Chapter3. MATRICES
AND DETERMINANTS
3.1 Introduction to Matrices: An array of rectangular entries in the form m horizontal lines(called rows) and n vertical (called columns) is called a
Matrix of order m by n. A matrix is written as an m x n matrix. Plural of matrix is Matrices.
Example: A = 1
6 8 -9
0 3
7 4
This is a matrix, having 2 rows and
4 columns. Its order is 2 x 4 and has 8 elements.
In general,
a11 a12 a13 .... a1n
A = a21 a22 a23 … a2n =
[aij]m x n
a31 a32 a33 … a3n
… … … … …
am1 am2 am3 … amn
3.2 Types of Matrices:
i)
Row Matrix: A matrix having only one row is called row matrix.
Example: A = [ 5 7 ]
is a row matrix of order 1 x 2
ii)
Column Matrix: A
matrix having only one column is called a Column matrix.
Example: A = 5
is a column matrix of order 2 x 1.
4
iii)
Zero or null Matrix: A matrix whose each element is
zero is called a zero or a null matrix.
Example: A = 0
0 is a null matrix.
0 0
iv)
Square Matrix: A
matrix in which number of rows is equal to number of column is called a square
matrix.
Example: A = 2
5 is a square matrix of the
order 2 x 2.
4 6
A
matrix of order n x n is called a square matrix.
v)
Diagonal Matrix: A
square in which every nondiagonal element is zero, is called a diagonal matrix.
Example: A
= 1 0 0 is a diagonal matrix.
0 4 0
0 0 6
It
is also written as A = diag [ 1 4 6 ]
vi)
Scalar
Matrix: In a scalar matrix every nondiagonal
element is zero and all diagonal elements are equal..
2 1
1
Example: A = 1 2
1 is a scalar matrix.
1 1
2
vii)
Unit Matrix: A scalar matrix in which every nondiagonal element
is zero and each diagonal element is 1.
Example: I3 = 1
0 0 is a unit matrix of the order 3.
0
1 0
0
0 1
It is also known as identity matrix.
viii)
Comparable
Matrix: Two matrices are said to be
comparable, if they are of the same order.
2 4 5 2 5 7
Example: A = 3 1 8 B = 1 6 3
1 3 7 3 6 2
A and B are comparable matrix of
order being 3 x 3.
ix)
Equal Matrix: Two
or more matrices are said to be equal matrices if they are of the same order
and are having same elements.
Example: x
3 = 4
3
5 9 y 9
Since the corresponding elements
of equal matrix are equal, therefore;
x = 4 ; y
= 5
3.3 Operations on Matrices
3.3.1 Addition of Matrices: If A and B are two matrices of the same order,
then A + B is the sum of the two matrices where each element is obtained by
adding corresponding elements of A and B.
Example: If A=
3 4 8
and B = 1
4 2
2 6 5 5 3
8
Then A + B =
3+1 4+4 8+2
= 4 8
10
2+5
6+3 5+8 7 9
13
If A
= [ 2 4 5 ]
and B =
4 5 6
1
2 3
As
A and B are not comparable (since both are having different orders) they cannot
be added.
Some results of
Addition of Matrices:
Theorem 1: Matrix addition is Commutative, i.e.
A+B
= B+A for all comparable matrices A and B
Proof: Let A = [ aij]m x n and B = [ bij]m x n
Then,
A+B = [ aij]m x n + [ bij]m x n
= [ aij + bij ]m x n (by definition of addition
matrices)
= [ bij+aij ]mxn (since addition of numbers is
commutative)
= [bij]mxn + [ aij]mxn = B+A
Hence,
A+B = B+A
Theorem 2: Matrix addition is associative. i.e
(A+B)+C
= A+(B+C) for all comparable matrices A,B,C.
Proof: Let A = [ aij]m x n ,
B = [ bij]m x n and C
= [ cij]m x n
Then,
(A+B)+C = ([ aij]m x n + [ bij]m x n) + [ cij]m x n
= [ aij + bij ]m x n + [ cij]m x n
=[( aij + bij)+ cij]m x n
=[aij + (bij + cij)]m x n
=[ aij]m x n + [(bij + cij)]m x n
= A+(B+C)
Hence,
(A+B)+C= A+(B+C)
Theorem 3: If A is an mxn matrix and O is an mxn null matrix, then
A+O
=O+A =A
Proof: Let A = [ aij]m x n and O = [ bij]m x n’
Where, bij = 0
Then,
A+O = [ aij]m x n + [ bij]m x n’
=
[ aij + bij ]m x n
= [ aij + 0 ]m x n
= [ aij]m x n
= A
Therefore,
A+O = A
Similarly,
O+A = A
Hence,
A+O=O+A=A
3.3.2 Negative of a matrix: For a given matrix A, its
negative is obtained by replacing each element with its corresponding additive
inverse.
Example: If A
= 3
2 4 5
then (-A) =
-3 -2 -4
-5
1
4 5 6+i -1 -4
-5 -6-i
3.3.3 Subtraction of matrices: For two matrices A and B of the same order, we define subtraction
of the two as A-B = A+(-B).
Example: If A=
3 4 8
and B = 1
4 2
2 6 5
5 3
8
Then
(- B) = -1 -4
-2
-5
-3 -8
Therefore,
(A-B) = A + (-B)
=
3 4 8
+ -1 -4
-2
2 6 5 -5
-3 -8
=
3+(-1) 4+(-4) 8+(-2)
2+(-5) 6+(-3) 5+(-8)
= 2 0
6
-3
3 -3
3.3.4 Scalar multiplication: Let A=[aij] be an mxn matrix and k be any
number called scalar.
Then
the matrix obtained by multiplying every element of A by k is called the scalar
multiple of A by k and is denoted by kA.
Thus,
kA = [k aij]m x n
Example: If
A = 5
6 7 find
3A
6
2 1
Solution: 3A = 3
5 6 7
6
2 1
=
3x5 3x6
3x7
3x6
3x2 3x1
=
15 18
21
18 6 3
3.3.5 Multiplication of matrices: Let A be a matrix of order mxn and B be a matrix of
order nxp.
Then
the product of the matrices A and B is of order mxp i.e. when we multiply two matrices the number of columns of the
first matrix should be equal to the number of rows of the second matrix.
Example: If A =
a11 a12 a13 and
B = b11 b12
a21 a22 a23 b21 b22
b31 b32
Solution: Let
AB = c11 c12
c21 c22
then,
c11
= a11b11 + a12 b21 + a13b31
c12
= a11b12 + a12b22 + a13b32
c21
= a21b11 + a22 b21 + a23b31
c22
= a21b12 + a22b22 + a23b32
So,
AB = a11b11 + a12 b21
+ a13b31 a11b12
+ a12b22 + a13b32
a21b11 + a22
b21 + a23b31 a21b12 + a22b22
+ a23b32
Properties of multiplication of
matrices
i) Matrix multiplication is not
commutative in general.
i.e. AB ≠ BA
ii) Matrix multiplication is associative
i.e.,
(AB)C = A(BC)
whenever
both sides are defined.
iii) Matrix multiplication is distributive
over matrix addition i.e.
(a) A (B + C) = AB + BC
(b) (A + B) C = AB + AC
Whenever
both sides of equality are defined.
3.4 Transpose of matrix: The
transpose of a matrix A is obtained by interchanging its rows and columns and
is denoted by A' or AT.
i)
If
A = [aij]mxn is a matrix of order mxn, the transpose of A = A' = [aji]nxm is a
matrix of order nxm.
ii)
(i,
j)th element of A = (j, i)th element of A’.
Example:
If A
= 2 5
7
-1
3 2
Then transpose of A = A’ = 2
-1
5 3
7 2
Properties of Transpose
i)
(AT)T
= A.
ii)
(A
+ B)T = AT + BT, A and B being of the same
order.
iii)
(KA)T
= KAT, k be any scalar (real or complex).
iv)
(AB)T
= BT AT; A and B being conformable for the product AB.
3.5 Symmetric and Skew-Symmetric matrix
3.5.1 Symmetric matrix:
A square matrix A = [aij]
is said to be symmetric if its (i, j)th element is the same as its (j, i)th element.
i.e. A = AT
Example: If
A = 2
4 5
4 5 1
5 1 3
Then,
AT
= 2 4 5
4 5 1
5 1 3
A = AT
Therefore,
A is symmetric matrix.
3.5.2 Skew symmetric matrix:
A square matrix A = [aij] is said to be skew-symmetric if
the (i, j)th
element of A
is the negative of the (j, i)th element of A
i.e. A T = -A
Example: A
= 2 4 -5
-4
5 1
5 -1 3
Then,
AT = 2
4 -5
-4
5 1
5
-1 3
Therefore, AT
= -A
Properties of Symmetric and
skew-symmetric matrix
i)
A square matrix A is said to be skew-symmetric
if A' = -A.
ii)
The
diagonal elements of a skew-symmetric matrix are all zero.
Examples:
iii)
It
may be checked that ½ (A +AT)
is symmetric and ½ (A - AT) is skew-symmetric for every square
matrix A.
iv)
A
= (A +AT)/2 + (A - AT)/2
That
is any square matrix can be expressed as the sum of a symmetric matrix and a
skew-symmetric matrix.
v)
A
matrix which is both symmetric and skew symmetric is a zero matrix.
3.6 Introduction to Determinants:
The
value of determinant of a(1x1) matrix [a] is defined as |a| = 1
Value
of determinant of order 2 is defined as
a11 a12 = (a11a22 – a21a12)
a21 a22
Value
of determinant of order 3 or more is determined by finding minor of aij
in |a| and co-factor in aij in |a|.
Minor of aij in |a| is defined as the value of the
determinant obtained by deleting the ith row and jth
column of |A| and is denoted by Mij.
Co-factor of aij in
|A| is defined
as
Cij = (-1)i+j . Mij
∆ = a11 a12 a13
a21 a22 a23
a31 a32 a33
M11
= a22 a23 = (a22a33 – a32a23)
a32 a33
M12
= a21 a23 = (a21a33 – a31a23)
a31 a33
M13 = a21 a22 = (a21a32 – a31a22)
a31 a32
M21 = a12 a13 = (a12a33 – a32a13)
a32 a33
Similarly, we can obtain the minor
of each one of the remaining elements.
Now
if we denote the co-factor of aij
by cij, then,
C11
= (-1)1+1. M11 = (a22a33 – a32a23);
C12
= (-1)1+2. M12 = -M12 = (a31a23
– a21a33)
C13
= (-1)1+3. M13 = M13 = (a21a32
– a31a22)
C21
= (-1)2+1. M21 = - M21 = (a32a13
– a12a33)
Similarly,
the co-factor of each one of the remaining elements of ∆ can be obtained.
3.6.1 Value of determinant:
The value of a determinant is the sum of the products of elements of a row (or
a column) with their corresponding co-factors.
3.6.2 Expansion of a determinant:
∆
= a11 a12 a13
a21 a22 a23
a31 a32 a33
= a11C11
+ a12C12 + a13C13
= a11M11
– a12M12 + a13M13
= a11 a22 a23 -
a12 a21 a23 + a13 a21` a22
a32 a33 a31 a33 a31 a32
= a11 ( a22a33
– a23a32) – a12 (a21a33
– a23a31) + a13 (a21a32
– a22a31)
Points to be noted:
i)
If
we expand a determinant by any row or column using minors, we keep in views,
the following symbols for a determinant of order 3.
+ - +
-
+ -
+ - +
ii)
If
a row or a column of a determinant consists of only zeros, the value of the
determinant is zero.
3.6.3 Properties of determinants:
i)
The
value of a determinant remains unchanged if its rows and columns are
interchanged.
ii)
If
two rows or columns of a determinant are interchanged, then the determinant retains
its absolute value but its sign is changed.
iii)
If
any two rows or columns of a determinant are identical then its value is zero.
iv)
If
each element of a row or column of a determinant is multiplied by a constant k
then the value of new determinant is k times the value of the original
determinant.
v)
If
any two rows or column of a determinant are proportional, then its value is
zero.
vi)
If
each element of a row or column of a determinant can be expressed as sum of two
or more terms, then the determinant can be expressed as the sum of two or more determinants.
vii)
If
any row or column of a determinant is added by the multiple of another row or
column, the value of the determinant remains same.
viii)
The
sum of the products of the elements of any row or column of a determinant with
co-factors of the corresponding elements of any other row or column is zero.
3.6.4 Applications of Determinants:
Area
of triangle in determinant form: For a triangle with given vertices (x1, y1), (x2, y2), (x3, y3), the area
is given by the formula:
∆ = ½
[x1 (y2 – y3)
– x2 (y1 – y3)
+ x3 (y1 – y2)]
= 1 x1
y1 1
2 x2 y2 1
x3 y3 1
Condition
of collinearity of three points:
Area of triangle = 0
Therefore,
½ [x1 (y2 – y3) – x2 (y1 – y3)
+ x3 (y1 – y2)]
= 0
Hence,
1 x1
y1 1
2 x2 y2 1 =
0
x3 y3 1
x1 y1 1
x2 y2 1 =
0
x3 y3 1
3.6.5 Solving system of linear equation:
Cramer’s Rule: The
Solution of system of linear equations
a1x + b1y = c1 (i)
a2x + b2y = c2 (ii)
is
given by x = (D1/D) and y
= (D2/D), where
a1 b1 c1 b1 a1 c1 D
= a1 b2 D1 = c2 b2 and D2 = a2 c2 and D≠0.
Proof:
Multiplying equation (i) by b2 and equation (ii) by b1and subtracting, we get;
(a1b2 – a2b1)
x – (b2c1 – b1c2)
= 0
↔ x = (b2c1 – b1c2)
(a1b2 – a2b1)
c1 b1
= c2 b2 = D1
D
a1 b1
a2 b2
Multiplying
(i) by a2 and (ii0 by a1, followed by subtraction, we get;
(a2b1 – a1b2)
y – (a2c1 – a1c2) = 0
↔
y = (b2c1 – b1c2)
(a1b2 – a2b1)
a1 c1
= a2 c2 = D2
D
a1 b1
a2 b2
3.7 Adjoint and Inverse of a Matrix:
3.7.1 Adjoint of a matrix: Let a square matrix A = [aij] of
order n and Aij denote the co-factor of aij in |A|, then
the adjoint of is defined as:
adjA = [Aij]
nxn
Hence,
adjA is the transpose of the matrix of corresponding co-factors of elements of
|A|.
If a11 a12 a13
A = a21 a22 a23
a31 a32 a33
Then,
A11 A12 A13 A11 A21 A31
adjA = A21 A22 A23 = A12 A22 A32
A31 A32 A33 A13 A23 A33
Example:
If A = 1
-2 3 , find adjA.
0
2 1
-4
5 3
Solution: 1 -2
3
|A|
= 0
2 1
-4 5
3
Then
the co-factors of elements of |A| are;
A11 = 2
1 = 1; A12 = -
0 1 = -4;
A13 = 0 2
= 8;
5 3 -4
3 -4
5
A21 = - -2
4 = 26 ; A22 =
1 4 = 19;
A23 = - 1 -2
= 3
5
3 -4 3 -4 5
A31 = -2
4 = -10 ; A32 = -
1 4 = -1 ;
A33 = 1 -2
= 2
2
1 0
1 0
2
T
1
-4 8 1 26
-10
Therefore,
adjA = 26 19
3 = -4
19 -1
-10
-1 2 8 3
2
Theorem1: If
a square matrix A is of order n, then prove that
A.(adjA) = (adjA).A
= |A|.In
Proof: a11 a12 a13 .... a1n
a21 a22 a23 ....
a2n
Let A= .... .... ....
.... ....
ai1 ai2 ai3 ....
ain
....
..... .... ....
....
an1 an2 an3 ....
ann
adjA= A11 A12 A13 ....
A1n
A21 A22 A23 ....
A2n
.... .... ....
.... ....
Ai1 Ai2 Ai3 ....
Ain
....
..... .... ....
....
An1 An2 An3 ....
Ann
= A11 A12 ....
Ak1 .... An1
A21 A22 ....
Ak2 .... An2
.... ....
.... .... ....
....
An1 An2 ....
Akn .... Ann
(i,
k)th element of A.(adjA) = a1 Ak1 +ai2Ak2
+....+ainAkn
|A|, when i = k
= 0, when i≠ k
Therefore,
|A| 0 0 .... 0
A.(adjA) = 0 |A| 0 .... 0
.... .... .... .... ....
0 0 0 .... |A|
1 0 0 .... 0
= |A|
0 1 0 .... 0
... ... ... .... ...
0 0 0 .... 1
= |A|.In
Thus, A.(adjA) = |A|.In
Similarly, (adjA).A = |A|.In
Therefore,
A.(adjA)
= (adjA).A = |A|.In
3.7.2 Inverse of a matrix: A non-zero square matrix A of order n is said
to be invertible if a square matrix B of order n also exist, such that AB = BA
= In.
Theorem 1: An
invertible matrix processes a unique inverse.
Proof: Let
A be an invertible square matrix of order n and matrix B and C be the inverse
of A.
Then,
AB = BA = In
And
AC = CA = In
Now,
BA = In
Therefore,
(BA)C = In.C
= C
And,
AC = In
B
(AC) = B.In = B
But we know that,
(BA)
C = B (AC)
Therefore,
B = C
Theorem2: A
square matrix A is invertible if A is non-singular, i.e.
|A| ≠
0
Proof: Let
A is an invertible square matrix of order n and another square matrix also
exists of the same order , such that,
AB
= BA =In
AB
= In
Therefore,
|AB|
= |In|
|A|.|B|
= 1
|A|
≠ 0
Therefore A is a non-singular
matrix. Thus A is invertible then A is non-singular.
Theorem3: (Cancellation
Law) If A B and C are square matrices of order n, then
AB
= AC.
If
A is non-singular, then B = C
Proof: As
A is a non-singular, A-1 exists.
Therefore,
AB = AC
A-1 (AB) = A-1 (AC)
(A-1
A) B = (A-1 A) C
(InB) = (InC) [since, A-1A
= In]
B = C
Theorem4: (Reversal
law) If A and B are invertible matrices of the same order, then show that AB is
also invertible and (AB)-1 = B-1A-1
Proof: Let A and B be the two invertible matrices, each of
order n.
Such that |A| ≠ 0 and |B| ≠ 0
|AB|
= |A|. |B| ≠ 0
This
shows that AB is non-singular
So,
(AB)-1(B-1A-1)
= A(BB-1)A-1
= (AIn) A-1
= AA-1 = In
Also,
(B-1A-1)(AB) = B-1(A-1A)B
= (BIn) B-1
= BB-1 = In
Therefore,
(AB)
(B-1A-1) = (B-1A-1)(AB) = In
Hence,
(AB)-1 = B-1A-1
Theorem5: If
A is invertible square matrix, then show that AT is also invertible
and
(AT)-1 = (A-1)T
Proof: Let A is an invertible square
matrix of order n.
Then,
|A|
≠ 0
Therefore,
|AT|
= |A| ≠ 0
Therefore AT is also
invertible.
Now,
AA-1 = A-1 A = In
(AA-1)T = (A-1
A )T= InT = In
(A-1)TAT = AT (A-1)T = In
Hence,
(AT)-1
= (A-1)T
Theorem6: If
A and B are invertible square matrix of the same order, then prove that
(adjAB)
= (adjA) (adjB)
Proof: let A and B be invertible square
matrix of order n.
Then,
|A| ≠
0 and |B| ≠ 0
|AB| =
|A|.|B| ≠ 0
We know that,
(AB)(adjAB)
= |AB| .In (i)
Also,
(AB).(adjA.adjB)
= A(B.adjB) .(adjA)
= A(|B| .In).(adjA)
= |B| . (A.adjA)
= |B|.|A|.In
= |A|.|B| .In
= |AB|.In
Thus,
(AB).(adjA.adjB)
= |AB|. In (ii)
So
from (i) and (ii) we get
(AB)(adjAB)
= (AB).(adjA.adjB) = |AB| In
As,
AB is invertible, then by cancellation law,
(adjAB)
= (adjA) (adjB)
Theorem7: If
A is invertible square matrix, prove that
(adjA)T = (adjAT)
Proof: let A be an invertible square
matrix of order n,
Then,
|A| ≠
0
Therefore,
|AT|
= |A| ≠ 0
This shows that AT is
non-singular and therefore invertible.
We know that,
A(adjA)
= |A| .In
Therefore,
[A.(adjA)]T
= (|A|.In)T
Or (adjA)T
. AT = |A|. InT
Or (adjA)T.AT
= |A|. In (i)
Also,
(adjAT).AT
= |AT|. In
Or (adjAT)
.AT = |A|. In (ii)
Thus,
from (i) and (ii) we get:
(adjA)T.AT
= (adjAT).AT
But AT being invertible, by cancellation law, we have
(adjA)T = (adjAT)
3.7.3 System of linear equation:
3.7.3.1 Linear equation in matrix notation: take the following system of
linear equation;
a11x+a12x+....+a1nxn
= b1
a21x+a22x+....+a2nxn
= b2
.... .... .... .... ... ... ... ....
... ... ... ... ....
.... .... ....
an1x+an2x+....+annxn
= bn
a11 a12 a13 ....
a1n
a21 a22 a23 ....
a2n
Let A= .... .... ....
.... ....
ai1 ai2 ai3 ....
ain
....
..... .... ....
....
an1 an2 an3 ....
ann
x1
X = x2
...
...
xn
and,
b1
B = b2
...
...
bn
The above system can also be
written in the form;
AX
= B
It can be rearranged as;
X
= A-1B
Let X = X1 and X = X2
be two solutions of AX = B
So,
AX1
= B and AX2 = B
Therefore,
AX1
= AX2
Since A being invertible, by
cancellation law, we have;
X1 = X2
Hence the given system of
equation has unique solution.
Criterion
for a given system of equation to have unique solution:
i)
If
|A| ≠ 0, then the system is consistent and has a unique solution and is
given by
X = A-1B
ii)
If
|A| = 0 and (adjA) B ≠ 0, then the system is inconsistent.
iii)
If
|A| = 0 and (adjA) B = 0 then the system is consistent and has infinitely many
solution.
Example: Using matrix method, solve the system of linear equation;
5x + 3y + 2 = 0
2x + 3y + 3= 0
Solution: the given set equations are
5x + 7y + 2 = 0
2x + 3y + 3= 0
Let,
A
= 5 7 X
= x and B= -2
2 3 y -3
The
given system is in the form of AX = B
|A|
= 5 7 = (15 – 14) = 1≠ 0
2 3
Thus
A is invertible
So,
the system is having a unique solution and is given by
X
= A-1B
The
co-factors of elements of |A| are:
A11
= 3, A12 = -2, A21 = -7 and A22 = 5
T
adjA
= 3
-2 =
3 -7
-7 5 -2
5
So,
A-1
= 1 .
adjA = 1/1 3
-7
|A| -2 5
Therefore,
X
= A-1B
x
= 3 -7 -2
y -2 5
-3
= 3
x (-2) – (-7)x(-3)
-2 x (-2) – (-5)x(-3)
x =
-27
y -11
hence, x = -27 and y = -11
3.7.3.2 Homogeneous System Of Linear Equation:
Consider
the following set of equation;
a11x1 + a12x2 + a13x3 = 0
a21x1 + a22x2 + a23x3 = 0
a31x1 + a32x2 + a33x3 = 0
Let a11
a12 a13 x1
0
A
= a21 a22 a23 X = x2 and B =
0
a31 a32 a33 x3
0
so
the given matrix is in the form of AX = 0
If
|A| ≠ 0 then A-1 exists
So,
AX = 0
A-1(AX) = A-1 0
(A-1A)X = 0
IX = 0
i.e. X = 0
Thus
in this case x1= 0, x2 = 0 and x3 = 0is the only solution.
This
type of solution is called as trivial
solution.