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Chapter 9

222

Matrices and Determinants

Chapter 9
Matrices and Determinants
9
...
Matrix algebra provides a clear and concise
notation for the formulation and solution of such problems, many of which
would be complicated in conventional algebraic notation
...
Hence we shall first explain a
matrix
...
2
Matrix:
A set of mn numbers (real or complex), arranged in a rectangular
formation (array or table) having m rows and n columns and enclosed by a
square bracket [ ] is called mn matrix (read “m by n matrix”)
...
Note that aij is the element in
the ith row and jth column of the matrix
...
e
...

Order of a Matrix:
The order or dimension of a matrix is the ordered pair having as
first component the number of rows and as second component the number
of columns in the matrix
...
In general
if m are rows and n are columns of a matrix, then its order is (m x n)
...


9
...


Row Matrix and Column Matrix:
A matrix consisting of a single row is called a row matrix or a
row vector, whereas a matrix having single column is called a column
matrix or a column vector
...


Null or Zero Matrix:
A matrix in which each element is „0‟ is called a Null or Zero
matrix
...
This
distinguishes zero matrix from the real number 0
...

0 0 0 0 
The matrix Omxn has the property that for every matrix Amxn,
A+O=O+A=A
3
...
A matrix A of order m x n can be written as Amxn
...
A square
matrix of order n x n, is simply written as An
...
Thus, the principal diagonal
contains elements a11, a22, a33 etc
...

Particular cases of a square matrix:
(a)Diagonal matrix:
A square matrix in which all elements are zero except those in the
main or principal diagonal is called a diagonal matrix
...

For example

1 0 0 
4 0


 0 2 and 0 1 0 


0 0 0 

are diagonal matrices
...
e
...
An identity matrix of order n is denoted by In
...


In general,

a12
    a1n 
 a11
a
a22
    a2n 
 21
A=                = [aij]mxn


             
 am1
am2
    amn 

is an identity matrix if and only if
aij = 0 for i ≠ j and aij = 1
for i = j
Note: If a matrix A and identity matrix I are comformable for
multiplication, then I has the property that
AI = IA = A i
...
, I is the identity matrix for multiplication
...


Equal Matrices:
Two matrices A and B are said to be equal if and only if they have
the same order and each element of matrix A is equal to the corresponding
element of matrix B i
...


Example 1: Find the values of x , y , z and a which satisfy the
matrix equation
=
Solution :

From (1)

By the definition of equality of matrices, we have
x + 3 = 0 ……………………………
...


The Negative of a Matrix:
The negative of the matrix Amxn, denoted by –Amxn, is the matrix
formed by replacing each element in the matrix Amxn with its additive
inverse
...
e
...

The sum Bm-n + (–Amxn) is called the difference of Bmxn and Amxn
and is denoted by Bmxn – Amxn
...
4 Operations on matrices:
(a) Multiplication of a Matrix by a Scalar:
If A is a matrix and k is a scalar (constant), then kA is a matrix
whose elements are the elements of A , each multiplied by k

 4 3
For example, if A =  8 2 then for a scalar k,


 1 0 
kA =

Chapter 9

227

Matrices and Determinants

5 8 4  15 24 12 
3 0 3 5 =  0
9 15

 
3 1 4   9 3 12 

Also,

(b)

Addition and subtraction of Matrices:
If A and B are two matrices of same order mn then their sum
A + B is defined as C, mn matrix such that each element of C is the sum
of the corresponding elements of A and B
...
e
...

Note: The sum or difference of two matrices of different order is not
defined
...

Then the product matrix AB has the same number of rows as A and the
same number of columns as B
...
The elements of AB are determined as follows:

Chapter 9

by

228

Matrices and Determinants

The element Cij in the ith row and jth column of (AB)mxn is found
cij = ai1b1j  ai2b2j + ai3b3j + ………
...
e
...
e
...

Similarly , c12 is obtained by multiplying the elements of the first
row of A i
...
, a11 , a12 by the corresponding elements of the second
column of B i
...
, b12 , b22 and adding the product
...


Note :

1
...
e
...

2
...
A matrix A can be multiplied by itself if and only if it is a square
matrix
...
A in such cases is written as A2
...
e
...
A2 = A3 , A2
...
In the product AB, A is said to be pre multiple of B and B is said
to be post multiple of A
...

 1 3 



Solution:

1 2 

AB = 

= 

 1 3  1 1  2  3 1  3

4 3
2 

= 
1

Chapter 9

229

 2 1  1

Matrices and Determinants

2

 2  1 4  3
2  3

BA = 
  1 3  = 1  1
1
1


 

1 7 

= 

0 5 
This example shows very clearly that multiplication of matrices in
general, is not commutative i
...
, AB  BA
...
We have

1 1
3 1 2  
 = 3  2  6 3  1  2 
AB = 
2
1

1  0  3 1  0  1

1 0 1  


3 1 
11 0 
= 

 4 0
Remark:
If A, B and C are the matrices of order (m x p), (p x q) and (q x n)
respectively, then
i
...
e
...

ii
...
e distributive laws holds
...
e, I is the identity matrix for multiplication
...
1
Q
...
1 Write the following matrices in tabular form:
i
...
B = [bij], where i = 1 and j = 1, 2, 3, 4
iii
...
No
...

ii
...


iv
...


Q
...


 b 21 b 22 
a 21 a 22 

equation X + A = B, where A = 
Q
...


X+ 
2

ii
...


iv
...
5
i
...


iii
...


231

Matrices and Determinants

Write each product as a single matrix:

1 1
3 1 1 

0 1 2  0 2 

 1 0 


1
[3 - 2 2]  2 
 
 2 
2
1

1
 1
 1

 1

2 1  1
1 2   1
0 1  1
2 5   2
1 3 1
2 4 1

 3 2 
,C=
0 

Q
...



 1 2 
 1 0
Show that if A = 
and B = 

 , then
0
1

1
2




(a)
(A + B)(A + B)  A2 + 2AB + B2
(b)
(A + B)(A – B)  A2 – B2

Q
...
6

1 4 

2 5 
1 3 
2 4 
2 1
1 2 
0 1

If A = 
,B=  4
2 1


(i)

Q
...


0  sin θ  1 0 0 
1
0   0 1 0
0 cos θ  0 0 1 
2 2

2 

Chapter 9

232

Matrices and Determinants

Answers 9
...
1(i)

(ii)
(iii)

Q
...
4

(i)

(iii)

Q
...
6

9
...
The determinants are defined only for square matrices
...

The determinant of the (2 x 2) matrix

 a11 a12 

a 21 a 22 

A= 

is given by det A = |A| =

a11 a12
a 21 a 22

= a11 a22 – a12 a21
Example 3:

3

1

If A = 
 find |A|
 2 3

Solution:
|A| =

3 1
= 9 – (–2) = 9 + 2 = 11
2 3

The determinant of the (3 x 3) matrix

 a11 a12
A = a 21 a 22

 a 31 a 32

a13 
a11 a12

a 23  , denoted by |A| = a 21 a 22
a 33 
a 31 a 32

a13
a 23
a 33

is given as, det A = |A|
= a11

a 22
a 32

a 23
a 21 a 23
a 21 a 22
– a12
+ a13
a 33
a 31 a 33
a 31 a 32

= a11(a22a33 – a23a32) – a12(a21a33 – a23a31) + a13(a21a32 – a22a31)
Note: Each determinant in the sum (In the R
...
S) is the determinant of a
submatrix of A obtained by deleting a particular row and column of A
...
We take the sign + or  , according
to (  1)i+j aij
Where i and j represent row and column
...
6
Minor and Cofactor of Element:
The minor Mij of the element aij in a given determinant is the
determinant of order (n – 1 x n – 1) obtained by deleting the ith row and
jth column of Anxn
...

a 31 a 32

The scalars Cij = (-1)i+j Mij are called the cofactor of the element aij
of the matrix A
...

The value of the determinant can be found by expanding it from
any row or column
...

Solution (a)
|A|

3 2 1
= 0 1 2
1 3 4

Chapter 9

235

=3

(b)

Matrices and Determinants

1 2
0 2
0 1
–2
+1
3 4
1 4
1 3

|A|

= 3(4 + 6) – 2(0 + 2) + 1 (0 – 1)
= 30 – 4 – 1
= 25

|A|

=3

1 2
2 1
0 1
–0
+1
3 4
3 4
1 3

= 3 (4 + 6) + 1 (–4 –1)
= 30 – 5
|A|
= 25
9
...
Interchanging the corresponding rows and columns of a
determinant does not change its value (i
...
, |A| = |A‟|)
...
(2)
Now again consider
|B|

a1 a 2
= b1 b 2
c1 c2

a3
b3
c3

Expand it by first column
|B|
= a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2)
which is same as equation (2)
so

a1
|B| = a 2
a3

b1
b2
b3

c1
c2
c3

or
|B| = |A|
2
...

For example if

Chapter 9

236

|A|

=

a1
a2
a3

Matrices and Determinants

b1
b2
b3

c1
c2
c3

b2
b1
b3

c2
c1
c3

Consider the determinant,
|B|

=

a2
a1
a3

expand by second row,
|B|
= –a1(b2c3 – b3c2) + b1(a2c3 – a3c2) – c1(a2b3 – a3b2)
= –(a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2))
The term in the bracket is same as the equation (2)
So

|B|

=

a1
– a2
a3

b1
b2
b3

c1
c2
c3

Or
|B|
= – |A|
3
...
For example
|A|

=

0
a2
a3

0
b2
b3

0
c2
c3

=
0(b2c3 – b3c2) –0(a2c3 – a3c2) +0(a2b3 – a3b3)
|A|
=
0
4
...
For example, if
|A|

=

a1
a1
a3

b1 c1
b1 c1
b 3 c3

= a1(b1c3 – b3c1) – b1(a1c3 – a3c1) + c1(a1b3 – a3b1)
= a1b1c3 – a1b3c1 – a1b1c3 + a3b1c1 + a1b3c1 – a3b1c1
|A|
=
0
5
...
For example if,

Chapter 9

237

|A|

=

Matrices and Determinants

a1
a2
a3

b1
b2
b3

c1
c2
c3

ka1 kb1 kc1
Consider a determinant, |B| = a 2
b2 c2
a 3 b 3 c3
|B| = ka1(b2c3 – b3c3) – kb1(a2c3 – a3c2) + kc1(a2b3 – a3b2)
= k(a1(b2c3 – b3c3) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2))
So

|B|

a1
= k a2
a3

b1
b2
b3

c1
c2
c3

Or
|B|
=
K|A|
6
...

For example, if
|A|

=

a1
a2
a3

b1
b2
b3

c1
c2
c3

Consider a matrix,
|B|

=

a1  ka 2
a2
a3

b1  kb2
b2
b3

c1 +kc2
c2
c3

= (a1+ka2)(b2c3–b3c2)–(b1+kb2)(a2c3–a3c2)+(c1+kc2)(a2b3–a3b2)
= [a1(b2c3 – b3c2) –b1(a2c3 – a3c2) +c1(a2b3 – a3b2)]
= [ka2(b2c3 – b3c2) –kb2(a2c3 – a3c2) +kc2(a2b3 – a3b2)]

a1
= a2
a3

b1
b2
b3

c1
a2
c2 + k a 2
c3
a3

b2
b2
b3

c2
c2
c3

Chapter 9

a1
= a2
a3

238

b1
b2
b3

Matrices and Determinants

c1
c2 + k(0) because row 1st and 2nd are identical
c3

|B| = |A|
7
...
For example, if
|A|

2 0 0
= 0 5 0
0 0 3

= 2(–15 – 0) – (0 – 0) + 0(0 – 0)
= 30, which is the product of diagonal elements
...
e
...
The determinant of the product of two matrices is equal to the
product of the determinants of the two matrices, that is |AB| =
|A||B|
...
(A)
and
|A|
= a11a22 - a12a21
|B|
= b11b22 - b12b21
|A| |B| = a11b11 a22 b22 + a12b21 a21 b12 - a11b12 a22 b21
- a12b22 a21 b11 …………………
...
H
...
The determinant in which each element in any row, or column,
consists of two terms, then the determinant can be expressed as the
sum of two other determinants

Chapter 9

239

a1  1
a 2  2
a 3  3

b1
b2
b3

c1
a1
c2 = a 2
c3
a3

Matrices and Determinants

c1
1
c2 +  2
c3
3

b1
b2
b3

b1
b2
b3

c1
c2
c3

Expand by first column
...
H
...
H
...
H
...
, then

a2

a

abc

1
b b2
L
...
S =
abc
c c2

abc
abc

Take abc common from 3rd column

a2 1

a
=

abc
b b2 1
abc
c c2 1

Interchange column first and third

1 a2

a

=  1 b2

b

1 c2

c

Again interchange column second and third

1 a

a2

= 1 b

b2

1 c

c2

= R
...
S
Example 6: Show that

1 a

a2

1 b b 2 = (b – c) (c – a) (a – b)
1 c

c2

Chapter 9

241

Matrices and Determinants

Solution:

a2

1 a

1 b b2

L
...
S =

1 c

c2

subtracting row first from second and third row

1

a2

a

0 b  a b2  a 2

=

0 ca

c2  a 2

from row second and third taking (b – a) and (c – a) common
...
H
...
H
...
H
...
H
...
H
...
H
...
8

242

Matrices and Determinants

Solution of Linear Equations by Determinants:
(Cramer’s Rule)
Consider a system of linear equations in two variables x and y,
a1x + b1y = c1
(1)
a2x + b2y = c2
(2)
Multiply equation (1) by b2 and equation (2) by b1 and subtracting,

we get
x(a1b2 – a2b1) = b2c1 – b1c2
x=

b2c1  b1c2
a1b2  a 2b1

(3)

Again multiply eq
...
(2) by a1 and subtracting, we
get
y(a2b1 – a1b2) = a2c1 – a1c2

a 2c1  a1c2
a 2b1  a1b2
a c  a 2c1
y= 1 2
a1b2  a 2b1
y=

(4)

Note that x and y from equations (3) and (4) has the same
denominator a1b2 – a2b1
...

The solutions for x and y of the system of equations (1) and (2) can
be written directly in terms of determinants without any algebraic
operations, as

c1
c
x= 2
a1
a2

b1
b2
and y =
b1
b2

a1 c1
a 2 c2
a1 b1
a 2 b2

This result is called Cramer‟s Rule
...
2
Q
...
2

Expand the determinants
(i)

1 2 0
3 -1 4
-2 1 3

(iii)

x 0 0
0 x 0
0 0 x

(ii)

a b 1
a b 1
1 1 1

Without expansion, verify that
(i)

-2 1 0
3 4 1 =0
-4 2 0

(iii)

a-b b-c c-a
b-c c-a a-b = 0 (iv)
c-a a-b b-c

(v)

x+1 x+2 x+3
x+4 x+5 x+6 =0
x+7 x+8 x+9

(ii)

1 2 1
1 2 1
0 2 3 = 0 2 3
2 -1 2
0 -5 0
bc ca ab
a3
1
a

b3
1
b

c3 = 0
1
c

Chapter 9

(vi)
Q
...
4

Q
...
6

Matrices and Determinants

a3
a1 a 2
= x b1 b2
b3
c3x+d3
c1 c2

a3
a1 a 2
b3 + b1 b2
c3
d1 d 2

a3
b3
d3

Show that
(i)

0 a b
a 0 c = 0
b c 0

a
b
c
a a+b
a+b+c = a 3
a 2a+b 3a+2b+c

(iii)

a-b-c 2a
2a
2b b-c-a 2b = (a + b + c)3
2c
2c c-a-b

(iv)

1
1
1
bc ca
ab = (b  c)(c  a)(a  b)
b+c c+a a+b

(ii)

Show that:
(i)

a a
a
a = (2a + )(  a) 2
a a

(ii)

a+
a
a

a
a+
a

a
a
a+

=

2

(3a + )

prove that:
(i)

= a + b + c  3 abc
3

3

3

Chapter 9

247

Matrices and Determinants

2

= λ (a + b + c + λ)

(ii)

(iii)
Q
...
8

0
Sin  = Sin (     )
Cos 

(ii)

0

Use Cramer‟s rule to solve the following system of equations
...
2
x3

Q
...
7

(i) x = -2 , 3

Q
...
9

248

Matrices and Determinants

Special Matrices:

1
...

It is denoted At or A
...


Symmetric Matrix:
A square matrix A is called symmetric if A = At
for example if

a b c 
A = b d e  ,


 c e f 

then

a b c 
A = b d e  = A


 c e f 
t

Thus A is symmetric
3
...

4
...

Example:
Solution:

Find k

If

A=

is singular

1
k-2
Since A is singular so 5
k+2 = 0
(k – 2)( k + 2) – 5 = 0
k2 – 4 – 5 = 0
k2 – 9 = 0 ⇨

K=+3

5
...
It is written as adj
...

For example , if

1 0 1
A = 1 3 1 


0 1 2 
Cofactor of A are:
A11 = 5,
A12 = - 2,
A21 = -1,
A22 = 2,
A31 = 3,
A32 = -2,
Matrix of cofactors is
C

Ct

5
=  1

 3
5
=  2

 1

2 1
2 1
2 3 
1 3 
2 2 
1 3 

A13 = +1
A23 = -1
A33 = 3

Chapter 9

250

Hence adj A = C

Matrices and Determinants

 5 1 3 
=  2 2 2 


 1 1 3 

t

Note: Adjoint of a 22 Matrix:
The adjoint of matrix

A =

is denoted by adjA is defined as

adjA =
6
...
e
...


 0 -2 -3
A = 1 3 3 


-1 -2 -2
Solution:
|A| = 0 +2 (–2 +3) – 3(–2 + 3) = 2 – 3
|A| = –1, Hence solution exists
...
11

252

Matrices and Determinants

Solution of Linear Equations by Matrices:
Consider the linear system:
a11x1 + a12x2 + ------ + a1nxn = b1
a21x1 + a22x2 + ------ + a2nxn = b2
……………
...

If now B  0 and A is non-singular then A-1 exists
...
Since X and A-1 B are
equal, each element in X is equal to the corresponding element in A-1 B
...


Chapter 9

253

Matrices and Determinants

If A is a singular matrix, then of course it has no inverse, and either
the system has no solution or the solution is not unique
...

Example 12: Use matrices to find the solution set of
4x + 8y + z = –6
2x – 3y + 2z = 0
x + 7y – 3z = –8
Solution:

4 8 1 
=  2 3 2 


1 7 3

Let

A

Since
So

|A|
= –32 + 48 + 17 = 61
A-1 exists
...
3
Q
...

(i)

Q
...
4

 1 2 1
 3 4 5 


 4 2 6 

Which of the following matrices are symmetric and skewsymmetric
(i)

Q
...
5

Find the solution set of the following system by means of matrices:

(i)

2x – 3y = –1 (ii)
x + 4y = 5

(iv)

–4x + 2y – 9z = 2
3x + 4y + z = 5
x – 3y + 2z = 8

Q
...
2

(i)
(iii)
(i)

Symmetric
Symmetric
8

(ii)

Skew-symmetric

x+y=2
(iii) x – 2y + z = –1
2x – z = 1
3x + y – 2z = 4
2y – 3z = –1
y–z =1
(v)

x + y – 2z = 3
3x – y + z = 0
3x + 3y – 6z = 8

Answers 9
...
3

1
7
Q
...
5

(iv)
(i)
(iv)

3 
7 

1 
7 

(ii)

(ii)

A-1 does not exist
...

2
...


4
...


256

Matrices and Determinants

Summary
If A = [aij], A = [bij] of order m x n
...

The product AB of two matrices A and B is conformable for
multiplication if No of columns in A = No
...

If A = [aij] is m x n matrix, then the n x m matrix obtained by
interchanging the rows and columns of A is called the transpose of
A
...

Symmetric Matrix:
A square matrix A is symmetric if At = A
...



c13 c23 c33 
And inverse of A is:

adj A
|A|

(ii)

A-1 =

6
...


Chapter 9

257

Matrices and Determinants

Short Questions
Write the short answers of the following:
Q
...
2:
Q
...

Define identity matrix
...


Q
...


Q
...


Q
...


Q
...
8:

Show that

Q
...
10: Evaluate -1 1 -3
2 4 -1 
1 2 3
Q
...
12: Find x and y if

Q
...
14: If

258
1 -1 2
A =3 2 5 
-1 0 4

Matrices and Determinants

and

2 1 -1
1 3 4  ,


-1 2 1

find A – B

Q
...
16: If A is non-singular, then show that (A-1)-1 = A
Q
...

Q
...
19: Define the minor of an element of a matrix
...
20: Define a co-factor of an element of a matrix
...
21: Without expansion verify that


  α + β 1
Q
...

3
2
1

1
5
4

-4
6
8

Q
...

3
2
1

1
5
4

-4
6
8

Q
...
25: If

A =

,

Then find k
...
26: If

A =

,

B=

,

Then find A – B

Q
...
28: If 4

3
k is singular ,

Q
...

5
A = 1

if

3
1

Answers
Q10
...


Q14
...


Q13
...
M 32 = 26 , C32 = – 26

Q22
...


K=+3

Q27
...


Q28
...
k = 3

Q26

k=6

Q29
...
1

Each questions has four possible answers
...


2
___1
...
The order of the matrix [1 2 3] is
(a) 1 x 3
(b) 3 x 1
(c) 3 x 3

(d) 1 x 3
(d) 2 x 3

0 0 

___3
...
Two matrices A and B are conformable for multiplication if
(a) No of columns in A = No of rows in B
(b) No of columns in A = No of columns in B
(c) No of rows in A = No of rows in B
(d) None of these
___5
...
In an identity matrix all the diagonal elements are:
(a) zero
(b) 2
(c) 1
(d) none of these

2 0

___7
...
If two rows of a determinant are identical then its value is
(a) 1
(b) zero
(c) – 1
(d) None of these

2 3 4 
___9
...
If all the elements of a row or a column are zero, then value of the
determinant is:
(a) 1
(b) 2
(c) zero
(d) None of these

Chapter 9

261

Matrices and Determinants

2

3

___11
...

6
m


(a) 6
(b) 3
(c) 8
(d) 9
___12
...
Matrix [aij]mxn is a row matrix if:
(a) i = 1
(b) j = 1
(c) m = 1
(d) n = 1
14
...
If A = [aij]mxn is a scalar matrix if :
(a) aij = 0  i  j
(b) aij = k  i = j
(c) aij = k  i  j
(d) (a) and (b)
16
...
Which matrix can be tectangular mayrix ?
(a) Diagonal
(b) Identity
(c) Scalar
(d) None
18
...
( A – B )2 = A2 – 2AB + B2 , if and only if :
(a) A + B = 0 (b)

AB – BA = 0

(c) A2 + B2 = 0 (d) (a) and c

20
...
1

(1) c
(2) a
(7) a
(8) b
(13) c (14) d
(19) b (20) d

(3) d
(9) a
(15) d

(4) a
(10) c
(16) d

(5) c
(11) d
(17) d

(6) c
(12) d
(18) a



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