scaling
operation.

Example 3.24. Find the reduced row echelon form of the matrix A.
 
0 0 2 3
A = 0 2 0 1 .
0 1 1 5
Solution:
Step 1: Here, the left most nonzero column is the second column.
Step 2: By row interchanging rule, we can obtain the pivot position as follows;
   
0 0 2 3 0 1 1 5
0 2 0 1 R1 ↔ R3 0 2 0 1
0 1 1 5 0 0 2 3

Step 3:
Now, the leading entry is 1, and to create zeros in all positions below the pivot, we use the
replacement rule:  
0 1 1 5
R2 → R2 + (−2)R1 0 0 −2 −9
0 0 2 3
Step 4:
Now we proceed to the second row. Here, the leading entry is −2. Using a scaling rule we
obtain a leading 1:
 
0 1 1 5
1 0 0 1 9 
R2 → (− )R2 2
2
0 0 2 3

123
And applying row replacement rule:
 
0 1 1 5
R3 → R3 + (−2)R2 0 0 1 29 
0 0 0 −6

And scaling R3 ,  
0 1 1 5
1
R3 ↔ (− )R3 0 0 1 29 
6
0 0 0 1
Step 5: Beginning with the rightmost pivot column, we create zeros above each pivot
element. That is, we start from the fourth column:
 
0 1 1 0
9
R1 → R1 + (−5)R3 , R2 → R2 + (− )R3 0 0 1 0
2
0 0 0 1

And using row replacement (to create zeros above the pivot element in the third column),
 
0 1 0 0
R1 → R1 + (−1)R2 , 0 0 1 0 .
0 0 0 1

Thus, the required matrix in reduced row echelon form is given by

 
0 1 0 0
à = 0 0 1 0 .
0 0 0 1

Exercise 3.5.

1. Determine which matrices are in reduced row echelon form.

   
  1 0 0 0 1 1 0 0 0 5
1 2 0
A= , B = 0 0 1 2 , C = 0 0 1 2 0 4
0 1 7
0 0 0 0 0 0 0 0 1 3

2. Give the row echelon form and also the reduced row echelon form of the following
matrices.
 
1 2 3    
2 1 −2 1 2 1 3 1 2 0 3
A= 3 0 0  , B = −3 2 1 0 , C = 2 1 2 2
    
3 2 1 1 1 1 0 3
3 2 1

124
3.6 Rank of matrix using elementary row operations
The ranks of matrices are useful in determining the number of solutions for linear systems.

Definition 3.20 (Rank of Matrix). Rank of an m×n matrix A, denoted by rank(A),

is the number of nonzero rows of the reduced row echelon form of A.

Example 3.25. Determine the ranks of the following matrices which, are in reduced row
echelon form.
     
1 0 0 1 0 0   1 1 0
1 0 0 2
A = 0 1 0 , B = 0 1 4 , C = , D = 0 0 0
0 0 1 2
0 0 1 0 0 0 0 0 0
Solution: Clearly, all the matrices are in reduced row echelon form. Hence, by Definition
3.20, we have rank(A) = 3 (since the number of nonzero rows in matrix A is 3). Similarly,
rank(B) = 2 (since the number of nonzero rows in matrix B is 2), rank(C) = 2 (since the
number of nonzero rows in matrix C is 2), and rank(D) = 1 (since the number of nonzero
rows in matrix D is 1).
 
1 1 2
Example 3.26. Find rank(A), where A = 2 2 5.
3 3 2
Solution: After a sequence of elementary row operations, we obtain the reduced echelon
form of A, which is given by  
1 1 0
Ae = 0 0 1 .
0 0 0
Thus, rank(A) = 2.

Remark. The matrix A and its transpose At have the same rank. That is

rank(A) = rank(At ).

Example 3.27. Verify that the given matrix A and its transpose At have the same rank.
   
1 1 2 1 0 0
A = 0 1 1  , and At = 1 1 0
0 0 0− 2 1 0
Solution: Observe that the matrix A is in its row echelon form, and hence its rank is 2.
Now, we apply elementary row operations to reduce matrix At into its row echelon form,
and and we get that  
1 1 0
0 1 0
0 0 0
Thus, rank(At ) = 2 = rank(A).

125
Exercise 3.6. Determine the rank of the following matrices.
   
  1 0 0 0 1 1 0 0 0 5
1 2 0
A= , B = 0 0 1 2 , C = 0 0 1 2 0 4
0 1 7
0 0 0 0 0 0 0 0 1 3
 
1 2 3    
2 1 −2 1 2 1 3 1 2 0 3
P =  , Q = −3 2 1 0 , R = 2
   1 2 2
3 0 0
3 2 1 1 1 1 0 3
3 2 1

3.7 Determinant and its properties

The determinant is a function that takes a square matrix as an input and produces a scalar
as an output. It has many beneficial properties for studying, matrices and systems of equa-
tions.
Definition
 3.21 (Determinant of 2 × 2 matrix). The determinant of a 2 × 2 matrix
a c
A= , denoted by det(A) (or |A|), is defined by the formula
d b

a c
det(A) = = ab − cd.
d b
 
5 2
Example 3.28. Find the determinant of a matrix A = .
3 4
Solution: Using Definition 3.21, the determinant of matrix A is given by
5 2
det(A) = = (5)(4) − (3)(2) = 14.
3 4
The determinant of a 3 × 3 matrix can be defined using the determinants of 2 × 2 matrices.
Definition 3.22 (Determinant of 3 × 3 Matrix). Let
 
a11 a12 a13
A = a21 a22 a23 
a31 a32 a33

be a 3 × 3 matrix, and Aij (for i, j = 1, 2, 3) be the 2 × 2 submatrix of A obtained

by deleting the ith -raw and the j th -column of A. Then determinant of A is denoted
by det(A) (or |A|), and is defined as:

|A| = (−1)1+1 a11 |A11 | + (−1)1+2 a12 |A12 | + (−1)1+3 a13 |A13 |
a22 a23 a a a a
= a11 − a12 21 23 + a13 21 22 .
a32 a33 a31 a33 a31 a32

126
 2 4 0
Example 3.29. Compute the determinant of a matrix A = 3 −1 2 .
2 1 1
Solution: Using Definition 3.22, the determinant is given by

2 4 0
−1 2 3 2 3 −1
det(A) = 3 −1 2 = 2 −4 +0
1 1 2 1 2 1
2 1 1

= 2(−1 − 2) − 4(3 − 4) + 0(3 + 2) = −6 + 4 + 0 = −2.

So far we discussed the determinants of 2 × 2 and 3 × 3 matrices. Next we define the
determinant of an n × n matrix for each positive integer n.

Definition 3.23 (Minors and Cofactors).

Let A = (aij )n×n , and Aij be the submatrix of A obtained by deleting the ith -raw
and j th -column of A for i, j = 1, 2, 3, ..., n. Then

(a) The minor for A at location (i, j), denoted by Mij (A), is the determinant of
the submatrix Aij . That is, Mij (A) = det(Aij ).

(b) The cofactor, denoted by Cij (A), for A at location (i, j) is the sighed determi-
nant of the submatrix Aij . That is, Cij (A) = (−1)i+j det(Aij ).

Remark. In Definition 3.23, the cofactor Cij (A) at location (i, j) can be computed
using the following formula:

det(Aij ), if i + j is even
Cij (A) =
−det(Aij ), if i + j is odd.

Example 3.30. Compute the matrix of cofactors for the given matrix.
 
  1 0 2
1 1
(a) A = (b) B = 1 1 3
−1 2
2 0 1

Solution: (a) The minors of A are

M11 (A) = 2, M21 (A) = 1, M12 (A) = −1, M22 (A) = 1,

and the cofactors are

C11 (A) = (−1)1+1 M11 (A) = (1)(2) = 2, C21 (A) = (−1)2+1 M21 (A) = (−1)(1) = −1,

C12 (A) = (−1)1+2 M12 (A) = (−1)(−1) = 1, C22 (A) = (−1)2+2 M12 (A) = (1)(1) = 1.

127
Thus, the matrix of cofactors for A is
 
2 1
[Cij (A)] = .
−1 1

(b) The minors of B are

1 3 0 2 0 2
M11 (B) = = 1, M21 (B) = = 0, M31 (B) = = −2,
0 1 0 1 1 3
1 3 1 2 1 2
M12 (B) = = −5, M22 (B) = = −3, M32 (B) = = 1,
2 1 2 1 1 3
1 1 1 0 1 0
M13 (B) = = −2, M23 (B) = = 0 and M33 (B) = = 1,
2 0 2 0 1 1
and the confactors are

C11 (B) = (−1)1+1 M11 (B) = 1, C21 (B) = (−1)2+1 M21 (B) = 0,
C31 (B) = (−1)3+1 M31 (B) = −2, C12 (B) = (−1)1+2 M12 (B) = 5,
C22 (B) = (−1)2+2 M22 (B) = −3, C32 (B) = (−1)3+2 M32 (B) = −1,
C13 (B) = (−1)1+3 M13 (B) = −2, C23 (B) = (−1)2+3 M23 (B) = 0,
and C33 (B) = (−1)3+3 M33 (B) = 1.
Thus, the matrix of cofactors for B is
 
1 5 −2
[Cij (B)] =  0 −3 0  .
−2 −1 1

Definition 3.24 (Determinants of n × n Matrix). The determinant of a square

matrix A = [aij ] of size n × n, denoted by det(A) (or |A|), is defined recursively as
follows: if n = 1 then det(A) = a11 ; otherwise, we suppose that determinants are
defined for all square matrices of size less than n and specify that
n
X
det(A) = ak1 Ck1 (A) = a11 C11 (A) + a21 C21 (A) + ... + an1 Cn1 (A), (3.1)
k=1

where Cij (A) is the (i, j)-th cofactor of A. The formula (3.1) is called a cofactor
expansion across the 1st column of A.

Example 3.31. Consider the matrices given in Example 3.30,

 
  1 0 2
1 1
A= and B = 1 1 3 .
−1 2
2 0 1

128
The cofactors of matrices A and B, respectively, are given by
 
  1 5 −2
2 1
and  0 −3 0  .
−1 1
−2 −1 1

Now, using Definition 3.24, we have

det(A) = a11 C11 + a21 C21 = (1)(2) + (−1)(−1) = 3, and

det(B) = b11 C11 + b21 C21 + b31 C31 = (1)(1) + (1)(0) + (2)(−2) = −3.
Example 3.32. Compute the determinant of matrix A:
 
1 1 0
A = 0 2 1
1 2 0

(a) by expanding the cofactrs across the 1st row

(b) by expanding the cofactrs across the 1st column

Solution: We have the matrix of cofactors Cij (A), given by

 
−2 1 −2
[Cij (A)] =  0 0 −1 .
1 −1 2

(a) Now, expanding the cofactors across the 1st row, we have

det(A) = a11 C11 (A)+a12 C12 (A)+a13 C13 (A) = (1)(−2)+(1)(1)+(0)(−2) = −1.

(b) Similarly, expanding cofactors across the 1st column, we have

det(A) = a11 C11 (A) + a21 C21 (A) + a31 C31 (A) = (1)(−2) + (0)(0) + (1)(1) = −1.

Observe that the determinant has the same value for expansions of cofactors across the 1st
row as well as the 1st column. This is true in general, i.e., the determinant value is the same
for the expansions of cofactors across any row or any column. This is briefly stated in the
following theorem.

129
 Theorem 3.7. The determinant of an n × n matrix A can be computed by cofactor
expansion across any row or any column. The expansion across ith row is
n
X
det(A) = aij Cij (A) = ai1 Ci1 (A) + ai2 Ci2 (A) + ... + ain Cin (A)
j=1

= (−1)i+1 ai1 |Ai1 | + (−1)i+2 ai2 |Ai2 | + ... + (−1)i+n ain |Ain |
and the expansion across j th column is
n
X
det(A) = aij Cij (A) = a1j C1j (A) + a2j C2j (A) + ... + anj Cnj (A)
i=1

= (−1)1+j a1j |A1j | + (−1)2+j a2j |A2j | + ... + (−1)n+j anj |Anj |

Remark. In Theorem 3.7, if the matrix A (for instance) is of size 3 × 3, then the
determinants can be easily computed as follows.

(i) The expansion across 2nd row is

|A| = −a21 |A21 | + a22 |A22 | + a23 |A23 |.

(ii) The expansion across 3rd column is

|A| = a13 |A13 | − a23 |A23 | + a33 |A33 |.

(iii) The sign + or − can be determined using the pattern.

 
+ − +
− + −
+ − +

(iv) The computation of determinants becomes easier by expanding the cofactors

across a row or column with the most zero entries.

Example 3.33. Compute the determinant of matrix A by expanding the cofactors across an
appropriate row or column.  
1 1 0
A = 0 2 1
1 2 0
Solution: Here, we observe that the 3rd column has more number of zero entries than any
other columns and rows. Thus, the determinant of A (by expanding the cofactors across the

130
3rd column) is given by
det(A) = a13 |A13 | − a23 |A23 | + a33 |A33 | = 0 − 1 + 0 = −1.

Properties of determinats: Let A be the square matrix of size n.

1. If an entire row (or an entire column) consists of zeros, then det(A) = 0.

2. If two rows (or columns) are equal, then det(A) = 0.

3. If one row (or column) is a scalar multiple of another row (or column), then
det(A) = 0.

4. If A, B and C, respectively, are the upper triangular, lower triangular, and

diagonal matrices, given by
     
a11 a12 a13 b11 0 0 d11 0 0
A =  0 a22 a23  , B = b21 b22 0  , D =  0 d22 0  ,
0 0 a33 b31 b32 b33 0 0 d33

then

det(A) = a11 a22 a33 , det(B) = b11 b22 b33 , and det(D) = d11 d22 d33 .

That is, the determinants of the triangular and diagonal matrices are simply
the products of the entries in the main diagonal.

Example 3.34. Determine the determinants of the following matrices.

       
1 2 3 1 1 0 1 1 0 1 1 3
A= 0  0 0 , B = 0 2 0 , C = −2 2 1 , D = −1 −1 −3
1 2 0 1 2 0 1 1 0 1 2 0
Solution: We have, det(A) = 0 (since the entire second row of matrix A consists of zeros),
det(B) = 0 (since the entire third column of matrix A consists of zeros), det(C) = 0 (since
the first and third rows of C are equal), and det(D) = 0 (since the second row of D is a
scalar multiple of the first row).
Example 3.35. Compute the determinants of the following matrices.
     
4 3 −6 3 0 0 4 0 0
A = 0 2 9  , B = 3 4 0 , D = 0 6 0
0 0 3 2 1 5 0 0 5
Solution: Using the properties of determinants, we have
4 3 −6 3 0 0
det(A) = 0 2 9 = (4)(2)(3) = 24, det(B) = 3 4 0 = (3)(4)(5) = 60, and
0 0 3 2 1 5

131
 2 0 0
det(D) = 0 3 0 = (2)(3)(5) = 30.
0 0 5

Theorem 3.8. For any square matrix A, det(A) = det(At ) (Transposition doesn’t
alter determinants).

Example 3.36. For the given matrix A, verify that det(A) = det(At ).
 
1 0 2
A = 2 −1 1
1 1 3
Solution: The transpose of matrix A is given by
 
1 2 1
At = 0 −1 1 .
2 1 3

Now, we have the determinants of A and At are

det(A) = 2, and det(At ) = 2.

Thus, det(A) = det(At ).

Theorem 3.9 (Effects of elementary row operations).

I. If matrix B is obtained from a square matrix A by interchanging any two rows

(i.e., Ri ↔ Rj ), then det(B) = −det(A). (Interchanging)

II. If matrix B is obtained from a square matrix A by multiplying the ith row by a
nonzero scalar α (i.e., Ri → αRi ), then det(B) = αdet(A). (Scaling)

III If matrix B is obtained from a square matrix A by adding scalar multiple of

one row to the other (i.e., Ri → Ri + αRj ), then det(B) = det(A). (Replace-
ment)
 
3 1 0
Example 3.37. Let A = 1 0 1  be the given matrix with det(A) = −2.
0 1 −1
(a) If a matrix B is obtained from A by interchanging the first and second rows
(i.e., R1 ↔ R2 ), then we have
1 0 1
det(B) = 3 1 0 = 2.
0 1 −1

132
 Thus, det(B) = −det(A). Here, we observe that if the row interchanging has been
made two times, then det(B) = (−1)2 det(A) = det(A). In general, if the row
interchanging has been made n times, then det(B) = (−1)n det(A). Thus, det(B) =
det(A) if n is even, and det(B) = −det(A) if n is odd.

(b) If a matrix B is obtained from A by multiplying the second row by 4

(i.e., R2 → 4R2 ), then we have
3 1 0
B = 4 0 4 = −8.
0 1 −1
Thus, det(B) = 4det(A). If each row of matrix A is multiplied by 4, then we have

det(B) = 43 det(A).

More generally, if A is an n × n matrix, and B is obtained by multiplying each row

of A by a nonzero scalar c, then we have det(B) = det(cA) = cn det(A).

(c) If a matrix B is obtained by replacing row 2 (i.e., R2 → R2 + 2R1 ), then

3 1 0
det(B) = 7 2 1 = 2. Thus, det(B) = det(A).
0 1 −1

Remark. Property (III) of determinants in Theorem 3.9 is particularly more inter-

esting, since it doesn’t change the determinant of the original matrix. This property
can be used to transform the given matrix into triangular matrix (upper or lower) for
which the computation of determinants is much easier than computing the determi-
nant of the original matrix directly, which is tedious and computationally inefficient.

Example 3.38. Compute the determinants of the matrices A and B using elementary row
operations.
 
  1 1 2 2
1 1 2 2 3 5 6
A = 2 3 1 , B = 

1 3 5 3

0 1 4
1 1 3 6
Solution:
(a) Consider the given matrix A. Applying the row replacement; R2 → R2 − 2R1 and
then R3 → R3 − R2 , we obtain the following upper triangular matrix.
 
1 1 2
à = 0 1 −3
0 0 7

Therefore, by Theorem 3.9 we have det(A) = det(Ã) = (1)(1)(7) = 7.

133
 (b) Similarly, by applying the row replacement

R2 → R2 − 2R1 , R3 → R3 − R1 , R4 → R4 − R1 ,

we obtain the following row equivalent matrix.

 
1 1 2 2
0 1 1 2
B̃ = 
0

2 3 1
0 0 1 4

Now, the determinant of the matrix B̃ (by expanding the cofactors across the 1st
column and using the determinant of matrix A computed above) is given by

1 1 2 2
1 1 2
0 1 1 2
B̃ = = 1 2 3 1 = (1)(7) = 7.
0 2 3 1
0 1 4
0 0 1 4

Therefore, by Theorem 3.9 we have det(B) = det(B̃) = 7.

Theorem 3.10 (Product Rule).

If A and B are two matrices for which the product AB is defined, then

det(AB) = det(A)det(B).
  
1 2 2 0
Example 3.39. Let A = and B = be the given matrices. Then verify that
3 −1 1 4

det(AB) = det(A)det(B).

Solution: Here, we have

 
4 8 1 2 2 0
AB = , det(AB) = −56, det(A) = = −7, and det(B) = = 8.
5 −4 3 −1 1 4

Thus,
det(A)det(B) = (−7)(8) = −56 = det(AB).

Definition 3.25 (Definition of rank using Determinant). Let A be an m×n matrix.

Then rank(A) = r, where r is the largest number such that some r × r submatrix of
A has a nonzero determinant.

134
  
1 0 2 −1
Example 3.40. Compute the the rank of matrix A = using determinants.
2 −3 2 0
 possible size of any square submatrix of A is 2 × 2. We
Solution: Observe that,the largest
1 0
have (say) a submatrix (which is obtained by deleting the last two columns of A)
2 −3
1 0
with = −3 6= 0. Therefore, rank(A) = 2.
2 −3
Exercise 3.7.

1. Compute the determinants of the following matrices using elementary row opera-
tions.  
  1 0 −2
1 2
A= , B =  5 −3 −1
3 4
−2 0 1

2. Compute the determinants of the following matrices by expanding cofactors across

any appropriate row or column.
 
  0 3 1 0 2
  1 3 0 1
1 3 0 −1 2 0 1
0 2 1 0 2
 
A = −1 2 0 , B =  ,C =  5 1 −1 3 3

 5 0 0 0  
6 1 2 0 0 1 0 0
 
4 1 1 2
4 1 1 0 1

3. Compute the matrix of cofactors for the given matrices.

 
  3 2 1 1
  1 0 −2
1 −2 −1 0 2 0
A= , B = −1 1 4  , C =  
2 3 4 1 −1 0
2 0 3
3 0 1 0

4. Determine the ranks of the following matrices using determinants.

   
 1
 2 0 1 0 0
1 2 3 0 1 3 2 1
 , C = 4
 1 1
A =  2 1 3 2 4 , B = 
2

1 0 2 1 0
−1 2 1 3 1
0 2 1 0 2 0

3.8 Adjoint and Inverse of a Matrix

The inverses of matrices are useful to solve linear systems. In this section, we define the
inverse of a matrix, we discuss different methods to compute an inverse, and the properties
of inverses.

135
 Definition 3.26 (Adjoint of a Matrix). Let A be an n×n matrix. If [Cij (A)] denotes
the matrix of cofactors for A, then the adjugate (or adjoint) matrix of A, denoted by
Adj(A), is defined by the formula

Adj(A) = [Cij (A)]t

That is, adjoint of matrix A is the transpose of the matrix of cofactors for A.

Example 3.41. Compute the adjoints of the given matrices.

 
  1 0 2
1 0
A= , and B = 1 1 3
−1 2
2 0 1

Solution: The matrix of cofactors for A is

 
2 1
[Cij (A)] = .
0 1

Thus, the adjoint of matrix A is

 
2 0
t
Adj(A) = [Cij (A)] = .
1 1

The matrix of cofactors for B is given by

 
1 5 −2
[Cij (B)] =  0 −3 0  .
−2 −1 1

Thus, the adjoint of matrix B is

 
1 0 −2
Adj(B) = [Cij (B)]t =  5 −3 −1 .
−2 0 1

Definition 3.27 (Inverse of a Matrix). Let A be an n×n square matrix. The inverse
of matrix A is an n × n matrix B such that

AB = In = BA,

where In is the n × n identity matrix. If such a Matrix B exists, then the matrix
A is said to be invertible (or nonsingular), and its inverse is denoted by A−1 (i.e.
B = A−1 ). A matrix that does not have an inverse is said to be noninvertible (or
singular).

136
Example 3.42. Consider the following matrices:
   
    2 3 1 2 −3 −2
2 3 −1 3
A= , B= , C = 1 2 0 , D = −1 2 1 .
1 1 1 −2
0 0 1 0 0 1

Then we have
       
2 3 −1 3 1 0 −1 3 2 3
AB = = = = BA
1 1 1 −2 0 1 1 −2 1 1

That is, the products AB and BA give us the identity matrix I2 . Therefore, matrix B is the
inverse of A i.e., A−1 = B.

Similarly, we have  
1 0 0
CD = 0 1 0 = DC.
0 0 1
Thus, the matrix D is the inverse of C i.e., C −1 = D.

Theorem 3.11. Let A be an n × n matrix. If A is invertible (non singular) then

det(A) 6= 0, and the inverse A−1 is given by the formula
1
A−1 = Adj(A).
det(A)

Example 3.43. Compute the inverse of the given matrix A.

 
1 0 0
A = 0 2 0
0 0 3

Solution: We have, det(A) = 6,

   
6 0 0 6 0 0
[Cij (A)] = 0 3 0 , and Adj(A) = [Cij (A)]t = 0 3 0 .
0 0 2 0 0 2

Therefore, by Theorem 3.11, we have

   
6 0 0 1 0 0
1 1
A−1 = Adj(A) = 0 3 0 = 0 21 0  .
det(A) 6
0 0 2 0 0 13

137
 Theorem 3.12 (Laws of Inverse). Let A, B, C be matrices of appropriate sizes so
that the following multiplications make sense, I is a suitably sized identity matrix,
and α a nonzero scalar. Then

i. If the matrix A is invertible, then it has one and only one inverse, A−1 .

ii. If A is invertible matrix of size n × n, then so is A−1 and hence, (A−1 )−1 = A.

iii If any two of the three matrices A, B, AB are invertible, then so is the third,
and moreover, (AB)−1 = B −1 A−1 .

iv If the matrix A is invertible, then so is αA. Moreover, (αA)−1 = α1 A−1 .

v If the matrix A is invertible, then so is At . Moreover (At )−1 = (A−1 )t .

vi Suppose A is invertible. If AB = AC or BA = CA, then B = C.

 
1 −1
Example 3.44. Let A = be the given matrix. Then we have
1 0
   
−1 0 1 −1 t 0 −1
A = , and (A ) = .
−1 1 1 1

Now,
   1 1
2 −2
(a) 2A = and (2A) = 2 1 2 = 12 A−1 . Thus, we have (2A)−1 = 12 A−1 .
−1
2 0 −2 0
   
1 1 t −1 0 −1
t
(b) A = and (A ) = . Thus, we have (At )−1 = (A−1 )t .
−1 0 1 1

Computation of Inverse Using Elementary Row Operations: Gauss-Jordan Elimina-

tion
Let A be an n × n invertible matrix and In be the identity matrix of size n × n.
   
a11 a12 . . . a1n 1 0 ... 0
 a21 a22 . . . a2n 
 , In = 0 1 ... 0

A=
 .

. ... .  . . ... .
an1 an2 . . . ann 0 0 ... 1

Then the inverse A−1 can be obtained using elementary row operations as follows.

138
 Finding the Inverse of a Matrix by Gauss-Jordan Elimination

1. Write the n × 2n matrix that consists of A on the left and the n × n identity
matrix In on the right to obtain [A|In ]. This process is called adjoining matrix
In to matrix A.

2. If possible, row reduce A to In using elementary row operations on the entire

matrix [A|In ]. The result will be the matrix [In |A−1 ]. If this is not possible,
then A is noninvertible (or singular).

3. Check your work by multiplying to see that AA−1 = In = A−1 A.

Example 3.45. Compute the inverses of the given matrices using Gauss-Jordan Elimination.
 
  1 0 0
1 −1
A= , B = 0 2 0
3 2
0 0 3
 
1 −1
Solution: Let A = . Then we have
3 2
   
1 −1 1 0 1 −1 1 0
[A|I2 ] = R2 → R2 + (−3)R1
3 2 0 1 0 5 −3 1

1 0 25 1
   
1 1 −1 1 0 5
R2 → R2 R1 → R 1 + R2 .
5 0 1 − 35 15 0 1 − 53 1
5

Therefore, the transformed matrix is

1 0 52 1
 
−1 5
[I2 |A ] =
0 1 − 35 1
5

2 1
 
−1 5 5
and hence, the inverse of matrix A is given by A = .
− 35 1
5
 
1 0 0
Similarly, for B = 0 2 0,
0 0 3
   
1 0 01 0 0 1 0 01 0 0
1
[A|I3 ] = 0 2 0 0 1 0 R2 → R2 0 1 0 0 21 0
2
0 0 30 0 1 0 0 30 0 1
 
1 0 01 0 0
1 0 1 0 0 1 0 
R3 → R3 2
3
0 0 1 0 0 13

139
Therefore, the transformed matrix is
 
1 0 01 0 0
[I3 |A−1 ] = 0 1 0 0 21 0 
0 0 1 0 0 13
 
1 0 0
Thus, A−1 = 0 21 0 .
0 0 13
Exercise 3.8.

1. For the given matrices A and B, compute the adjoint matrices.

 
  1 0 −2
1 −2
A= , B = −1 1 4 
2 3
2 0 3

2. Compute the inverse of the given matrix (if it exists).

 
  1 0 −2
1 2
A= , B = 0 1 2 
4 −2
0 1 3

3. Compute the inverse (if it exists) of the given matrix using elementary row opera-
tions.  
  1 1 2 1
  1 2 3
4 1 0 2 0 0 
A= , B = 1 3 4 , C = 

0 2 1 −2

2 3
1 4 4
0 3 2 1

3.9 System of Linear Equations

Consider an oil refinery that produces gasoline, kerosene and jet fuel form light crude oil
and heavy crude oil. The refinery produces 0.3, 0.2 and 0.4 of gasoline, kerosene and jet
fuel, respectively, per barrel of light crude oil. And it produces 0.2, 0.4 and 0.3 of gasoline,
kerosene and jet fuel, respectively, per barrel of heavy crude oil. This is shown in Table 2.
Note that 10% of each of the crude oil is lost during the refining process.

Table 2

Gasoline Kerosene Jet fuel

Light crude oil 0.3 0.2 0.4
Heavy crude oil 0.2 0.4 0.3

140
Suppose that the refinery has contracted to deliver 550 barrels of gasoline, 500 barrels of
kerosene, and 750 barrels of jet fuel. The problem is to find the number of barrels of each
crude oil that satisfies the demand.

If l and h represent the number of barrels of light and heavy crude oil, respectively, then
the given problem can be expressed as a system of linear equations

0.3l + 0.2h = 550

0.2l + 0.4h = 500
0.4l + 0.3h = 750

The given linear system has three equations and two unknowns. The matrix
 
0.4 0.2
0.2 0.4
0.4 0.4

is known as the coefficient matrix of the system, and the right side of the system is a matrix
 
550
500 .
750
 
l
With the column vector of unknowns , the above information can be organized in
h
matrix form    
0.3 0.2   550
0.2 0.4 l = 500 .
h
0.4 0.3 750
Example 3.46. Consider the following system of two equations and two unknowns x, y

ax + by = b1
.
cx + dy = b2

If we interpret (x, y) as coordinates in the xy-plane, then each of the two equations repre-
sents a straight line, and (x∗ , y ∗ ) is a solution if and only if the point P with coordinates
x∗ , y ∗ lies on both lines. In this case, there are three possible cases: there exists only one
solution if the lines intersect (see Figure 1 a), there are infinitely many solutions if the lines
coincide (see Figure 1 b) and the system has no solution if the lines are parallel (see Figure
1 c).

141
 (a) (b) (c)

x+y=1 x+y=1 x+y=1

2x−y=0 2x+2y=2 x+y=0

Figure 1: In this figure: (a) represents the case where the lines intersect (b) represents the
case where the lines coincide (c) represents the case where the lines are parallel

Let us briefly discuss the three different cases: In part (a) the linear system is given by

x+y =1
2x − y = 0.

This system has only solution, namely (x, y)=( 13 , 32 ).

In part (b) the linear system is given by

x+y =1
2x + 2y = 2.

This system has infinitely many solutions. In fact, the point (α, 1 − α) is a solution for each
real number α.

And finally, in part (c) the linear system is given by

x+y =1
x + y = 0,

which has no solutions, since the expressions in the left side of the two equations are the
same, but different values in the right side of the two equations.

142
 Definition 3.28. A linear system (or system of linear equations) of m-equations in
n-unknowns x1 , x2 , x3 , ..., xn is a set of equations of the form

a11 x1 + a12 x2 + ... + a1n xn = b1

a21 x1 + a22 x2 + ... + a2n xn = b2
(3.2)
................................................
am1 x1 + am2 x2 + ... + amn xn = bm ,

where aij ’s (for i = 1, 2, 3, ..., m and j = 1, 2, 3, ..., n), are given numbers, called
the coefficients of the system, and b1 , b2 , b3 , ..., bm on the right side are also numbers.

A solution of (3.2) is a set of numbers x1 , x2 , x3 , ..., xn that satisfies all the m-equations
simultaneously.

Matrix Form of a Linear System

From the definition of matrix multiplication, we see that the m-equations of (3.2) may be
written as a single vector equation
Ax = b, (3.3)
where      
a11 a12 . . . a1n x1 b1
 a21 a22 . . . a2n   x2   b2 
    
A= . , x = and b = . ,

.. .. .
 ..  ..   .. 
   
. ... . 
am1 am2 . . . amn xn bm
are known as the coefficient matrix, the column vector of unknowns and the column vector
of numbers, respectively. We assume that the coefficients aij are not all zero, so that A is
not a zero matrix. Note that x has n components, whereas b has m components.

For the system of linear equations in (3.2), precisely one of the statements below is true:

1. It admits a unique Solution: There is one and only one vector x = (x1 , x2 , x3 , ..., xn )
that satisfies all the m-equations simultaneously (the system is consistent).

2. It has infinitely Many Solutions: There are infinitely many different values of x
that satisfy all the m-equations simultaneously (the system is said to be consistent).

3. Has no Solution: There is no vector x that satisfies all equations simultaneously, or

the solution set is empty (the system is said to be inconsistent).

3.9.1 Gaussian Elimination

Gaussian elimination, also known as row reduction, is used for solving a system of linear
equations. It is usually understood as a sequence of elementary row operations performed

143
on the corresponding matrix of coefficients.

Consider the linear system given in (3.2). The augmented matrix which represents the
system is given by
 
a11 a12 . . . a1n b1
 a21 a22 . . . a2n b2 
[A|b] = 
 .
.
. ... . . 
am1 am2 . . . amn bm
Then, the idea here is, we solve the linear system whose augmented matrix is in row echelon
form, which is row equivalent to the original system. And, we have the following theorem
on the row equivalent linear systems.

Theorem 3.13. Row-equivalent linear systems have the same set of solutions.

Thus, if the augmented matrix is initially in row echelon form, then we simply solve it by
using back substitution. If it is not, then first rewrite it as a row equivalent system whose
augmented matrix is in its row echelon form, and then apply Theorem 3.13.
Example 3.47. Rewrite the following linear system as a row equivalent system, and then
solve it.
x1 − x2 = 1
x1 + 2x2 = 4.
Solution: Here, the augmented matrix of the given system is
 
1 −1 1
[A|b] = ,
1 2 4

which has row echelon form (after a sequence of elementary operations)

 
1 −1 1
[A|b] =
g .
0 1 1

Thus, the row equivalent system is

x1 − x2 = 1
x2 = 1.

Clearly, solving the above linear system (whose augmented matrix is in row echelon form)
is much easier than solving the original system. The only solution of the linear system
(represented by an augmented matrix in row echelon form) is (x1 , x2 ) = (2, 1). And, hence
by Theorem 3.13, a vector (x1 , x2 ) = (2, 1) also solves the original linear system.

144
 Gaussian Elimination:

(a) Write the augmented matrix for the linear system.

(b) Use elementary row operations to rewrite the matrix in row echelon form.

(c) Write the system of linear equations corresponding to the matrix in row eche-
lon form, and use back-substitution to find the solution.

Example 3.48. Consider an oil refinery’s problem which is given as a system of linear
equations
0.3l + 0.2h = 550
0.2l + 0.4h = 500
0.4l + 0.3h = 750
where l and h represent the number of barrels of light and heavy crude oil, respectively.
The augmented matrix of the given linear system is
 
0.3 0.2 550
[A|b] = 0.2 0.4 500 ,
0.4 0.3 750

where    
0.3 0.2 550
A = 0.2 0.4 , and b = 500 .
0.4 0.3 750
And the matrix in row echelon form is given by
 
0.1 0.2 250
[A|b]
g =  0 0.1 50  .
0 0 0

Now, rewriting the given linear system as row equivalent system we have

0.1l + 0.2h = 250

0.1h = 50.

The only solution of the above system (in row echelon form) is (l, h) = (1500, 500), which
is also a solution for the original system. Thus, an oil refinery needs 1500 barrels of light
crude oil and 500 barrels of heavy crude oil in order to satisfy the demand.
Example 3.49. Solve the given linear system by using the method of Gaussian elimination.

x1 + 2x2 + x3 = 2
x1 − x2 − 2x3 = −1.

145
Solution: The augmented matrix representing the given system is
 
1 2 1 2
[A|b] = .
1 −1 −2 −1

Now, by replacing R2 (i.e., R2 → R2 − R1 ), we obtain

 
1 2 1 2
0 −3 −3 −3

and by Scaling R2 (i.e., R2 → (− 31 )R2 ), we have

 
1 2 12
.
0 1 11

The last matrix is in its row echelon form, and hence the row equivalent system is

x1 + 2x2 + x3 = 2
x2 + x3 = 1.

In this case, the system has infinitely many solutions, and the set of solutions is be given by

{(1 − α, α, 1 − α) : α ∈ R}.

Example 3.50. Solve the following system of linear equations using the method of Gaussian
elimination.

4x2 + 3x3 = 8
2x1 − x3 = 2
3x1 + 2x2 = 5
Solution: The augmented matrix of the given system is
 
0 4 3 8
[A|b] = 2 0 −1 2
3 2 0 5

Applying the following elementary row operations:

R1 ↔ R3 (Interchanging R1 and R3 )
 
3 2 0 5
2 0 −1 2
0 4 3 8

R2 ↔ R3 (Interchanging R2 and R3 )
 
3 2 0 5
0 4 3 8
2 0 −1 2

146
R3 → R3 + (− 23 )R1 (Replacing R3 )
 
3 2 0 5
0 4 3 8 
0 − 3 −1 − 34
4

R3 → R3 + 13 R2 (Replacing R3 )
 
3 2 05
0 4 3 8  .
0 0 0 34
The last matrix is in row echelon form, and hence the row equivalent system is given by

3x1 + 2x2 = 5
4x2 + 3x3 = 8
0 = 43
We observe that the last equation in the linear system above is a contradiction to the fact
that 0 6= 34 . Consequently, the given linear system has no solution.

Theorem 3.14. Consider the system of linear equations in (3.2). If A and b are the
matrices of coefficients and the column vector of numbers, respectively. Then the
following statements are true.

(i) If rank(A) = rank([A|b]) = number of unknowns, then the linear system has
only one solution.

(ii) If rank(A) = rank([A|b]) < number of unknowns, then the linear system has
infinitely many solutions.

(iii) If rank(A) < rank([A|b]), then the linear system has no solution.

Remark.
(a) From Theorem 3.14, we observe that the linear system (3.2) has no solution if an
echelon form of the augmented matrix has a row of the form [0, 0, ..., 0 b] with b
nonzero.

(b) A linear system has unique solution when there are no free variable, and it has in-
finitely many solutions when there is at least one free variable.
Example 3.51. Use matrix rank to determine the number of solutions for the system.

x 1 + x2 + x3 = 1 x1 + x2 + 2x3 = 3 x1 + 2x2 + 3x3 = 1

(a) 2x2 + 4x3 = 2 , (b) 2x2 + 2x3 = 4 (c) 2x2 + 2x3 = −2
2x1 + 7x3 = 5 x2 + x3 = 2 −2x2 − 2x3 = 3
Solution:

147
(a) We have a linear system
x 1 + x2 + x3 = 1
2x2 + 4x3 = 2
2x1 + 7x3 = 5
and the augmented matrix given by
 
1 1 11
[A|b] = 0 2 4 2 .
0 2 75

After a sequence of elementary row operations, we obtain its row echelon form
 
1 1 11
g = 0 1 2 1 .
[A|b]
0 0 11

From the transformed matrix, we can see that the matrix A in its row echelon form is
 
1 1 1
e = 0 1 2 .
A
0 0 1

Thus, we have rank(A) = rank([A|b]) = number of unknowns. Hence, the given

linear system has only one solution.

(b) We have a linear system

x1 + x2 + 2x3 = 3
2x2 + 2x3 = 4
x2 + x3 = 2
In this case, the augmented matrix and its row echelon form, respectively, are given
by    
1 1 23 1 1 23
[A|b] = 0 2 2 4 and [A|b]g = 0 1 1 2
0 1 12 0 0 00
The matrix A in its row echelon form is
 
1 1 2
A
e = 0 1 1 .
0 0 0

Here, the matrices A,

e and [A|b]
g have only two nonzero rows. Thus,
rank(A) = rank([A|b]) < number of unknowns. Therefore, by Theorem 3.14, the
given system has infinitely many solutions.

148
 (c) Here, we have a linear system

x1 + 2x2 + 3x3 = 1
2x2 + 2x3 = −2
−2x2 − 2x3 = 3.

The augmented matrix [A|b] and its row echelon form [A|b],
g respectively, are given
by    
1 2 3 1 1 2 3 1
[A|b] = 0 2 2 −2 and [A|b]g = 0 1 1 −1 .
0 −2 −2 3 0 0 0 1
Here, the number of nonzero rows of the row echelon form of A and that of [A|b] are
2 and 3, respectively. Therefore, the given linear system has no solution.

Exercise 3.9. Solve the following linear systems using the method of Gaussian elimination.

−x1 + x2 = 4
(a)
−2x1 + x2 = 0
x1 + x2 = −1
(b) x1 − x2 = 0
−2x1 + x2 = 3

x1 + 2x2 + x3 = 0
(c) 4x1 + 5x2 + 6x3 = 3
7x1 + 8x2 + 9x3 = 6.

x1 + 2x2 + x3 = 0
(d)
2x2 + 3x2 − 2x3 = 0

3.9.2 Cramer’s rule

Cramer’s Rule is a method for solving linear systems where the number of equations and
the number of unknowns are equal. Cramer’s rule relies on determinants. Consider the
following linear system of n-equations in n-unknowns x1 , x2 , x3 , ..., xn

a11 x1 + a12 x2 + ... + a1n xn = b1

a21 x1 + a22 x2 + ... + a2n xn = b2
(3.4)
................................................
an1 x1 + an2 x2 + ... + ann xn = bn

which has a matrix notation

Ax = b.

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Let us define the determinants
a11 a12 . . . a1n a11 a12 . . . a1(j−1) b1 a1(j+1) . . . a1n
a21 a22 . . . a2n a21 a22 . . . a2(j−1) b2 a2(j+1) . . . a1n
D= . .. .. , Dj = .. .. .. .. .. .
.. . ... . . . ... . . . . . . ..
am1 am2 . . . amn an1 an2 . . . an(j−1) bn an(j+1) . . . ann
(3.5)
for j = 1, 2, 3, ..., n. Here, D is the determinant of the coefficient matrix A, and for each j
Dj represents the determinant of a matrix which is obtained from A after replacing the j-th
column by the column vector b.

Theorem 3.15 (Cramer’s rule).

(a) If a linear system (3.4) of n-equations in the same number of unknowns

x1 , x2 , x3 , ..., xn , has a nonzero coefficient determinant D = |A|, then the
system has precisely one solution. This solution is given by
D1 D2 Dn
x1 = , x2 = , ..., xn =
D D D
where D and Dj for j = 1, 2, 3, ..., n are defined in (3.5).

(b) If the system (3.4) is homogeneous and D 6= 0, then it has only the trivial
solution x1 = 0, x2 = 0, x3 = 0, ..., xn = 0. If D = 0 the homogeneous
system also has nontrivial solutions.

Example 3.52. Use Cramer’s rule to solve the system of linear equations.

4x1 − 2x2 = 10
3x1 − 5x2 = 11

Solution: Here, the coefficient matrix A and the column vector b, respectively, are
   
4 −2 10
, and .
3 −5 11

And the determinants D, D1 , D2 are

4 −2 10 −2
D= = (−20) − (−6) = −14, D1 = = (−50) − (−22) = −28,
3 −5 11 −5

4 10
D2 = = (44) − (30) = 14.
3 11
Therefore, by Theorem 3.15, the unique solution of the given linear system is
 
D1 D2
(x1 , x2 ) = , = (2, −1).
D D

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Example 3.53. Solve the following system of linear equations using Cramer’s rule

2x1 − x2 = 0
−x1 + 2x2 − x3 = 0
−x2 + x3 = 1

Solution: With the coefficient matrix

   
2 −1 0 0
A = −1 2 −1 , and column vector b = 0 ,
0 −1 1 1

the determinants D, D1 , D2 and D3 are computed as follows;

2 −1 0 0 −1 0 2 0 0
D = −1 2 −1 = 1, D1 = 0 2 −1 = 1, D2 = −1 0 −1 = 2
0 −1 1 1 −1 1 0 1 1

and
2 −1 0
D3 = −1 2 0 = 3.
0 −1 1
Thus, by Theorem 3.15, the only solution of the given linear system is
 
D1 D2 D3
(x1 , x2 , x2 ) = , , = (1, 2, 3).
D D D

Remark. Cramer’s rule doesn’t work if the determinant of the coefficient matrix is zero or
the coefficient matrix is not square.
Exercise 3.10. Solve the following linear systems using Cramer’s rule (if possible).

4x1 − 2x2 = 10
(a)
3x1 − 5x2 = 11

−x1 + 2x2 − 3x3 = 1

(b) 2x1 + x3 = 0
3x1 − 4x2 + 4x3 = 2.

x1 =7
(c) 2x2 =8
3x3 = 24.

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3.9.3 Inverse method
The Inverse method is one of the important methods to solve a linear system with n equa-
tions in n unknowns.
Example 3.54. Consider a linear system

x−y =1
x + y = 3.

Using matrix notation, it can be rewritten as

    
1 −1 x 1
= . (3.6)
1 1 y 3

And if we denote the coefficient matrix by A, then we have

   1 1

1 −1 −1
A= , and A = 2 1 2
1 .
1 1 −2 2

Now, multiplying (from the left) both sides of equation (3.6) by A−1 , we have
 1 1      1 1  
2 2
1 −1 x 1
= 2 1 21 .
− 12 12 1 1 y −2 2 3

And using the fact A−1 A = I2 , we have

        
1 0 x 2 x 2
= . This implies =
0 1 y 1 y 1

Thus, (x, y) = (2, 1) is the only solution of the given system of linear equations. This
shows the usefulness of the matrix inverse to solve linear systems.
Consider the following linear system with n-equations in n-unknowns x1 , x2 , x3 , ..., xn ;

a11 x1 + a12 x2 + ... + a1n xn = b1

a21 x1 + a22 x2 + ... + a2n xn = b2
(3.7)
................................................
an1 x1 + an2 x2 + ... + ann xn = bn .

The matrix notation of the linear system (3.7) is

Ax = b,

where      
a11 a12 ... a1n x1 b1
 a21 a22 ... a2n   x2   b2 
A= . , x =  .  and b =  .  .
     
.. ..
 .. . ... .  ..  .. 
an1 an2 . . . ann xn bn

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 Theorem 3.16 (Inverse Method). If A is an invertible matrix, then for each b ∈ Rn ,
the linear system Ax = b has a unique solution, which is given by

x = A−1 b.

Example 3.55. Solve the following system of linear equations using matrix inverse method.

2x1 − x2 = 1
3x1 + 2x2 = 12

Solution: The matrix of coefficients A, the inverse A−1 , and the column vector b, respec-
tively, are given by
   2 1  
2 −1 −1 7 7
1
A= , A = 3 2 , and b = .
3 2 −7 7 12

Thus, by Theorem 3.16, the only solution of the given linear system is
   2 1    
x1 −1 1 2
= A b = 7 3 72 =
x2 − 7 7 12 3
.
Example 3.56. Use matrix inversion to solve the following linear system.

2x1 + 3x2 + x3 = 1
x1 + 2x2 = −2
x3 = 3

Solution: The coefficient matrix A, the column vector b and the inverse A−1 , respectively,
are given by
     
2 3 1 1 2 −3 −2
A = 1 2 0 , b = −2 , A−1 = −1 2 1 .
0 0 1 3 0 0 1
Thus, by Theorem 3.16, the unique solution of the given linear system is
      
x1 2 −3 −2 1 2
x2  = A−1 b = −1 2 1  −2 = 0 .
 
x3 0 0 1 3 3

Exercise 3.11. Solve the following linear systems using the method of matrix inversion (if
possible).

3x1 + 4x2 = −4
(a)
5x1 + 3x2 = 4

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 4x1 − x2 − x3 = 1
(b) 2x1 + 2x2 + 3x3 = 10
5x1 − 2x2 − 2x3 = −1.

3x1 = 12
(c) 4x2 = 16
5x3 = 20.

Review exercises
1. For every square matrix A, show that A + At is symmetric.

2. Given matrices
 
3 0    
4 −1 1 4 2
A = −1 2 , B= , C=
0 2 3 1 5
1 1

(i) Compute the products A(BC), (AB)C, and verify that A(BC) = (AB)C.
(ii) Compute the products α(AB), (αA)B, A(αB), and verify that

α(AB) = (αA)B = A(αB).

3. A fruit grower raises two crops, apples and peaches. The grower ships each of these
crops to three different outlets. In the matrix
 
125 100 75
A=
100 175 125

aij represents the number of units of crop i that the grower ships to outlet j. The
matrix B = $3.5 $6.00 represents the profit per unit. Find the product BA and
state what each entry of the matrix represents.

4. A corporation has three factories, each of which manufactures acoustic guitars and
electric guitars. In the matrix
 
70 50 25
A=
35 100 70

aij represents the number of guitars of type i produced at factory j in one day. Find
the production levels when production increases by 20%.

5. Find the value of x for which the matrix is equal to its own inverse
     
3 x 2 x x 2
(a) (b) (c)
−2 −3 −1 −2 −3 4

154
  
cos(θ) sin(θ)
6. If A = , then
−sin(θ) cos(θ)

i. show that A = A−1

 
n cos(nθ) sin(nθ)
ii. show that A = .
−sin(nθ) cos(nθ)
   
cos(θ) sin(θ) cos(φ) sin(φ)
7. If A = , and B = , then show that
−sin(θ) cos(θ) −sin(φ) cos(φ)
 
cos(θ + φ) sin(θ + φ)
AB = .
−sin(θ + φ) cos(θ + φ)

 
1 1 0
8. Determine the values of α for which the matrix A = 1 0 0  is invertible and
1 2 α
−1
find A .

9. Show that if A is invertible, then so is Am for every positive integer m; moreover,

(Am )−1 = (A−1 )m .

10. If A and B are n × n matrices with A is invertible, then show that

(A + B)A−1 (A − B) = (A − B)A−1 (A + B).

11. Solve the following systems of linear equations using Gaussian elimination

x1 − x2 + 2x3 = 4
x1 + x3 = 6
(a)
2x1 − 3x2 + 5x3 = 4
3x1 + 2x2 − x3 = 1

x1 − 2x2 + 3x3 = 9
(b) −x1 + 3x2 = −4
2x1 − 5x2 + 5x3 = 17
2x1 + x2 − x3 + 2x4 = −6
3x1 + 4x2 + x4 = 1 = 2
(c)
x1 + 5x2 + 2x3 + 6x4 = −3
5x1 + 2x2 − x3 − x4 = 1

12. Use Cramer’s rule (if possible) to solve the following linear systems.

x1 + 2x2 = 5
(a)
−x1 + x2 = 1

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 4x1 − x2 − x3 = 1
(b) 2x1 + 2x2 + 3x3 = 10
5x1 − 2x2 − 2x3 = −1
4x1 − 2x2 + 3x3 = −2
(c) 2x1 + 2x2 + 5x3 = 16
8x1 − 5x2 − 2x3 = 4

13. Use matrix inversion method (if possible) to solve the following linear systems.

2x1 + 3x2 + x3 = −1
(a) 3x1 + 3x2 + x3 = 1
2x1 + 4x2 + x3 = −2

2x1 + 3x2 + x3 = 4
(b) 3x1 + 3x2 + x3 = 8
2x1 + 4x2 + x3 = 5
4x1 − 2x2 + 3x3 = 0
(c) 2x1 + 2x2 + 5x3 = 0
8x1 − 5x2 − 2x3 = 0

156