Unit II
Matrices and Determinants
UNIT II
Matrices and Determinants
Introduction
Now days, Matrices and the algebraic operations defined on them play an important role in various
disciplines. In this unit we study on matrices and some algebraic operations on matrices, determinants and
how to use the concept on matrices and determinants in solving systems of linear equations.
Our discussion focuses on matrices and some algebraic operations on matrices, some special types of
matrices, Elementary row operations, determinants of any order, Matrix inversions and finally on how to
solve systems of linear equations. The exercises at the end of each subtopic are organized in such a way
that the reader can get enough practice on how to apply most of the central ideas of the topic.
2.1 Basic concepts and Operations on Matrices
This section is devoted to introduce the definition of matrices and some related terms, and some of the
algebraic operations on matrices such as: matrix addition, scalar multiplication and matrix multiplication.
2.1.1 Definitions and Notations
Matrices are presented in different forms. However, throughout this unit we use the representation used in
definition 2.1.
Definition 2.1 A rectangular array of the form
 a11
 a 21
 ....
a
 m1
a12
a 22
....
am2
.... a1n 
.... a 2 n 
.... .... 
.... a mn 
................
................
................
................
where the aij called entries or elements are scalars in  or functions, is called
a matrix. If the entries or elements are scalars in , then it is called a matrix
over  or simply a matrix if  is implicit.
The above matrix is also denoted by a ij  m  n , i = 1, 2, 3, … , m, j = 1, 2, 3, … , n or simply a ij  m  n .
The m-horizontal n-tuples, a 11, a 12, ..., a 1n  ,
a 21, a 22, ..., a 2n , . . . , a m1, a m2, ..., a mn 
are
called rows of the matrix and the n-vertical m-tuples,
 a 1n 
 a 12 
a 
a 
 22  …  2 n  are called columns of the matrix.
 ... 
 ... 
 a m2 
 a mn 
In the double subscript notation a ij means the entry in the i th row and j th column. A matrix with m
 a 11 
a 
 21 
 ... 
 a m1 
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Matrices and Determinants
rows and n columns is called an m by n matrix, or m  n matrix. The pair of numbers (m, n) is called the
size or shape or order of the matrix.
Example 1 Find the order of each of the following matrices.
1
 2 1 3

A= 
 and B =  2
3
2
1


 3
2
3
4
3
4
1
4
1 .
2
Solution Number of rows of A = 2 and number of columns of A = 3, and number of rows of B = 3 and
number of columns of B = 4.
Therefore, (2, 3) is the order of A and (3, 4) is the order of B.
Definition 2.2 Sub-matrix
A matrix obtained by omitting some rows or columns or both from a matrix.
 0 2 3
 2 3
 0 2 3


Example 2 A = 
 and B =   3 4 1 are sub-matrices of C =  2 1 2  .
 1 2


  3 4 1 
Note that: A matrix with one row is referred to as a row matrix (or row vector) and with one column as
a column matrix ( or column vector).
2
Example 3 A = 1, 0, 1 is a row matrix and B =  1  is a column matrix.
 0 
Definition 2.3 Equality of Two Matrices
Two matrices A and B are said to be equal, written A = B if and only if they have
the same size and corresponding entries are equal.
 0 2 3
 2 3
 2 3
Example 4 Let A = 
and C = 
 ,B= 
 . Justify that A  B and A  C.

1
2


 4 2
  3 4 1
Solution A  B because they have different sizes and A  C because one pair of corresponding entries are
not equal.
Example 5 Determine the values of x and y for which
 5 7  2 x  y x  y 
 1 8 =  x  y 4 y  x

 

Solution From the definition of equality of matrices, solving
2x  y = 5 and x + y = 7
we get: x = 4 and y = 3.
Therefore, x = 4 and y = 3.
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Unit II
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2.1.2 Operations on Matrices
Addition and Scalar Multiplication
Definition 2.4 Addition of Matrices
Let A = a ij  and B =  b ij  be two matrices with the same size. The sum of A and B,
written A + B, is the matrix obtained by adding the corresponding entries.
That is if A = a ij  and B =  b ij  are m  n matrices, then A + B = a ij  b ij  m  n .
1 2
 2 4 
and B = 
 . Find A + B.

 3 11 
3 4
Example 6 Let A = 
 1 2
Solution Both A and B are 2  2 matrices and hence A + B = 
3  ( 3)
 3
2  ( 4)
 3
= 

4  11 
 0
 2
.
15 
 2
Therefore, A + B = 
.
 0 15 
Note that: Matrices of different sizes cannot be added.
 2 3

0 2 3
Example 7 Let A = 
 and B =   3 4 1 . A and B have different sizes and hence the sum of A
 1 2


and B is not defined.
Definition 2.5 Scalar Multiplication
The product of any m  n matrix A = ( a ij ) and any scalar c, written c A, is the
matrix c A = (c a ij ) , of size (m, n), obtained by multiplying each entry in A by c.
In particular if c =  1, then c A =  A.  A is called the negative of A.
If A and B are matrices of the same size, then A  B is called the difference of A and B.
2 6 0
Example 8 Let A = 
1  and B =
4 8

2

 1  2 4

3  . Find A + B, 2A and A  B.
2 0

2

2 1 6  ( 2) 0  4 
3 4 4
1 3 = 
Solution A + B = 
,
 
6 8 2
4  2 8  0

2 2

 2 . 2 2 . 6 2 . 0  4 12 0
2A =  .

.
. 1 = 
2 4 2 8 2   8 16 1 
2

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2  1 6  ( 2)
and A  B = 
4  2

0  4  1 8  4
1 3 = 
  2 8 1
2 2
80
3 4 4
4 12 0
1 8  4
; 2A = 
and A  B = 


.
6 8 2 
 8 16 1 
2 8 1
Therefore, A + B = 
Definition 2.6 Transpose of a Matrix
The transpose of an m  n matrix A, written A t , is an n  m matrix whose
a
ji
th entry is the
5
Example 9 Let A = 
 0
a ij
th
2
3

1 1 
entry of A for j = 1, 2, 3, … , n and i = 1, 2, 3, … , m.
 2

and B = 1

 0
3 
2  . Find A t and B t .
3 
1
3
1
Solution Now the i th row of A becomes the j th column of A t for i = 1, 2.
0
 5

Hence A =
1  .
 2
 3
1 
Similarly the i th row of B becomes the j th column of B t for i = 1, 2, 3.
t
 2 1

Hence B = 1
3

 3
2
0
 5

t
Therefore, A =
1  and
 2
 3
1 
t

.


0
1
3
2

B = 1

 3
t
1
3
2
0
1
3

.


Definition 2.7 The Zero Matrix
A matrix whose entries are all zero, denoted by 0, is called a zero matrix.
Example 10
0
0
 0 0
0
0  ,  0  , 
 and  0
0
0


 0  
0
0
0
are examples of zero matrices.
0 
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Matrices and Determinants
Properties of matrices under matrix addition and scalar Multiplication
1. Let V be the set of all m  n matrices over . Then for any matrices A, B, C  V and any ,   .
i) A + B = B + A
iv) A + ( A) = 0 m  n
ii) A + (B + C) = (A + B) + C
iii) A + 0 m  n = A
v) (A + B) =  A +  B
vi) ( + ) A =  A +  A
vii) (  ) A =  ( A) and viii) 1  A = A and 0  A = 0 m  n
2. Let A and B be matrices of the same size and let   . Then
ii) ( A) t =  A t
i) ( A  B ) t = A t + B t
 0 1
1
Example 11 Let A = 
3
2
2 1
and B = 

0
 1 3
0
. Find A t + B t .
2 
2 0
Solution A t + B t = ( A  B ) t . Now A + B = 
 4 2
 2
Therefore, A t + B t =  0
 2
iii) ( A t ) t = A
2
.
2
4
 2  .
2 
Matrix Multiplication
Definition 2.8 The product C = A B (in this order) of an m  n matrix A = ( a ij ) and
an r  p matrix B = (b ij ) is defined if and only if n = r. i.e. the number of
columns of A is equal to number of rows of B, and is then defined as the
m  p matrix C = (c ij ) with entries:
n
c ij =
 a ik b kj
k 1
i.e. the entry in the i th row and j th column of the product is the dot product
of the i th row of A and the j th column of B.
0
Example 12 Let A =  2
 1
1
2
7 and B = 
3
4
6
. Find A B.
1 
Solution A 3  2 and B 2  2 . Hence A B 3  2 .
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Unit II
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

and A B =


0 . 2  1. 3
2.2  7.3
1 .2  4 . 3
 3
Therefore, A B =  25

 14
0 . 6  1 .1
2 . 6  7 .1
1 . 6  4 .1
  3
 =  25
 
  14
1
19  .
10 
1
19  .
10 
 9
 2
3
and B =
0 
Example 13 Let A = 
4 
. Show that AB  BA.
5 
1
2

Solution A 2  2 and B 2  2 . Hence AB and AB are well defined.
 15  21
8
 2

 17
 and B A =  8


Now A B = 
3
.
6 
Therefore, AB  BA.
Note that: In general matrix multiplication is not commutative.
1
 1
and B = 

2
 1
1
2
Example 14 Let A = 
1 
. Show that AB = 0 2  2 while BA  0 2  2 .
1 
Solution A 2  2 and B 2  2 . Hence AB and AB are well defined.
1
 1
.
1 
 1
while BA  0 2  2 .
Now A B = 0 2  2 and B A = 
Therefore, AB = 0 2  2
Note that: In general A B = 0 does not necessarily imply A = 0 or B = 0 or B A = 0.
Properties of Matrix Multiplication
If the sizes of the matrices are such that the stated operations can be performed, then for any   
i) ( A) B =  (A B) = A ( B)
ii) A (B C) = (A B) C
iii) (A + B) C = A C + B C
iv) C (A + B) = C A + C B
4
Example 15 Let A =  0
 1
9
 1
2  and B = 
 2
6 
 14 35 
t
Solution ( A B ) =  4 6 
 11 20 
t
v) ( A B ) t = B t A t
2
. Show that ( A B ) t = B t A t
3 
 14 4 11 

 35 6 20 
= 
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Unit II
Matrices and Determinants
 14
  1 2   4 0 1
and B t A t = 
 = 
· 
 2
3   9 2 6
4 11 
.
 35 6 20 
Therefore, ( A B ) t = B t A t .
Exercises 2.1
1. Construct a 3  2 matrix A = ( a ij ) such that
a ij
 j  2i, for i  j


2

i  2 j , for i  j
In exercise 2-8 use the information given below and determine which of the following matrix operations
re defined, find the order of the resulting matrix whenever defined.
Given: Suppose A, B, C, D and E are matrices with the following order
Matrix
A
B
C
D
E
size
45
45
52
42
5 4
2. BA
3. AC + D
4. AE + B
5. E (A + B)
6. E (AC)
7. E t A
8. ( At  E) D
In exercise 9-13 use the information given below to find
 3

Given: A =   1
 1

9. AB
0

2 , B =
1 
 1 5

1 4 2

 and C =   1 0
3 1 5
 3 2

11. ( 2 A t + B) t
10. AB + 2C
2

1 .
4 
12. C t + A B
13. ( A t  3B) C t
In exercises 14-16, compute each of the following operations, where
1  2
2 2 3
 2 1
A = 3
and C = 
.
4 , B = 

1 4 5 
1 4


0  4
14. AB
15. 3C  B At
16. (AB)C
17. If  and  are two roots of the equation 2 x2  x  1  0 , then find
 
 2 2


   2 2 
+

      
without computing the values of  and .
18. Determine the values of x, y and z for which
 x
1

 2  4
y   3
1 
 2
= 

2
 13
z 
x  y 
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Unit II
Matrices and Determinants
19. Verify that for any angles  and 
 cos 

 sin 
 cos (   )  sin (   ) 
=

cos    sin (   )
cos (   ) 
 sin    cos 
 
cos    sin 
 sin  
2.2 Special Matrices
Definition 2.9 Square matrix is a matrix with the same number of rows as columns.
An n  n matrix is called a square matrix of order n, or an n-square matrix.
1
Example 16 Let A = 
3
0
2
1
and
B
=

4
 3
2
5
4
1 
0  . Then A is a 2-square matrix while B is a 3-square
1 
matrix.
Definition 2.10 Symmetric Matrix
A square matrix A is called a symmetric matrix if and only if A = At .
i.e. Whenever a ij is an entry of A, a ij = a ji .
0
2
Example 17 
0
Example 18 
3
1

2
 0
and


1
 3
 1
2
 2
and

1
 3
0
1
2
0
1
2
3
2  are examples of symmetric matrices.
4 
3
2  are examples of matrices that are not symmetric matrices.
4 
Definition 2.11 Skew – Symmetric Matrix
A square matrix A is said to be a skew - symmetric matrix if and only if A t =  A.
0
Example 19 
2
 0
Example 20 
 2

2 
and 

0 


 0
 2
 2
and

1 
 3
0
2
3
2
0
4
2
0
4
3
4
0

 are examples of skew – symmetric matrix.


3
4  are examples of matrices that are not skew – symmetric.
0 
Note that: Any n-square zero matrix is both symmetric and skew-symmetric.
Remark: Any square matrix A can be written as the sum of the symmetric matrix R and the skew
symmetric matrix S where
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Unit II
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R=
1
1
(A + A t) and S =
(A  A t).
2
2
Example 21 Express each of the following matrices as a sum of symmetric and skew-symmetric matrices.
1
Solutions i) A
t
1
2
4 
i) A = 
3
ii) B =  4
 5
2
6
3
3
5
2




5
1


 
3
 1

 0
2
2 .
and hence R = 

 S=  1
5
4 


0
4
 2

 2

4 5
1 3 4
 0 1



1
and
hence
R
=
and
S
=
6 3
0
3 6 4

 4 4 2 
 1 1
5 2 
1
= 
 2
1
ii) B t =  2
 3
1
1 .
0
Definition 2.12 Idempotent and Nilpotent matrices
An n- square matrix A is called an idempotent matrix if A2 = A and
a Nilpotent matrix if Am = 0 n  n for some non-negative integer m.
 0.5
0.25 
 . Then show that matrix A is idempotent.

0.5 
Example 22 Let A = 

0.25
Solution applying the method of multiplying two matrices we get:
 0.5


0.25
0.25 


0.5 
 0.5


0.25
0.25   0.5
 = 
 
0.5  0.25
0.25 


0.5 
Therefore, matrix A is idempotent.
 2
Example 23 Let A = 

4
1 
 . Then show that matrix A is Nilpotent.

2
Solution applying the method of multiplying two matrices we get:
 2


4
1 


2
 2


4
1 
0
 = 


0
2
0
0 
Therefore, matrix A is Nilpotent.
Definition 2.13 A square matrix with non-negative entries and row sum all equal
to one is called a Stochastic matrix.
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Example 23 The matrix
 0.3 0.5 0.2 


A = 0.2 0.3 0.5


 0.5 0.4 0.3 
is an example of stochastic matrix.
Definition 2.14 Triangular Matrix
A square matrix whose elements a ij are all zero for i  j (for i  j) is called
Upper ( lower) triangular matrix.
 1
Example 24 
 0
1
Example 25 
3
1
2
and  0
4 
 0
1
0
and  3
4 
 0
3
0
0
0
5
4

 are examples of upper triangular matrices.


0
0  are examples of lower triangular matrices.
2 
0
4
2
In a square matrix A = ( a ij ) of order n, a 11 , a 22 , …, a nn are called the diagonal elements (or are on
the main diagonal) of the square matrix A.
Note that: The transpose of an upper triangular matrix is a lower triangular matrix and vice versa.
Definition 2.15 The sum of the diagonal elements of a square matrix A is called the
trace of A, written tra (A).
Note that: For any two square matrices A and B of the same size and any   :
ii) tra ( A) =  tra (A).
i) tra (A + B) = tra (A) + tra (B).
1
Example 26 Let A =  2
 3
i) tra (A + B)
2
4
5
3
5
6

 and B =


1
4

 5
ii) tra (A) + tra (B) and
2
Solutions i) A + B =  6
 8
4
10
8
6
10
8
2
6
3
3
5
2

 . Then find:


iii) tra ( A) and  tra (A) for any   .

 and hence tra (A + B) = 20.


ii) tra (A) + tra (B) = 11 + 9 = 20.
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 
iii)  A =  2 
 3
2
4
5
3
5
6

 and hence tra ( A) = 11  and tra (A) = 11 , for any   .


Definition 2.16 Diagonal Matrix
A square matrix whose non-diagonal elements are all equal to zero is called a
diagonal matrix.
0
0
1

Example 27 
 and  0
0
1


 0
0 0
0 0  are examples of diagonal matrices.
0 1 
Note that: i) Any diagonal matrix is both upper and lower triangular matrix.
ii) Any n-square zero matrix is a diagonal matrix.
Definition 2.17 Unit Matrix
A diagonal matrix whose diagonal elements are all equal to one is called a
unit matrix.
If the order of a unit matrix is n, then we can denote the unit matrix by I n .
1
Example 28 
0
1
0
= I and  0
2
1 
 0
0 0
1 0  = I are examples of unit matrices.
3
0 1 
Exercises 2.2
1. If A and B are square matrices such that AB = BA, then show that ( A  B) 2 = A2  2 AB  B 2 .
3 1 1 


2. If A =  1
1  1  , then show that A 2 – 3A + 2 I 3 = 0
1  1 1 


3. Show that any square matrix A can be written as the sum of symmetric and skew- symmetric matrices.
Use the result to write A as the sum of symmetric and skew-symmetric matrices.
0
A =  1
 2
3
2
1
2
3  .
4 
 1
 1
4. If A and B are 2-square matrices such that A = At , B =  B t and A + B = 
3
, then
 5 
find the two matrices A and B.
5. Prove that for any two square matrices A and B of the same size and any   .
i) tra (A + B) = tra (A) + tra (B).
ii) tra ( A) =  tra (A).
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 1
6. Given: Any non-singular 2-square matrix A can be written as A = LU, where L =  l
 11
u
(Lower triangular matrix) and U =  11
 0
0 
1 
u 12 
(Upper triangular matrix). Find L and U
u 22 
for which A = LU where
 1
 1
A= 
3
.
 5 
In exercise 7 and 8, refer the information given below.
Given: Two small cottage industries 1 and 2 were opened to produce chairs and tables. Matrices A and B
represent the production capacities and the cost of production in birr per week respectively .
chairs
A =  42
 30

Tables
6   and B =  3690  1
1
 3600 
12  

 2
2
7. Calculate the cost of production of a single chair and a table.
8. If the total profit of the two cottage industries 1 and 2 is birr 1,560.00 and 1,500.00 respectively, then
calculate the selling price of a single chair and table.
2.3 Elementary Row Operations and Echelon Forms
Elementary Row Operations
E1 : Interchange the i th and the j th rows.
R i  Rj
E 2 : Replace the i th row by a non-zero scalar multiple of itself.
R i  k R i , where k  0.
E3 : Replace the i th row by k times the j th row plus the i th row.
R i  k Rj + R i .
E1 , E 2 and E3 are called Elementary row operations.
Note that: We can apply E 2 and E3 in one step.
R i  k R j + k  R i , where k , k   K and k   0.
Definition 2.18Row Equivalent Matrices
Matrix A is said to be row equivalent to matrix B if B can be obtained
from A by applying a finite sequence of elementary row operations on A.
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1
3
and B =
5 6 
3
1 2 3
 


6
0 3 6
Example 29 Let A = 
4
1
Solution A = 
4
2
5
2
1
0

R2
2
3
3
. Show that A and B are row equivalent.
6 
 4 R1  R 2 .
Therefore, A is row equivalent to B.
Definition 2.19 A matrix A = (a ij) is an echelon matrix, or in echelon form if the number
of zeros preceding the first non-zero entry of a row increases row by row until only
zero rows remain. That is if there exist non-zero entries
a 1j ,a
1
, ... , a
where j  j  j  ... j
1
2
3
r
2j2
rj r
with the property that a ij = 0 for for i  r, j  j and for i  r.
i
, ... , a
We call a 1 j , a
the distinguished elements of the echelon matrix A.
rj r
1 2j2
 1

Example 30 A = 0

 0
0
1
0
2
3  , B =
0 
 1
 0

 0
4 
0  , C =
2 
3
1
0
2
0

 0
1
0
0
0
3
0
3
0
0
4
0 
2 
are examples of echelon matrices while
 1

E= 0

 0
0
1
2
0
0  and F =
0 
 1
 0

 0
2
0
0
3
0 
1 
are examples of matrices that are not echelon matrices.
In particular, an echelon matrix is called a row reduced echelon matrix if the distinguished elements are:
i) The only non-zero entries in their respective columns.
ii) Each equal to 1.
 1

Example 31 A = 0

 0
0
1
0
2
3  , B =
0 
 1
 0

 0
0
1
0
4
0  and D =
0 
1
0

 0
4
0
0
0
1
0
3
0
0
0
0 
1 
are examples of row-reduced echelon matrices.
Note that: The zero matrixes 0 for any number of rows or columns are row reduced echelon matrices.
Algorithm that Row Reduces a Matrix to an Echelon Form
Step 1 Suppose the j 1 column is the first column with a non-zero entry. Interchange the rows so that this
non-zero entry appears in the first row, that is, so that a 1 j  0.
1
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Step 2 For each i  1, apply the operation
R i   a ij R1 + a 1 j R i .
1
1
Repeat steps 1 and 2 with the sub-matrix formed by all the rows excluding the first. Continue the process
until the matrix is in echelon form.
Example 32 Find two different echelon forms of
1
3
.
7
A= 
2
Solution By step 1, since a 11 = 1  0 and j1 =1 no need to interchange row 1 and row 2.
By step 2, for the 2nd row apply the operation
R2   2 R1 + R2
and hence R2  (0, 1).
1
0
3
. But this is an echelon matrix.
1 
 1 0
Again if we apply the operation R1   3 R2 + R1 we get 
 which is an echelon matrix.
 0 1
Thus A is row equivalent to 
 1 0
 0 1  are two different echelon forms of matrix A.


4
 1 3

Example 33 Reduce A = 2
7
8  to an echelon form.

 5 2
3 
4
 1 3
3
4 
 1

  0 1
 R2  R2  2R1
Solution
0
2
7
8

 R  5R  R


3
1
3


0
13
17
 5 2

3 
1
Therefore, 
0
3
and
1
 1
  0
 0
 1
Therefore,  0
 0
3
1
0
4
0
17
3
1
0
4
0
17




R3  R3  13 R2

 is an echelon form of matrix A.


1
Example 34 Reduce A =  4
 2
2
0
5
1
2
1
0
1
2
4
3
1

 to an echelon form.


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1
Solution  4
 2
2
0
5
1
2
1
0
1
2
4
3
1

1
  0



 0
1
  0
 0
1
Therefore,  0
 0
1
6
 30
2
8
0
1
Example 35 Reduce A =  4
 2
1
Solution  4
 2
1
2
1
2
8
1
2
5
0
1
3
6
2
8
0
0
1
2
1
6
 30
4
13
7




R2  4 R1  R2
R3  2 R1  R3
4 
13 
61 
0
1
7
0
1
7
4 
13  is an echelon form of matrix A.
61 
2
8
1
1
2
1
0
1
2
0
1
2
4
3
1
0
2
1
4
3
1
1

  0



 0
1
  0
 0
1
Therefore,  0
 0
1
6
3
2
8
9
4
7
13

 to an echelon form.


2
0
5
2
5
0
R3  9R2  8R3
1
6
3
1
3
6
0
1
2
0
2
1
4
13
7
4
7
13








R2  4 R1  R2
R3  2 R1  R3
R2  R3

 is an echelon form of matrix A.


Note that: i) echelon form of a non-zero matrix is not unique.
ii) The row-reduced echelon form of a matrix is unique.
Rank of a Matrix
Definition 2 20 If A is an m  n matrix and B is an echelon matrix that is row equivalent
to matrix A , then the number of non-zero rows (or number of columns that contains
distinguished elements) of matrix B is called rank of A, written rank (A).
Theorem 2.1 Row-equivalent matrices have the same rank.
Theorem 2.2 A matrix and its transpose have the same rank.
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Example 36 Find the row reduced echelon form of matrix A and determine its rank, where


A= 



1
2
0
0
2
5
3
2
0
2
5
6
15
10
8
18
3 
6 
0 

6 
Solution Now reduce matrix A to row reduced Echelon form.


A= 



1
2
0
0
2
5
3
2
0
2
5
6
15
10
8
18
3 
6 
0 

6 


~ 



1 2
0
1


~ 



0
0
5
10
0
3
0
2
15
18
10
8
0
3
0
2


~ 



1 2
0 1
0
0
0
12
0
12


~ 



1 2
0 1
0
3
0
2
0
0
0
0
12
0
12
0


~ 



1
0
0
1
6
3
4
2
0
0
0
0
12
0
12
0
1
0
0
2
3
0
1
0
1
0
0
0
1
1
0
0
0
0
0
0
0
0


Therefore, the row reduced echelon form of A is 


3 
0 
0 

0 
R 2  2 R1  R2
R 4   2 R1  R4
3 
0 
0 

0 
3 
0 
0 

0 
R3  R3  5R2
R4  10 R2  R4
R3  R4
3 
0 
R  R1  2 R2
0  1

0 
1
R1  R1  R3

2

1

R2  R 2  R3

4

1
R3 
R

12 3
1
0
0
2
0
1
0
1
0
0
1
1
0
0
0
0
3 
0 
 and rank (A) = 2.
0 
0 
Note that: For any m  n matrix A, rank (A)  min. m, n.
Exercises 2.3
1. Give an example of an upper triangular matrix that is not an echelon matrix?
2. Give an example of an echelon matrix that is not an upper triangular matrix?
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In exercise 3-8, reduce the given matrices to an echelon forms and determine their ranks.
0 2

3. A =  1
3
 2 1

1
6. A =  0
 4
1

4
0 
0

4. A =  1
2

3
3 
6 
2
1
2
2
3
1
0
7. A =  1
 2
0

6
1 
1
4
0
0
5. A =  0
 0
1
4 
0 
2
3
1
3
4
7
4
1
3
0 2
8. A =  1 3
 2 1
1
4
0
6
2
8
0
6  .
1 
In exercise 9 and 13, reduce the given matrices to row reduced echelon forms.
1
9. A =  0
 4
3
3 
6 
2
1
2
 2
12. A =  1
 1
0
10. A =  1
 2
1 
2 
1 
1
3
2
 0
13. A =  2
 1
1
4 
0 
2
3
1
1
2
0
3
1
5
0
2
3
 2 1
1
4
0
11. A =  1
9
8 
8 


14. A = 



0
6  .
1 
1
5
2
9
3
6
1
2
2
8
6
6
1 
3 
2 

1 
2.4 Inverse of a Matrix
Definition 2.21 If A is a square matrix and if a matrix B of the same size can be found such
that A B = B A = I, then A is said to be invertible and B is called the inverse of A.
3
1
Example 37 Let B = 
1
0
Then A B = 
5
 5
 2
and A = 
.

3
2
 1
0
 1 0
and B A = 
.

1
1
0
Hence, B is the inverse of A.
Notation If A is invertible, then the inverse of A is denoted by A  1 .
Theorem 2.3 The inverse of an invertible matrix is unique.
Theorem 2.4 The matrix
a
 c
A= 
b
d 
is invertible if and only if ad  bc  0, in which case the inverse of
matrix A is given by the formula:

1  d
 
A  1 = 
 ad  bc    c
b 
a 
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Unit II
Matrices and Determinants
7
3
2
. Find A  1 .
1
Example 38 Let A = 
 1
 3
Solution Now 7  6 = 1  0 and hence by theorem 2.4 A  1 = 
 1
 3
Therefore, A  1 = 
2 
.
7 
2 
.
7 
Definition 2.22 An n-square matrix A is called non-singular if rank (A) = n.
Otherwise A is called singular.
1
Example 39 Let A =  4
 7
2
5
8
3
6  . Show that A is singular.
9 
Solution We need to show that rank (A)  3.
1
A =  4
 7
2
5
8
3
1

6    0
 0
9 
1

 0

 0
2
3
6
3 
6 
12 
2
3
0
3
6 
0 
R2  4 R1  R2
R3  7 R1  R3
R3  2 R2  R3
Hence rank (A) = 2  3.
Therefore, A is singular.
Existence and Determination of Inverse
(Gauss-Jordan Method)
To check whether a given square matrix A is invertible or not we need to find a finite sequence of
elementary row operations that reduces A to an echelon matrix. If rank (A) is equal to the order of matrix A,
then we can conclude that matrix A has an inverse; otherwise matrix A is singular. Whenever A is
invertible find a finite sequence of elementary row operations that reduces A to the unit matrix I n and then
perform these sequences of elementary operations on I n to obtain A  1 .
1

Example 40 Determine whether matrix A = 2

 1
2
5
0
3
3
8

 is invertible or not, if invertible find A 1.


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Unit II
Solution
Matrices and Determinants
1
2
1
2
5
0
3
3
8
1
0
0
0
1
0
0
0 
1
1
0
0
2
1
2
1
0
0

2
1
0
3
3
5
3
3
1
1
2
1
1
2
5
0
1
0
0
1
2
0
0
1
0
0
1
R2  R2  2R1
R3  R3  R1
R3   2 R2  R3
Now rank (A) = 3 and hence A is non-singular.
1
2
1
2
5
0
3
3
8
1
0
0
0
1
0
0
0
1

1
 0
0
Therefore,
A 1 =
  40 16
 13  5

 5  2
 1
Example 41 Let A =  2
  1
1
0
0
0
1
0
0
1
0
9
3
1
0
0
1
2
1
2
5
2
5
 40
13
5
16
5
2
0
0
1
9
3
1
R1   2 R2  R1
R1   9 R3  R1
R2  R2  3R3
9 
 3 
1 
4
 1  . Determine whether matrix A is invertible or not. If A is invertible
5 
6
4
2
find A  1 .
1
Solution 2
1
6
4
2
4
1
5
1
0
0
0
1
0
0
0 
1
1
0
0
6
8
8
4
9
9
1
2
1
0
1
0
0
0
1

1
0
0
6
8
0
4
9
0
1
2
1
0
1
1
0
0
1
R2  R2  2R1
R3  R3  R1
R3  R2  R3
Now we get a zero row and hence A cannot be row equivalent to I3 .
Therefore, A is a singular matrix.
Theorem. 2.5 If R is the row reduced echelon form of an n -square matrix A,
then either R has a row of zeros or R is the identity matrix I n .
Theorem. 2.6 If A and B are invertible matrices of the same size, then:
AB is invertible and ( A B)  1 = B  1 A  1 .
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Unit II
Matrices and Determinants
Theorem. 2.7 If A is an invertible matrix, then:
i) A  1 is invertible and ( A  1)  1 = A.
ii) For any natural number n, A n is invertible and ( A n )
1
= ( A  1) n .
iii) For any non-zero scalar k, the matrix kA is invertible and (kA)  1 =
1
A 1 .
k
iv) A t is also invertible and ( A t )  1 = ( A  1 ) t .
Theorem. 2.8 An n-square matrix is invertible if and only if its row reduced
echelon form is I n .
Exercises 2.4
In exercises 1-3, determine the value(s) of t for which each of the following matrices fails to be invertible.
t 1 0 

1. 0 t
t 

15 17 t  1
 1
2.  4
 t  2
2
t
8
t
 2

3.
3
 0
 1  3t
3 
6 
t  4 
t3 
4 
4 
In exercises 4-9, determine whether the following matrices are invertible or not. If invertible find the
inverse.
1 2 3
5. 3 1 0
1 1 1
1 2 0  1 
1 3
 2
1  1 2
0 



7.
8.
3
7 
 0
1 3 0
2
 1  3
4 


1
1 0 3
2. Give an example of a two-square matrix A such that A 2 = I 2
1
 3
4. 

  8  4
1
6. A =  2
 3
2
 1
 1
3
9. 
 2
4

1
 1
3
5 
6 
2
4
5
3
3
3
1
1
2 
3

1
3. For each of the following matrices find ( A n )  1 and (2 A)  1 .
1
 2

i) A =  0
3
 1 3

3
7
4








ii) A = 



1 2
1 1
1
1
3
0
0 1 

2
0
0
2

3
1 
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Unit II
Matrices and Determinants
2.5 The Determinant of a Matrix
The determinant of a square matrix of order n is called a determinant of order n.
Second Order Determinant
Definition 2.23 The determinant of a second order matrix A = ( aij ) is denoted and
defined by:
D = det (A) =
4
2
a11
a12
a21
a22
= a11 a22  a12 a21
3
. Find A .
5 
Example 42 Let A = 
= (4  5) – (2 3) = 14.
Solution By definition 2.21, A
Therefore, A = 14.
Third Order Determinant
Definition 2.24 The determinant of a third order matrix A= ( aij ) is defined by:
det (A) = A
=
= a11
a 11
a 12
a 21
a 22
a 13
a 23
a 31
a32
a33
a 22
a 23
a 32
a 33
 a12
a 21
a 23
a 31
a 33
+ a13
a 21
a 22
a 31
a 32
= a11 a22 a33  a11 a23 a32 + a12 a23 a31  a12 a21a33
+ a13 a21 a32  a13 a22 a31
 1
Example 43 Let A =  2
 1
3
6
0
0
4 . Then find A .
2 
6
0
Solution By definition 2.22 A = 
Therefore, A
4
=  12.
2 
=  12.
 1
Example 44 Let A =  2
 0
3
6
0
Solution By definition 2.22
Therefore, A
4
 2
3 

2
 1

 , then find A .


3
1
A = 2
 = 0.
2
6


0
4
2
= 0.
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41
Unit II
Matrices and Determinants
Determinant of any Order n
Definition 2.25 A determinant of order n is a scalar associated with an n-square matrix
A = ( aij ) defined:
for n = 1 by det (A) = a11 and for n  2 by det (A) =
n
or det (A) =
 a ij c ij
n
 a ij c ij
, where i = 1, 2, … , n.
j 1
, where j = 1, 2, … , n
i 1
where cij = (1) i  j M ij called the cofactor of aij and M ij is the determinant of the submatrix of A obtained from A by deleting the i th row and the j th column from A called the
minor of aij .
 2
Example 45 Let A =  2
 1
0
4 . Find A .
2
3
6
0
Solution The minors are:
M11 =
M 21 =
M 31 =
6
4
0
3
2
0
0
2
3
0
6
4
= 12, M 12 =
= 6,
2
4
1
2
M 22 =
= 12 ,
M 32 =
= 8, M 13 =
2
0
1
2
2
0
2
4
2
6
1
0
=4,
M 23
=8,
M 33 =
=6
1
3
1
0
2
3
2
6
= 3
=6
and the cofactors are:
C12 =  8,
C22 = 4 ,
C32 =  8 ,
C11 = 12,
C 21 =  6,
C31 = 12 ,
C13 = 6
C23 =  3
C33 = 6
Therefore, A = 0.
Hence the cofactors of det (A) are:

 M ij if i  j is even.
c ij  
 M ij if i  j is odd.


n
i j
Therefore, det (A) =
(1)
a ij M ij , where i = 1, 2, … , n
j 1
n
i j
or det (A) =
(1)
a ij M ij , where j = 1, 2, … , n
i 1


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Unit II
Matrices and Determinants
Remark: The determinant of any triangular matrix is the product of all the entries on the main diagonal.
Properties of the Determinant
Theorem 2.9 The determinants of an n-square matrix and its transpose are equal.
2
3 
1

Example 46 Let A = 0  2
4  . Then show that A = At .

 3
2
0 
0
3 
1

Solution At = 2  2
2  .

 3
4
0 
2
3  1
2
3 
1



Now 0  2
4    0 2
4  R3  3R1  R3

 3
2
0   0
4
9 
2
3 
1

 0 2
4  R3  2R2  R3

 0
0
17 
Thus, A = 34.
1
2

 3
0
2
4
3 
3 
1 2
R2  2 R1  R2


4 
2 0 2
R3   3R1  R3
 0 4  9 
0 
2
3 
1

 0 2
4  R3  2R2  R3

 0
0
17 
Thus, At = 34.
Therefore, A = At .
Theorem 2.10 If one of the rows (or columns) of an n-square matrix A is a multiplied by
a scalar k, then the determinant of the new matrix is k times det (A).
Further more det (kA) = k n det (A).
1

Example 47 Let A = 0

 0
2
2
2
3
4
0

 . Then find det (2A).


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Unit II
Matrices and Determinants
Solution By theorem 2.11, det (2A) =  2 4
1
3
0
4
=  64.
Therefore, det (2A) =  64.
Theorem 2.11 If every entries in a row (or column) of an n-square matrix is zero,
then its determinant is zero.
 5

Example 48 Let A = 15

 0
1
3
0
8
6
0

 5
 and B =  15



 10
1
3
5

.


0
0
0
Then show that det (A) = det (B) = 0.
Solution By theorem 2.12, det (A) = det (B) = 0.
Therefore, det (A) = det (B) = 0.
Theorem 2.12 If each entries in a row (or column) of an n-square matrix is the
sum of m terms, then the determinant can be written as the sum of
m determinants.
 5

Example 49 Let A = 15

 10
1
3
4
8
6
2

 . Then show that A can be written as the sum of B and C,


where
1

i) B = 4

 3
 5

ii) B = 5

 10
1
4
3
1
1
4

 4
 and C =  12



 6
8 
 5
8  and C =  10
 10
2 
8
2
6
3
4
8 
6  .
2 
1
2
4
8
2
2
1




Solution Using the definition of determinants we get:
A = 5 (6  24)  (30  60)  8 (60  30) = 180,
i) B = (24  6)  (24  6)  8 (0 ) = 0 andC = 4 (6  24)  (24  36)  8 (48 18 ) = 180.
Therefore, A= B+C.
ii) B= 5 (2  32)  (10  80)  8 (20  10 ) = 0 andC = 5 (4  8)  (20  20)  8 (40  20 ) = 180.
Therefore, A= B+C.
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Unit II
Matrices and Determinants
Theorem 2.13 If any two rows (or columns) of an n-square matrix are interchanged,
then the value of the determinant is multiplied by  1.
 5
Example 50 Let A =  15

 10
1
3
4
8
6
2

 and B =


 5
 10

 15
8 
2  . Then show thatA =  B.
6 
1
4
3
Solution By theorem 2.14, matrix B is obtained from matrix A by interchanging the 2 nd and the 3 rd rows
of matrix A.
Therefore, A =  B.
Theorem 2.14 If corresponding entries in two rows (or two columns) of an n-square
matrix are proportional, then its determinant is zero.
 1
Example 51 Let A =  2
 10
3
6 . Then show that det (A) = 0.
2
2
4
4
Solution By theorem 2.15, R2 = 2 R1 and hence det (A) = 0.
Therefore, det (A) = 0.
Theorem 2.15 If a matrix obtained from an n-square matrix by adding to the elements
of its i th row (or column) a scalar multiple of the corresponding elements of
another row (or column), then the determinant remains unchanged.
 5

Example 52 Let A = 15

 10
1
3
4
8
6
2

 and B =


 5
 15

 15
1
3
5
8
6  . Then show that det (A) = det (B).
10 
Solution Matrix B is obtained from matrix A by replacing the 3 rd row of A by the sum of the first and the
third rows of A.
Therefore, by theorem 2.16, det (A) = det (B).
Theorem 2.16 For any two n-square matrices A and B
AB  A B
 9
 2
Example 53 Let A = 
3
and B =
0 
1
2

4 
. Then evaluate det (A), det (B) and det (A B).
5 
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45
Unit II
Matrices and Determinants
 21 
.
8 
 15
 2
Solution Now A B = 
Therefore, det (A) = 6, det (B) = 13 and det (A B) = 78.
Adjoint of a Matrix
Definition 2.26 The Adjoint
Let A = ( aij ) be an n-square matrix and c ij be the cofactor of aij . Then
the adjoint of A is denoted and defined by:
Adj A = (cij ) t
Example
Theorem 2.17 For any n-square matrices A,
A · (adj A) = (adj A ) ·A = det (A) · I n .
Thus if det (A)  0
Adj A
det ( A)
A 1 =
Note that: If A and B are n-square matrices, then
ii) If det (A)  0, then det adj A = (det ( A)) n  1
i) adj A B = adj B · adj A
Inverse of a Matrix Using the Determinant
Example 50 Find A  1 where
1

A= 2

 1
3
3 
8 
2
5
0
Solution Det (A) =  1
and  11 = 40;
 12 =  13;
 13 =  5

21 =  16;

22 = 5;


31 =  9,

32 = 3;

Therefore,
A 1
  40
 13
=

 5
16
5
2
23 = 2
33
9
3
1
= 1.

.


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46
Unit II
Matrices and Determinants
Example 51 Find A  1 , where
1
A=  2

 3
3
5 
6 
2
4
5
Solution Det (A) =  1
and  11 =  1;
 12 = 3;
 13 =  2

21 = 3;

22 =  3;


31 =  2,

32 = 1;

 1
Therefore, A  1 =   3
 2
3
3
1
23 = 1
33
= 0.
2
1 .
0
Exercises 2.5
1. Evaluate the determinant of each of the following matrices by first reducing it to an echelon form.
 1

 5
ii) B = 
1
1

1
 2

2. If A is an n- square matrix, show that |kA| = k n |A|
1

i) A =  3
1

2
3

0
1 
2
9
3
6
2
8
6
6
1 

3 
2 

1 
3. Let A be 4-square matrix with det (A) = 8, then find det (2A).
0 
t 1


4. Let A =  0
t
t  . Determine the value(s) of t for which A fails to be invertible.
15 17 t 1


5. Find the inverse of each of the following matrices by the adjoint method, if it has any.
2
i) A = 
4
 3

0 
cos θ
ii) B = 
 sin θ
2
 3

iii) C =  4
0
 7 2

 sin θ 

cos θ 
1

7
5 
In exercise 15 - 20, evaluate the determinants of the following matrices.
0
15. A =  1
 2
2
3
1
0 2
18. A =  1 3
 2 1
1
4 
0 
1
4
0
0
6 
1 
 1
16. A =   2
 5
3
4
2
 0
19. A =  2
 1
1
2
0
0
1 
2 
3
1
5
9
8 
8 
 2
17. A =  1
 1
2
 1
 5
9
20. 
 1
2

8
 2
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1 
2 
1 
1
3
2
3
6
6
6
1
3 
2 

1
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Unit II
Matrices and Determinants
In exercise 21 - 26, reduce each of the following matrices into an echelon form and then use properties of
determinants to evaluate their determinants from the steps you used.
0
21. A =  1
 2
1
4 
0 
2
3
1
0 2
24. A =  1 3
 2 1
22. A =   2
3
4
2
 0
25. A =  2
 1
1
2
0

1
 5
0
6 
1 
1
4
0
0
1 
2 
3
1
5
9
8 
8 

2
23. A =  1
26.






1
5
1
2
 1
2
9
2
8
1 
2 
1 
1
3
2
3
6
6
6
1
3 
2 

1
2.6 System of Linear Equations
A linear system of m equations in n unknowns x1 , x2 , x3 , ... , xn is a set of equations of the form:
a x1  a x2  a x3  ...  a xn = b1
11
12
13
1n
a x1  a x2  a x3  ...  a
21
22
23
x
2n n
= b2
…
a
x 
m1 1
a
x
m2 2
 am3 x3  ...  amn xn = bm
where a ij ’s are given scalars called coefficients of the system and the bi ’s are given scalars.
If the bi ’s are all zero, then this system is called a homogeneous system of linear equations and if at least
one bi is not zero, then the system is called a non-homogeneous system of linear equations.
A solution of this system is a set of an n-tuple of scalars ( 1 ,  2 ,  3 , ... ,  n ) that satisfies all the m
equations.
Note that: If the system is homogeneous, then it has at least the trivial solution
x1  x2  x3  ...  xn  0 .
In matrix notation the above system of linear equation can be written as
 a11
a
 21
 ...

a m1
a12
a 22
...
...
...
am2
...
...
a1n 
a 2n 
... 

a mn 
 x1   b1 
 x  b 
 2   2
. . . . . .
   
 x n  b m 
Or more compactly as:
AX=B
Where A = (a )
ij m  n
called the coefficient matrix X = ( x1 , x2 , x3 , ... , xn ) t and B =
(b1 , b2 , b3 , ... , bm ) t .
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Unit II
Matrices and Determinants
The matrix


Ã= 



a 11
a 12
..........
a 1n
a 21
a 22
...........
a 2n
....
....
...........
....
a m1 a m2 ...........
a mn
b1 
b 2 
.... 

b m 
is called the augmented matrix associated to the above system of linear equations.
Definition 2.27The augmented matrix à of a given system of linear equations
expressed in echelon form the unknowns x j where the j th column does
not contain a distinguished element are called free variables.
2.6.1 Homogeneous System of Linear Equations
In a homogeneous system of m linear equations in n unknowns given by
A X = 0 m 1
Where A is the m  n coefficient matrix and X = ( x1 , x2 , x3 , ... , xn ) t .
There are two possibilities:
i)
If rank (A) = n., then the system has only the trivial solution.
ii) If rank (A)  n, then the system has a non-zero (non-trivial) solution.
Theorem 2. 18 A homogeneous system of linear equations with more unknowns
than equations has a non- trivial solution.
Let r = rank (A). To solve such system of linear equations assign any arbitrary value for each of these
n  r free variables and using back substitution solve for the non-free variables in terms of the values
of these free variables.
Example 51 Solve:
x + 2y  3z + w = 0
x  3y + z  2 w = 0
2x + y  3z + 5 w = 0
Solution Reduce the augmented matrix to row reduced echelon form.
1
1
2
2
3
1
3
1
3
1
2
5
0
0
0
~
1
0
0
2
5
3
3
4
3
1
3
3
0
0
0
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R2  R1 + R2
R3  2R1 + R3
49
Unit II
Matrices and Determinants
3
4
3
~
1
0
0
2
5
0
~
5
0
0
0
5
0
~
15
0
0 15
0
0
7
4
3
0
0
3
1
3
 24
0
0
0
R3  3R2 5R3
1
 3
 24
0
0
0
R1  2R2 + 5R1
165
105
 24
0
0
0
R1   7R3 + 3R1
R2 4R3 + 3R2
Now w is the free variable. Let w = t, then x =  11t, y =  7t, z =  8t and w =t.
Therefore, ( 11t,  7t, 8t, t): t   is the solution set of system.
Example 52 Solve:
x+y z=0
2 x + 4y  z = 0
3x + 2y  2z = 0
Solution Reduce the augmented matrix to row reduced echelon form.
1
2
3
1
4
2
1
1
2
0
0
0
1
0
0
~
~
~
1
1
1
1
2
1
0
0
0
1
0
0
0
0
0
2
0
1
3
0
0
1
0
0
0
0
0
6
0
0
3
0
0
R2  2R1 + R2
R3  3R1 + R3
R3  R2 + 2R3
R2   R3 + 3R2
Therefore, x = y = z = 0 is the only solution.
2.6.2 Non-homogeneous System of Linear Equations
In a non-homogeneous system of m linear equations in n unknowns given by:
AX=B
where A is the m  n coefficient matrix, X = ( x1 , x2 , x3 , ... , xn ) t and B = (b1 , b2 , b3 , ... , bm ) t .
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Unit II
Matrices and Determinants
Let à be the augmented matrix associated with the above system.
There are two possibilities:
i)
rank (A)  rank (Ã).
In this case the system is inconsistent and has no solution.
ii) rank (A) = rank (Ã).
If rank (A) = n, then the system has a unique solution given by
X = A 1 B
If rank (A) = r  n, then the system has more than one solution.
Example 53 Solve the following system of linear equations.
a) 2x + y  2z + 3w = 1
b) x + 2y  3z = 1
c) x + 2y  2z + 3w = 2
3x + 2y  z + 2w = 4
x + 3y + z = 11
3x + 3y + 3z  3w = 5
2x + 5y  4z = 13
2x + 4y  3z + 4w = 5
5x + 10y  8z + 11w = 12
Solutions a) Reduce the augmented matrix to row reduced echelon form:
2
1
2
3
1
3
3
2
3
1
3
2
3
4
5
2
0
0
~
~
~
1
1
3
2
4
12
3
5
15
1
5
7
2
0
6
8
4
0
0
1
0
4
0
5
0
5
8
1
0
0
0
1
0
3
4
0
4
5
0
0
0
1
Now rank (A) = 2  3 = rank (Ã)
R2   3R1 + 2R2
R3  3R1 + 2R3
R1   R2 + R1
R3  3R2 + R3
1
1
R3 + R1
4
2
1
R3   R3
8
R1  
Therefore, the system has no solution.
b) Reduce the augmented matrix to row reduced echelon form:
1
2
3
1
1
2
3
5
1
4
11
13
~
1
2
3
1
0
0
1
1
4
2
10
11
R2   R1  R2
R3   2 R1  R3
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51
Unit II
Matrices and Determinants
~
~
1
0
11
19
0
0
1
0
4
2
10
1
2
0
0
0
1
0
R1   2 R2  R1
R3   R2  R3
 49
12
1
0
0
2
R1   11R3  2 R1
R2  R2  2 R3
Now rank (A) = 3 = rank (Ã) = the number of unknowns.
Therefore, ( 
49
1
, 12,  ) is the only solution for the system.
2
2
c) Reduce the augmented matrix to row reduced echelon form:
1
2
5
2
4
10
2
3
8
3
4
11
2
5
12
~
~
1
2
2
3
2
0
0
0
0
1
2
2
4
1
2
1
0
0
2
0
0
0
1
0
1
2
0
R2   R1 + R2
R3  2R1 + R3
4
1
0
R1  2R2 + R1
R3  2R2 + R3
Now rank (A) = rank (Ã) = 2  4 = the number of unknowns and hence y and w are free variables. Thus
z 2w = 1  z = 2w + 1 and x + 2y  w = 4  x = 4 + w  2y.
Therefore, (4 + w  2y, y, 2w + 1, w) is the solution for any y, w  k.
Cramer’s Rule
Theorem 2.19 If the determinant of the coefficient matrix A of a system of n linear
equations in n unknowns x1 , x2 , x3 , ... , xn is not zero, then the system
has precisely one solution. This solution is given by the formulas
xj =
det( A j )
det ( A)
for j = 1, 2, 3, … , n
where Aj is the matrix obtained from A by replacing the j th column of A by the
column with entries b1 , b2 , b3 , ... , bm .
Note that: If the system is homogeneous and det (A)  0, then it has only the trivial solution
x1  x2  x3  ...  xn  0 .
and if det (A) = 0, then the homogeneous system also has non-trivial solution.
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Unit II
Matrices and Determinants
Example 54 Solve the following by Cramer’s rule.
a)  x + 3y  2z = 7
b) 2x + 5y + 3z = 1
+ 3z =  3
3x
 x + 2y + z = 2
2x + y + 2z =  1
x+y+z=0
 1
Solutions a) Let A =  3

 2
2 
3  .
2 
3
0
1
Then det (A) =  3; det ( A ) =  6; det ( A ) =  3 and det ( A ) = 9.
1
3
2
Therefore, x= 2; y = 1andz =  3.
 2

b) Let A = 1

 1
5
2
1

 .


3
1
1
Then det (A) = 3; det ( A ) =  3; det ( A ) = 0 and det ( A ) = 3
1
3
2
Therefore, x =  1; y = 0 and z = 1.
Example 55 Solve the following by Cramer’s rule, if possible.
b)  x + 3y  2z = 0
a) 2x + 5y + 3z = 0
 x + 2y + z = 0
3x +4z = 0
x+y+z =0
2x + 3y + 2z = 0
 2

Solutions a) Let A = 1

 1
3
1  .
1 
5
2
1
Then det (A) = 3
Therefore, (0, 0, 0) is the only solution for the system.
 1

b) Let A = 3

 2
3
0
3
2
4  .
2 
Then det (A) = 0. Thus the system has a non-trivial solution and the solution can be obtained using GaussJordan elimination method.
1
3
2
3
0
3
2
4
2
0
0
0
~
1
0
0
3
9
9
2
2
2
0
0
0
R1   R1
R2   3R1  R2
R3  2 R1  R3
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Unit II
Matrices and Determinants
~
1
0
0
1
0
0
4
3
2

9
0
0
0
0
1
R  R1
3 2
1
R2  R2
9
R3   R 2  R3
R1 
Now rank (A) = 2  3 = The number of unknowns and hence z is the only free variable. Thus y =
x=
2
z and
9
4
z.
3
Therefore, (
4 2
z, z, z) is the solution for any z  k.
3 9
The LU-Factorization
We can apply the concept of LU- Factorization for solving systems of n linear equations in n unknowns,
written
AX = B
Where A is the n-square coefficient matrix, X = ( x1 , x2 , x3 , ... , xn ) t and B = (b1 , b2 , b3 , ... , bm ) t if and
only if A is non-singular.
Whenever the coefficient matrix for the given system of n linear equations in n unknowns is non-singular the
rows of the coefficient matrix can be reordered so that the resulting matrix A can be uniquely expressed as:
PA = LU
(1)
Where L is a lower triangular matrix whose diagonal elements are all equal to one, U is an upper triangular
matrix and PA stands for the permutation of the rows of matrix A. In the process of finding an echelon form
of matrix A, there might be permutation of rows. In such cases, rearrange the linear equations in such a way
that the rows of the coefficient matrix need no more permutation of rows in finding an echelon form of the
coefficient matrix.
If A 0, then we can express AX = B as:
LUX = B  LY = B and UX = Y.
Now solve LY = B for Y and then solve UX = Y for X.
This method of solving such systems of linear equations is called Doolittle’s Method.
2
5
Example 56 Let A = 
 1
 l 21
Solution Let L = 
1
. Find the two matrices L and U for which (1) holds true.
3 
0
 u11
and U = 

1
 0
u12 
.
u 22 
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54
Unit II
Matrices and Determinants
2
5
1  1
=
3   l 21
Then 

u11 = 2, u 12 = 1,
 1
 2.5
0
1 
1 0
Example 57 Let A =  3
2

 1 0
Therefore, L = 
 1

Solution L = l21

 l31
0
1
 3

 1
0

2
0
2
0
1
 u11
= 

3
 u11 l 21

 u 22 
u12
u12 l 21
2
1  . Find the two matrices L and U for which (1) holds true.
1 
0
0  and U =
1 
0
1
l32
1

Then 3

 1
0   u11
u12 
2
 



1  0
u 22 
5
5
1
and u 22 = .
21 =
2
2
2
1


and U = 
.
0
0
.
5


2  1
1  =  l21
1   l31
 u11
 0

 0
0
0 
1 
0
1
l32
2   u11

1  =   21 u11
1    31 u11
u12
u 22
0
 u11
 0

 0
u13 
u 23  .
u33 
u12
u 22
0
u13 
u 23 
u33 
u12
u13
 21 u12 
u 22
 31 u12   32
u 22
 21 u13 
 31 u13   32
u2 3
u 2 3  u 33





 21 = 3,  31 = 1,  32 = 0, u11 = 1, u 12 = 0, u 13 = 2, u 22 = 2, u 23 =  5 and u 33 =  1
1

Therefore, L = 3

 1
0
0  and U =
1 
0
1
0
1
0

 0
0
2
0
2
 5  .
1 
Example 58 Solve: x + y + z = 5
x + 2y + 2z = 6
x + 2y + 3z = 8
Solution First find the LU decomposition of the coefficient matrix A, where
1

A= 1

 1
1
2
2
1
2 
3 
Perform a method similar to example 57 and obtain:
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Unit II
Matrices and Determinants
1
L = 1

 1
0
0  and U =
1 
0
1
1
1
0

 0
1
1
1
1
1  .
1 
Now applying the Doolittle’s Method we get:
1
1

 1
1
 1

 1
0
0 
1 
0
1
1
0
1
1
0
0 
1 
1
0

 0
 y1
y
 2
 y3
1
1 
1 
1
1
0
x
y

 z
 5 
 = 6 
  
  8 
 5 
1
 =  6  and  0
  

  8 
 0
1
1
0
1
1 
1 
x
y

 z
  y1 
 = y  .
  2 
  y3 
Now solve LY = B for Y and then solve UX = Y for X.
1
LY = B   1

 1
0
1
1
1

and hence, UX = Y  0

 0
0
0 
1 
 y1
y
 2
 y3
1
1
1 
1 
1
0
 5 
 = 6  Y=
  
  8 
x
y

 z
5
1

 2
 5 
 = 1  X=
  
  2 




 4
 1

 2




Therefore, x = 4, y =  1 and z = 2.
Exercises 2.6
1. Determine the conditions on a, b and c so that the following system of linear equations has a solution.
x – 2y – 3z = a
3x – y + 2z = b
x + 3y + 8z = c
2. Determine the value(s) of k such that the system
x – 3z = – 3
2x + ky – z = – 2
x + 2y + k z = 1 has
i) unique solution
ii) no solution
iii) infinite solutions
3. Solve the following systems of equations by the Gauss-Jordan elimination method.
i)
x + 2y – 8z = 0
2x – 3y – 5z = 0
3x – 2z = 0
iii)
w+x+y+z=–1
2w – x + y = 1
2x + 3z = – 1
– w + 2y + z = – 2
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x + 3y = – 7
iv) w – 2x + y – 4z = 1
3x + 5y + 4z = -9
w + 3x + 7y + 2z = 2
ii)
2x – y + 7z = 7
w – 12x – 11y – 16z = 5
4. Solve by Cramer's Rule. If not possible, use the Gauss-Jordan method.
i)
2x – y + z = 5
iii) 2x + 5y – 3z = 0
x + z = –1
x – 2y + z = 0
– 3x + y + 4z = 0
7x + 4y – 3z = 0
x – 4y + z = 0
iv) x + 3y + z = 2
– 4x + 7y – 3z = 5
2x + 3y – 4z = 7
ii)
– 2x – y + 8z = – 9
3x + 7y – z = 8
5. The currents I1 , I 2 , I 3 and I 4 in an electric network satisfy the system of linear equations:
3 I1 + 2 I 3 – I 4 = 60
2 I1 – I 2 + 4 I 3 = 160
4 I 2 + I 3 – 2 I 4 = 20
5 I1 – I 2 – 2 I 3 + I 4 = 0. Find I 3
6. Consider the cubic equation f (x) = ax 3  bx 2  cx  d . If the points (0, 10), (1, 7), (3, 11) and
(4,  14) are on the graph of f, find the coefficients a, b, c and d.
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DEFENCE ENGINEERING COLLEGE
Department of Basic and applied courses.
Applied Mathematics I (Math 201)
Worksheet II
(Matrices and Determinants)
January 2008
1. Find all values of a, b and c for which matrix A is symmetric, where
2
A = 3

 0
2a  b  c
a  2b  2c
2
5
ac
7
.


2. Suppose A, B, C, D and E are matrices with the following order
Matrix
A
B
C
D
E
size
45
45
52
42
5 4
determine which of the following matrix operations are defined, find the order
of
the resulting matrix whenever defined.
a) BA
b) AC + D
c) AE + B
e) E (AC)
f) E t A
g) ( At  E) D
d) E (A + B)
3. Construct a 3  2 matrix A = ( a ij ) such that
a ij
 j  2i, for i  j


2

i  2 j , for i  j
4. Given three matrices, if possible
1  2
 2
A = 3
4 , B = 
1
0  4
Compute: a) AB
b) 3C
1
2 2 3
and C = 
.

4
1 4 5 
 B At
c) (AB)C
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5. Prove that any square matrix can be expressed as a sum symmetric and skew
symmetric matrices. Use the result to write matrix A as a sum symmetric
matrix
R and skew symmetric matrix s, where
 1 0  6
A =  2 4 7


 3 5 2 
6. Reduce each of the following matrices to row-echelon form and reduced rowechelon form and determine their ranks.
 1  2
 2  4
b) 
 3
6

 4  8
 2 3  4
a)  1  5
1
 4 6  2
3
6
 9

12
1
 2
 3
0 
c) 
 1  4


2
 8
7. Find the inverses of each of the following matrices using the Gauss-Jordan
Method.
1 2 3
b) 3 1 0
1 1 1
1
 3
a) 

  8  4
8. Find the inverses of matrix A by the determinant method, where
2 3 1
A = 8 4 1
6 7 0
9. Find the determinants of the following matrices.
cos 
a) 
 sin 
 sin  
cos  
 3  2 8
b)  5 0 7
1 2 1
1
0
c) 
0

0
4
8
8  2

0  7
2
3
4 2
0
0
0 
t 1


10. Let A =  0 t
t  . Determine the value(s) of t for which A fails to be
15 17 t 1


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Unit II
Matrices and Determinants
invertible.
11. Reduce each of the following matrices into an echelon form and then use
properties of determinants to evaluate their determinants from your steps.
1

a) A =  3
1

2
1
1
 1

 5
b) B = 
1

 2

3

0
1 
2
9
3
6
2
8
6
6
1 

3 
2 

1 
12. Solve the following systems of linear equations by the Gauss-Jordan Method.
a) x + y  z = 3
b) x + 2y + 3z + w = 8
3x + y  2z = 0
3x  4 y + 8z + 2w =  1
2x  y + 3z =  1
13. a) Determine the conditions on a, b and c so that the following system of
linear equations has solution.
x – 2y – 3z = a
3x – y + 2z = b
x + 3y + 8z = c
b) Determine the value(s) of k such that the system
x – 3z =  3
2x + ky – z = – 2
x + 2y + k z = 1 has
i) unique solution
ii) no solution
iii) infinite solutions
14. Solve the following systems of linear equations by Cramer’s rule. If not
possible
use the Gauss-Jordan Method.
a)  x + y  2z = 7
b) x  2y + z = 5
3x + z =  3
x+z=4
2x + y + 2z =  1
 2x + 4y  2z =  10
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Matrices and Determinants
Defence Engineering College
Department of Basic and Applied Courses
Applied Mathematics I (Math. 201)
Work Sheet II
On Matrices and Determinants
May, 2007
1. Given matrices
 3

A =  1
 1

0
 1


1 4 2
 , C =   1
2  , B = 
3 1 5
 3
1 

ii) AB + 2C
iii) 2 A t
i) AB
5
0
2
B

2

1  , find:
4 
t
iv) C t AB
2. a) Find the missing entries “ * ” (elements) in the following product:
2 0

* 2
0 *

3

b) If B =  1
1

1
1
1
0

0
2 
2
5
2
* * 0
 4



0 * * =   2
 0 0 2
 0



1

 1  , then show that B
1 
2
– 3B + 2I3 = 0
c) If  and  are two roots of the equation 2x
 

 2 2

 
 + 
 

0

2 
5 
2
 x  1 = 0, then find
2 2 

 
3. Show that any square matrix A can be written as a sum of symmetric and
skew-symmetric matrices. Use the result to write the following matrix A as a
sum of symmetric and skew-symmetric matrices.
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 0

 1
A= 
2

 1

1
2
1
1
3
1
1
2
2 

0
4 

1 
4. a) If A and B are square matrices such that AB = BA, then show that
(A + B)2 = A2 + 2AB + B2.
b) Give an example to show that (A + B)2 = A2 + 2AB + B2 doesn't hold in
general.
5. a) Give an example of two matrices A and B such that AB = 0 but neither A =
0
nor B = 0.
b) If A is an m x n matrix and A(BA) is defined, then determine the size of B.
c) Under what condition does the following statement hold, for the
matrices A, B, C? AB = AC  B = C.
d) Is an upper triangular matrix an echelon matrix? Is the converse true?
6. Reduce each of the following matrices to its echelon form and determine its
rank and also find their corresponding row reduced echelon forms.
0 2

i) A =  1 3
 2 1

1

4
0 
0

ii) B =  1
2

2
3
1
1
4
0
0

6
1 
7. Find the inverse of each of the following matrices (if it has any) by using
elementary row operations (Gauss- Jordan method).
1
 2

i) A =  0
3
 1 3

3
7
4








ii) B = 



1 2
1 1
1
1
3
0
0 1 

2
0
0
2

3
1 
8. Evaluate the determinant of each of the following matrices by first reducing it
to its echelon form.
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3
1
 1 2

9
6
3
 5
ii) B = 
1
1
2 6
2

1
 2
8
6
1

9. a) If A is an n- square matrix, show that |kA| = k n|A|
1

i) A =  3
1

2
3

0
1 
3 2
 2

1
 3 2
b) Let A = 
3
2
3

 2
4
0

direct evaluation of the







4 

2 
. If det (A) =  286, then find det (2A) without
4 

5 
determinant.
0 
t 1


10. Let A =  0 t
t  . Determine the value(s) of t for which A fails to be
15 17 t 1


invertible.
11. Find the inverse of each of the following matrices by the adjoint method, if it
has any.
2
i) A = 
4
 3

0 
2
 3

iii) C =  4
0
 7 2

 cosθ
ii) B = 
 sinθ
 3

5
iv) D = 
2

 2

1

7
5 
 sinθ 

cosθ 
2
2
5
8
4
3
7
5
4 

5 
3 

8 
12. a) Determine the conditions on a, b and c so that the following system of
linear equations has solution.
x – 2y – 3z = a
3x – y + 2z = b
x + 3y + 8z = c
b) Determine the value(s) of k such that the system
x – 3z =  3
2x + ky – z = – 2
x + 2y + k z = 1 has
i) unique solution
ii) no solution
iii) infinite solutions
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13. Solve the following systems of equations by the Gauss-Jordan elimination
method.
i) x + 2y – 8z = 0
iii)
x1 + x2 + x3 + x4 = –1
2x – 3y – 5z = 0
2x1 – x2 + x3 = 1
3x – 2z = 0
2x2 + 3x4 = –1
– x1 + 2x3 + x4 = -2
ii)
x + 3y = – 7
iv) x1 – 2x2 + x3 – 4x4 = 1
3x + 5y + 4z = -9
x1 + 3x2 + 7x3 + 2x4 = 2
2x – y + 7z = 7
x1 – 12x2 – 11x3 – 16x4 = 5
14. Solve by Cramer's Rule. If not possible, solve by Gauss-Jordan method.
i) 2x – y + z = 5
iii)
x + z = –1
2x + 5y – 3z = 0
x – 2y + z = 0
– 3x + y + 4z = 0
7x + 4y – 3z = 0
ii) x – 4y + z = 0
iv)
– 4x + 7y – 3z = 5
x1 + 3x2 + x3 = 2
2x1 + 3x2 – 4x3 = 7
– 2x1 – x2 + 8x3 = – 9
3x1 + 7x2 – x3 = 8
15. a) The currents I1, I2, I3 and I4 in an electric network satisfy the system of
equations:
3I1 + 2I3 – I4 = 60
2I1 – I2 + 4I3 = 160
4I2 + I3 – 2I4 = 20
5I1 – I2 – 2I3 + I4 = 0. Find I3
b) Consider the cubic equation f(x) = ax3 + bx2 + cx + d. If the points (0,
10),
(1, 7), (3, 11) and (4, 14) are on the graph of f, find the coefficients
a, b, c and d.
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Unit II
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DEFENCE UNIVERSITY COLLEGE
COLLEGE OF ENGINEERING
DEPARTMENT of BASIC and APPLIED COURSES
Math 201, Group Assignment I
July 2004.
1. Let a and b be non-zero vectors in 3, and c =
b
a +
a
b . Show that if c  0 , then
c bisects the angle formed by a and b .
2. Let π 0 and π 1 be two parallel planes, given respectively by 2x – 3y + 4z = 2 and 2x – 3y + 4z = 6.
Let l be the line with symmetric equations
x5
y3
z


3
2
6
.
Assuming that the points of intersection are Q 0 and Q 1, determine | Q 0 Q 1 | without finding
either Q 0 or Q 1 .
3. Given: Any non-singular 2-square matrix A can be expressed as A = LU, where
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 1
L= 
 l 11
 2
If A = 
 8
0
u
(Lower triangular matrix) and U =  11

1
 0

3
,
5 
u 12 
u 22 
(Upper triangular matrix).
then find the two matrices L and U for which A = LU holds true.
4. Let A be an n-square invertible matrix.
i) Show that adj (adj A) = (det (A)) n - 2 A.
ii) Find all possible conditions for which adj (adj A) = A.
DEFENCE ENGINEERING COLLEGE
Department of Basic and applied courses.
Applied Mathematics I (Math 201)
Worksheet II
(Matrices and Determinants)
November 2007
1. Given: The Matrices
  2  3 4 1
 1  2 2 1
 0 2 1 4






A 1
6
0  2  , B    2 0 3  2  and C   2  3 3 0 
 1  2  3  4
 6 9 1 6
3 2 2 0 






Find a) A + B
b) (A – B) + C
c) 2A – 3B + 4C
2. Given: The matrices
 2 3 5 


A   1 4
5 
 1 3 4 


5
 1 3

B   1 3 5
 1 3
5






 2 2 4 


C  1 3
4 
 1 2 3 


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Matrices and Determinants
1

1
E 
0

1

D = 1 2 3 4
i) find a) DE
b) EF
c) DF
ii) show that a) AB = BA = 0
0 1 2 

1 2 5
0 1 2

2 1  1
0
 
  1
F  
4
 
5
 
d) EDt
b) AC = A, CA = C
3. a) find the values of the variables x and y so that the matrix
x
3 
 2


 y  x 0 4  x  becomes symmetric.
 3
y
5 

b) Express the matrix B, where
 0 1 1 2


 1 1 3 5
B 
 2 3 1 5


 1 1 2 1


as a sum of a symmetric and a skew symmetric matrix
4. Reduce the following matrices to row reduced echelon form and also determine the rank of
each.
1 3 4 6 


2 4 1 2 
a) A  
3 7 3 8


 5 11 2 10 


1

3
b) B  
2

1

2 3

6 10 
5 7

2 4 
5. Find the inverses of the following matrices using elementary row operations
1
1

 2 1
ii) 
0
2

1 0

 1 1 3


i)  2  4 7 
 4  9 2


1 1

3 0
0 3

2 1 
6. a) Find the determinant of the following
1

0
i) A  
0

0

2 1 2

2 5 1
,
0 4 3

0 0 3 
 1  2 1  1


 0 1 0 1 
ii) C  
0 1 1 2 


 2 0

0
3


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 0 0 0 


0  0 0
where θ, β, α and δ are non  zero real numbers
iii) B  
0 0  0


0 0 0  


a bc 1
b) Give your justification, without expanding that b c  a 1  0 .
c ab 1
7. Find the adjoint and then the inverses of each of the following matrices.
 2 1 1


b) B   1 0 1 
 3 1 4


1 1

a) A  
0 4
8. Solve the following systems of linear equations.
x  y  2z  1

a)  2 x  z  2
x yz 3

 x  2 y  4 z  3w   2

b)  3x  y  3z  2w  0
 x  3 y  z  4w   8

 x  4y  z  0
c) 
 4 x  7 y  3z  5
2 x  4 y  2 z  1

d) 2 x  5 y  2 z  1
x  y  z 1

1. Find all values of a, b and c for which matrix A is symmetric, where
2

A = 3
0

a  2b  2c
5
ac
2a  b  c 

2
.
7


2. Suppose A, B,C, D and E are matrices with the following order
Matrix
A
B
C
D
E
size
45
45
52
42
5 4
Determine which of the following matrix operations are defined, find the order
of
the resulting matrix whenever defined.
a) BA
b) AC + D
c) AE + B
e) E (AC)
f) E t A
g) ( At  E) D
d) E (A + B)
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3. Construct a 3  2 matrix A = ( a ij ) such that
a ij

 j  2i ,


2

i  2 j ,
for i  j
for i  j
4. Given three matrices, if possible
1  2
 , B =  2 1 and C = 2 2 3 .
A = 3
4
1 4 5 
1 4






0  4
Compute
b) 3C  B At
a) AB
c) (AB)C
5. Prove that if A is a square matrix, then A + At is symmetric and A  At is skew
symmetric.
6. Express matrix A as a sum of symmetric and is skew symmetric matrices,
where
 1 0  6
A =  2 4 7


 3 5 2 
7. Reduce each of the following matrices to row-echelon form and reduced rowechelon form and determine their ranks.
 2 3  4
a)  1  5
1
 4 6  2
 1  2
 2  4
b) 
 3
6

 4  8
3
6
 9

12
1
 2
 3
0 

c)
 1  4


2
 8
8. Find the inverses of each of the following matrices using the Gauss-Jordan
Method.
1
 3
a) 

  8  4
1 2 3
b) 3 1 0
1 1 1
9. Find the inverses of matrix A by the determinant method, where
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Unit II
Matrices and Determinants
2 3 1
A = 8 4 1 


6 7 0
0 
t 1


10. Let A =  0 t
t  . Determine the value(s) of t for which A fails to be
15 17 t 1


invertible.
10. Find the determinants of the following matrices.
cos 
a) 
 sin 
1
0
c) 
0

0
 3  2 8
b)  5 0 7
1 2 1
 sin  
cos  
4
8
8  2

0  7
2
3
4 2
0
0
9. Reduce each of the following matrices into an echelon form and then use
properties of determinants to evaluate their determinants from your steps.
1

i) A =  3
1

2
1
1
 1

 5
ii) B = 
1

 2

3

0
1 
2
9
3
6
2
8
6
6
1 

3 
2 

1 
11. Solve the following systems of linear equations by the Gauss-Jordan Method.
a) x + y  z = 3
b) x + 2y + 3z + w = 8
3x + y  2z = 0
3x  4 y + 8z + 2w =  1
2x  y + 3z =  1
12. a) Determine the conditions on a, b and c so that the following system of
linear equations has solution.
x – 2y – 3z = a
3x – y + 2z = b
x + 3y + 8z = c
b) Determine the value(s) of k such that the system
x – 3z =  3
2x + ky – z = – 2
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x + 2y + k z = 1 has
i) unique solution
ii) no solution
iii) infinite solutions
12. Solve the following systems of linear equations by Cramer’s rule. If not
possible
use the Gauss-Jordan Method.
a)  x + y  2z = 7
b) x  2y + z = 5
3x + z =  3
x+z=4
2x + y + 2z =  1
 2x + 4y  2z =  10
13. Find all values of a, b and c for which matrix A is symmetric, where
2

A = 3
0

a  2b  2c
5
ac
2a  b  c 

2
.
7


DEFENCE ENGINEERING COLLEGE
Department of Basic and applied courses.
Applied Mathematics I (Math 201)
Worksheet II
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71
Unit II
Matrices and Determinants
(Matrices and Determinants)
January 2008
1.
2. Suppose A, B,C, D and E are matrices with the following order
Matrix
A
B
C
D
E
size
45
45
52
42
5 4
Determine which of the following matrix operations are defined, find the order
of
the resulting matrix whenever defined.
a) BA
b) AC + D
c) AE + B
e) E (AC)
f) E t A
g) ( At  E) D
d) E (A + B)
3. Construct a 3  2 matrix A = ( a ij ) such that
 j  2i, for i  j

a ij  
2

i  2 j , for i  j
4. Given three matrices, if possible
1  2
2 2 3
 2 1
A = 3
and C = 
.
4 , B = 

1 4 5 
1 4


0  4
Compute
a) AB
b) 3C  B At
c) (AB)C
5. Prove that any square matrix can be expressed as a sum symmetric and skew
symmetric matrices. Use the result to write matrix A as a sum symmetric
matrix
R and skew symmetric matrix s, where
 1 0  6
A =  2 4 7
 3 5 2 
6. Reduce each of the following matrices to row-echelon form and reduced rowechelon form and determine their ranks.
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 1  2
 2  4
b) 
 3
6

 4  8
 2 3  4
a)  1  5
1

 4 6  2
1
 2
 3
0 
c) 
 1  4


2
 8
3
6
 9

12
7. Find the inverses of each of the following matrices using the Gauss-Jordan
Method.
1
 3
a) 

  8  4
1 2 3
b) 3 1 0


1 1 1
8. Find the inverses of matrix A by the determinant method, where
2 3 1
A = 8 4 1 


6 7 0
9. Find the determinants of the following matrices.
cos 
a) 
 sin 
 3  2 8
b)  5 0 7
1 2 1
 sin  
cos  
1
0
c) 
0

0
4
8
8  2

0  7
2
3
4 2
0
0
10. Solve the following systems of linear equations by the Gauss-Jordan Method.
a) x + y  z = 3
b) x + 2y + 3z + w = 8
3x + y  2z = 0
3x  4 y + 8z + 2w =  1
2x  y + 3z =  1
11. Solve the following systems of linear equations if possible by Cramer’s rule;
otherwise by the Gauss-Jordan Method.
a)  x + y  2z = 7
b) x  2y + z = 5
3x + z =  3
x+z=4
2x + y + 2z =  1
 2x + 4y  2z =  10
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Matrices and Determinants
0 
t 1


invertible. Let A =  0 t
t  . Determine the value(s) of t for which A fails
15 17 t 1


to be
invertible.
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