1. J.Baskar Babujee
Department of Mathematics
Anna University, Chennai-600 025.
MA6151MA6151

2. CONTENT
1.1 INTRODUCTION
1.2 TYPES OF MATRICES
1.3 CHARACTERISTIC EQUATION
1.4 EIGEN VECTORS
1.5 PROBLEMS
1.6 CAYLEY HAMILTON THEOREM
2

3. 1.7 DIAGONALISATION OF A MATRIX
1.8 REDUCTION OF A MATRIX TO DIAGONAL
FORM
1.9 ORTHOGONAL TRANSFORMATION OF A
SYMMETRIC MATRIX TO DIAGONAL FORM
1.10 QUADRATIC FORMS
1.11 NATURE OF A QUADRATIC FORM
3

4. 1.12 REDUCTION OF QUADRATIC FORM TO
CANONICAL FORM
1.13 INDEX AND SIGNATURE OF THE QUADRATIC
FORM
1.14 LINEAR TRANSFORMATION OF A
QUADRATIC FORM.
1.15 REDUCTION OF QUADRATIC FORM TO
CANANONICAL FORM BY ORTHOGONAL
TRANSFORMATION
4

5. 1.1 INTRODUCTION:-
A matrix is defined as a rectangular array (or
arrangement in rows or columns) of scalar subject
to certain rules of operations.
If mn numbers (real or complex) or functions
are arranged in the columns (vertical lines) then A
is called an m n matrix. Each of the mn numbers
is called an element of the matrix.
×
Unit 1 MATRICES

6. An m n matrix is usually written as
A matrix is usually denoted by a single capital letter A, B or C
etc.
11 12 13 1
21 22 23 2
31 32 33 3
1 2 3
....
....
....
.... .... .... .... ....
....
n
n
n
m m m mn
a a a a
a a a a
a a a a
a a a a
 
 
 
 
 
 
  
×

7. 7
Thus, an m n matrix A may be written as
A = , where i = 1, 2, 3, … , m ;
j = 1, 2, 3, … , n
In Algebra, the matrices have their largest
application in the theory of simultaneous
equations and linear transformations.
×
ija  

8. E.g., The set of simultaneous equations
3333232131
2323222121
1313212111
bxaxaxa
bxaxaxa
bxaxaxa
=++
=++
=++

9. may be symbolically represented by the equation
Where A = , X = , B =










333231
232221
131211
aaa
aaa
aaa
A X = B










3
2
1
x
x
x










3
2
1
b
b
b

10. 





−
−
720
135
1.2 TYPES OF MATRICES
(1) Real Matrix :- A matrix is said to be
real if all its elements are real numbers.
E.g., is a real matrix.

11. (2) Square Matrix:- A matrix in which the number
of rows is equal to the number of columns is
called a square matrix, otherwise, it is said to be
rectangular matrix.
i.e., a matrix A = is a square matrix if m = n
a rectangular matrix if m n
ij m n
a
×
  
≠

12. A square matrix having n rows and n columns is
called “ n – rowed square matrix”,
is a 3 – rowed square matrix
The elements of a square matrix are
called its diagonal elements and the diagonal
along with these elements are called principal or
leading diagonal.
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
 
 
 
  
33,2211 aa,a

13. The sum of the diagonal elements of a
square matrix is called its trace or spur.
Thus, trace of the n rowed square matrix
A= isija  
∑
=
=++++
n
1i
ijnn332211 aa....aaa

14. (3) Row Matrix :-
A matrix having only one row and any
number of columns,
i.e., a matrix of order 1 n is called a row
matrix.
[2 5 -3 0] is a row matrix.
×
Example:-

15. (4) Column Matrix:-
A matrix having only one column
and any number of rows,
ii.e., a matrix of order m 1 is called a column
matrix.
is a column matrix.










− 1
0
2
×
Example:-

16. (5) Null Matrix:-
A matrix in which each element is zero
is called a null matrix or void matrix or zero matrix.
A null matrix of order m n is denoted by
=
16
× nmO ×
Example :-






0000
0000
42O ×

17. (6) Sub - matrix :-
A matrix obtained from a given matrix A by
deleting some of its rows or columns or both is
called a sub – matrix of A.
is a sub – matrix of






41
03










−
−
2461
7053
5210
Example:-

18. (7) Diagonal Matrix :-
A square matrix in which all non – diagonal
elements are zero is called a diagonal matrix.
i.e., A = [a ] is a diagonal matrix if a = 0 for i j.
is a diagonal matrix.
ij nn× ij ≠
Example:-










−
000
010
002

19. (8) Scalar Matrix:-
A diagonal matrix in which all the diagonal
elements are equal to a scalar, say k, is called a
scalar matrix.
i.e., A = [a ] is a scalar matrix if
is a scalar matrix.










200
020
002
ij nn× 


=
≠
=
jiwhenk
jiwhen0
aij
Example :-

20. (9) Unit Matrix or Identity Matrix:-
A scalar matrix in which each diagonal
element is 1 is called a unit or identity matrix. It is
denoted by .
i.e., A = [a ] is a unit matrix if
is a unit matrix.
nI
ij nn× 


=
≠
=
jiwhen1
jiwhen0
aij
Example






10
01

21. (10) Upper Triangular Matrix.
A square matrix in which all the elements
below the principal diagonal are zero is called an
upper triangular matrix.
i.e., A = [a ] is an upper triangular matrix if a = 0
for i > j
is an upper triangular
matrix









−
300
510
432
ij nn× ij
Example:-

22. (11) Lower Triangular Matrix.
A square matrix in which all the elements
above the principal diagonal are zero is called a
lower triangular matrix.
i.e., A = [a ] is a lower triangular matrix if a = 0
for i < j
is a lower triangular
matrix.
ij nn× ij
Example:-









−
023
065
001

23. (12) Triangular Matrix:-
A triangular matrix is either upper
triangular or lower triangular.
(13) Single Element Matrix:-
A matrix having only one element is
called a single element matrix.
i.e., any matrix [3] is a single element matrix.

24. (14) Equal Matrices:-
Two matrices A and B are said to be equal iff
they have the same order and their corresponding
elements are equal.
i.e., if A = and B = , then A = B
iff a) m = p and n = q
b) a = b for all i and j.
qpij]b[ ×nmij]a[ ×
ij ij

25. (15) Singular and Non – Singular Matrices:-
A square matrix A is said to be singular if |
A| = 0 and non – singular if |A| 0.
A = is a singular
matrix since |A| = 0.










−
−
−
110
432
432
≠
Example :-

26. 1.3 CHARACTERISTIC EQUATION
If A is any square matrix of order n, we
can form the matrix , where is the nth
order unit matrix.
The determinant of this matrix equated to zero,
i.e.,
IλA − I
0
λa...aa
............
a...λaa
a...aλa
λA
nnn2n1
2n2221
1n1211
=
−
−
−
=− I

27. is called the characteristic equation of A.
On expanding the determinant, we get
where k’s are expressible in terms of the elements a
The roots of this equation are called Characteristic
roots or latent roots or eigen values of the matrix A.
0k...λkλkλ1)( n
2n
2
1n
1
nn
=++++− −−
ij

28. 1.4 EIGEN VECTORS
Consider the linear transformation Y = AX ...(1)
which transforms the column vector X into the
column vector Y. We often required to find those
vectors X which transform into scalar multiples of
themselves.
Let X be such a vector which transforms into
X by the transformation (1).
λ

29. Then Y = X ... (2)
From (1) and (2), AX = X AX- X = 0
(A - )X = 0 ...(3)
This matrix equation gives n homogeneous linear
equations
...
(4)
I⇒
λ
λ λ
λ







=−+++
=++−+
=+++−
0λ)x(a...xaxa
................
0xa...λ)x(axa
0xa...xaλ)x(a
nnn2n21n1
n2n222121
n1n212111
⇒ I

30. These equations will have a non-trivial solution only
if the co-efficient matrix A - is singular
i.e., if |A - | = 0 ... (5)
Corresponding to each root of (5), the
homogeneous system (3) has a non-zero solution
X = is called an eigen vector or latent
vector
λI
Iλ












4
2
1
x
...
x
x

31. 31
Properties of Eigen Values:-Properties of Eigen Values:-
1.The sum of the eigen values of a matrix is the
sum of the elements of the principal diagonal.
2.The product of the eigen values of a matrix A is
equal to its determinant.
3.If is an eigen value of a matrix A, then 1/ is
the eigen value of A-1
.
4.If is an eigen value of an orthogonal matrix, then
1/ is also its eigen value.
λ λ
λ
λ

32. 32
PROPERTY 1:-PROPERTY 1:- If λ1, λ2,…, λn are the eigen values of
A, then
i. k λ1, k λ2,…,k λn are the eigen values of the matrix
kA, where k is a non – zero scalar.
ii. are the eigen values of the inverse
matrix A-1
.
iii. are the eigen values of Ap
, where p
is any positive integer.
n21 λ
1
,...,
λ
1
,
λ
1
p
n
p
2
p
1 λ...,,λ,λ

33. 33
Proof:-
i. Let λr be an eigen value of A and Xr the
corresponding eigen vector.
Then, by definition,
Multiplying both sides by k,
(kA)Xr = (kλr)Xr
Then k λr is an eigen value of kA and the
corresponding eigen vector is the same as that of
λr, namely Xr.
AXr = λrXr

34. 34
ii. Pre multiplying both sides of AXr = λrXr by A-1
A-1
(A Xr) = A-1
(λr Xr )
Xr = λr (A-1
Xr )
=> A-1
Xr = (Xr)
Hence is an eigen value of A-1
and the
corresponding eigen vector is the same as that of
λr , namely Xr.
rλ
1
rλ
1

35. 35
iii. Pre multiplying both sides of AXr = λrXr by A
A(A Xr) = A(λr Xr )
A2
Xr = λr (AXr )
= λr (λr xr)
= λr
2
Xr.
Similarly, we can prove that A3
Xr = λr
3
Xr, …,
Ap
Xr = λr
p
Xr, where p is any positive integer. Hence λr
p
is any eigen value of Ap
and the corresponding
eigen vector is the same as that of λr, namely Xr.

36. 36
THEOREM :-
A matrix A is singular if and only if 0 is an eigen
value of A.
1.5 PROBLEMS
1. Find the sum and product of the eigen values of
the matrix
without finding the eigen values.










−−
−
−−
=
021
612
322
A

37. 37
Solution:-
Sum of the eigen values of A = sum of its diagonal
elements.
= - 2 + 1 + 0
= -1.
Product of the eigen values of A = | A |
=
= 45.
021
612
322
−−
−
−−

38. 38
2. Two eigen values of the matrix
are equal to 1 each. Find the third eigen value.
Solution:- Let a be the third eigen value of A.
Since sum of the eigen values = sum of the diagonal
elements,
1 + 1 + a = 2 + 3 + 2
a = 5
Therefore, the third eigen value of A is 5.










=
221
131
122
A

39. 39
3. The product of two eigen values of the matrix
is 16. Find the third eigen value.
Solution:-
Let a be the third eigen value of A.
Since product of the eigen values = | A |
16a =
Therefore, a = 2.










−
−−
−
=
312
132
226
A
312
132
226
−
−−
−

40. 40
4. Find the sum of the eigen values of the inverse of
Solution:-
The eigen values of the lower triangular matrix
A is 1, -3, 2. Then the eigen values of A-1
are
Sum of the eigen values of A-1
=
=










−=
250
032
001
A
.
2
1
,
3
1
,1 −
.
2
1
3
1
1 +
−
+
6
7

41. 41
5. If , find the eigen values of 3A, A-1
and – 2A-1
.
Solution:-
The eigen values of A are 2, -1, 4.
The eigen values of 3A are 3×2, 3×(-1), 3×4
i.e., 6, -3, 12










−=
400
210
572
A

42. 42
The eigen values of A-1
are
4
1
1,,
2
1
i.e.,
4
1
,
1
1
,
2
1
−
−
2
1
2,1,i.e.,
4
1
21),2)((,
2
1
2
are2AofvalueseigenThe 1
−−






−−−





−
− −

43. 43
6. Find the eigen values and eigen vectors of the
matrix
A =
Solution:- The characteristic equation of the given
matrix is






−
−
45
21
0|IA| =λ−

44. or
Thus,
Corresponding to =6, the eigen vectors are given by
(A – 6 ) X1 = 0
1,6
0652
−=λ=>
=−λ−λ=>
0
45
21
=
λ−−
−λ−
λ
I
0
x
x
25
25
or0
x
x
645
261
or
2
1
2
1
=











−−
−−
=











−−
−−
the eigen values of A are 6, -1.

45. We get only one independent equation - 5x1 - 2x2 = 0






−
=∴
−=
=
=
−
=⇒
5
2
kXarevectorseigenThe
5kx
2kx
(say)k
5
x
2
x
11
12
11
1
21

46. Corresponding to = -1, the eigen vectos are given
by ( A + ) X2 = 0
λ
I






=∴
==
==⇒
=−⇒






=











−
−
⇒
1
1
kXarevectorseigenThe
kx,kx
(say)k
1
x
1
x
0xx
0
0
x
x
55
22
22
2221
2
21
21
2
1

47. 7. Find the eigen values and eigen vectors of the
matrix A =
Solution:- The characteristic equation of the given
matrix is










−−
−
−−
021
612
322
0|IA| =λ−
0
21
612
322
=
λ−−−
−λ−
−λ−−

48. Thus,
53,3,λ
05)3)(λ3)(λ(λ
015)2λ3)(λ(λ
it.satisfies3λtrial,By
04521λλλ
0λ)]1(143[6]2λ2[12]λ)λ(1λ)[2(
2
23
−−=⇒
=−++⇒
=−−+∴
−=
=−−+⇒
=−+−−−−−−−−−−⇒
the eigen values of A are -3, -3, 5.

49. Corresponding to = - 3, the eigen vectors are
given by
λ
( )









−
+










=









 −
=
∴
−===
=−+
=




















−−
−
−−
=+
0
1
2
k
1
0
3
k
k
k
k2k3
0x3x2
0
x
x
x
321
642
321
or0XI
21
1
2
21
32
3
2
1
1
1
2112213
1
X
bygivenarevectorseigenThe
2k3kxthenkx,kxLet
xequationtindependenoneonlygetWe
3A

50. Corresponding to λ = 5, the eigen vectors are
given by (A – 5 I )X2 = 0.
052
032
0327
0
0
0
521
642
327
321
321
321
3
2
1
=−−−
=−−
=−+−⇒










=




















−−−
−−
−−
⇒
xxx
xxx
xxx
x
x
x

51. 51










−
=
−===∴
=
−
==⇒
−−
=
+
=
1
2
1
kX
kx2kx,kx
(say)k
1
x
2
x
1
x
22
x
53
x
6-10
x
32
333,231
3
321
321
bygivenarevectorseigentheHence
,equations.twofirstFrom

52. 8. Find the eigen values and eigen vectors of the
matrix
Solution:- The characteristic equation is










−
−−
−
=
342
476
268
A
0|IA| =λ−
.arevalueseigenHence
3,150,λ
045λ18λλ
0
λ342
4λ76
26λ8
23
15,3,0
=⇒
=−+−⇒
=
−−
−−−
−−
⇒

53. 53
Eigen vector X1 corresponding to λ1= 0 is given by
(3)...03xx42x
(2)...04x7x6x
(1)...02x6x8x
0
0
0
x
x
x
342
476
268
x
x
x
Xwhere0,I)Xλ(A
321
321
321
3
2
1
3
2
1
111
=+−
=−+−
=+−⇒










=










−
−−
−
⇒










==−

54. 54
Solving equations (1) and (2) by cross-multiplication,
we get
which satisfy equation (3) also.
Required eigen vector corresponding to λ1 = 0 is
131211
11
321
321
2kx,2kx,kx
0kwhere(say)k
2
x
2
x
1
x
3656
x
3212
x
1424
x
===
≠===⇒
−
=
+−
=
−










=










=
2
2
1
2
2 1
1
1
1
1 k
k
k
k
X
∴

55. 55
Eigen vector X2 corresponding to λ2= 3 is given by
(6)...0x42x
(5)...04x4x6x
(4)...02x6x5x
0
0
0
x
x
x
042
446
265
x
x
x
Xwhere0,I)Xλ(A
21
321
321
3
2
1
3
2
1
222
=−
=−+−
=+−⇒










=










−
−−
−
⇒










==−

56. 56
Solving equations (4) and (5) by cross-
multiplication, we get
which satisfy equation (6) also.
Required eigen vector corresponding to λ2 = 3 is
232221
22
321
321
2kx,kx,kx
0kwhere(say)k
2-
x
1
x
2
x
3620
x
2012
x
824
x
−===
≠===⇒
−
=
+−
=
−
2










−
=










−
=
2
1
2
2
2
2
2
2
2
2 k
k
k
k
X
∴

57. 57
Eigen vector X3 corresponding to λ3= 15 is given by
(9)...012xx42x
(8)...04x8x-6x
(7)...02x6x7x-
0
0
0
x
x
x
12-42
48-6
267-
x
x
x
Xwhere0,I)Xλ(A
321
321
321
3
2
1
3
2
1
333
=−−
=−−
=+−⇒










=










−
−−
−
⇒










==−

58. 58
Solving equations (7) and (8) by cross-multiplication,
we get
which satisfy equation (9) also.
Required eigen vector corresponding to λ3 = 15 is
333231
33
321
321
kx,kx,kx
0kwhere(say)k
1
x
2-
x
2
x
20
x
40-
x
40
x
=−==
≠===⇒
==
22










−=










−=
1
2
2
2
2
3
3
3
3
3 k
k
k
k
X

59. 59
9. Find the eigen values and eigen vectors of the
matrix
Solution:- The characteristic equation is










−
−−
−
=
312
132
226
A
0|IA| =λ−
8.2,2,arevalueseigenHence
82,2,λ
03236λ12λλ
0
λ312
1λ32
22λ6
23
=⇒
=+−+−⇒
=
−−
−−−
−−
⇒

60. 60
Eigen vector corresponding to is
given by
221 == λλ
0xx2x
0xx2x
02x2x4x
0
0
0
x
x
x
112
112
224
x
x
x
Xwhere0,I)Xλ(A
321
321
321
3
2
1
3
2
1
111
=+−
=−+−
=+−⇒










=










−
−−
−
⇒










==−

61. These equations are equivalent to a single
equation … (1)
Let x3 = 2 k3 and x2 = 2 k2 then from (1)
2x1 – 2 k2 + 2 k3 = 0
=> x1 = k2 – k3
Required eigen vector corresponding to
is
02 321 =+− xxx
∴
21 2 λλ ==









−
+










=









 −
=
2
0
1
0
2
1
2
2 32
3
2
32
1 kk
k
k
kk
X

62. Similarly, the eigen vector corresponding to = 8 is
given by,
3λ
....(4)05xx2x
....(3)0x5x2x
....(2)02x2x2x
0
0
0
x
x
x
512
152
222
0)X8-(A
x
x
x
Xwhere0I)Xλ-(A
321
321
321
3
2
1
3
3
2
1
333
=−−
=−−−
=+−−⇒










=




















−−
−−−
−−
⇒
=⇒










==

63. Solving equations (2) and (3) by cross-multiplication,
we get
which satisfy equation (4) also.
Required eigen vector corresponding to λ3 = 8 is
131211
1
321
321
,,2
)(
112
6612
kxkxkx
sayk
xxx
xxx
=−==
==
−
=⇒
=
−
=
∴










−=










−=
1
1
22
1
1
1
1
3 k
k
k
k
X

64. 64
Show that if λ1,λ2, … λn are the latent roots of the
matrix A, then A3
has the latent roots
Solution:- Let λ be a latent root of the matrix A.
Then there exists a non – zero vector X such that
....,,, 33
2
3
1 nλλλ
Example 1:-

65. 65
A X = λ X … (1)
=> A2
(AX) = A2
(λ X)
=> A3
X = λ (A2
X) [using (1)]
But A2
X = A ( A X) = A (λ X)
= λ (AX) = λ (λX) = λ2
X
Therefore, A3
X = λ (λ2
X) = λ3
X
=> λ3
is a latent root of A3.

66. 66
Therefore, If λ1,λ2, … λn are the latent roots of the
matrix A, then are the latent roots of
A3
.
If λ1,λ2, … λn are eigen values of A then find eigen
values of the matrix (A – λI)2
.
Solution:- (A – λI)2
= A2
– 2 λAI + λ2
I2
= A2
– 2 λA + λ2
I
33
2
3
1 ...,,, nλλλ
Example 2:-

67. 67
Eigen values of A2
are
Eigen values of 2 λA are 2 λ λ1,2 λ λ2, …, 2 λ λn.
Eigen values of λ2
I are λ2
Eigen values of ( A – λI)2
are
.λ...,λ,λ 2
n
2
2
2
1
.λ)(λ,...,λ)(λ,λ)(λor
.λ2λλλ,...,λ2λλλ,λ2λλλ
2
n
2
2
2
1
2
n
2
n
2
2
2
2
2
1
2
1
−−−
+−+−+−
∴

68. 68
Find the eigen values and eigen vectors of the matrix
Solution:- The characteristic equation is










=
500
620
413
A
5.2,3,arevalueseigenHence
52,3,λ
0λ)λ)(5λ)(2
0
λ500
6λ20
41λ3
=⇒
=−−−⇒
=
−
−
−
⇒
3(
0|IA| =λ−
Example 3:-

69. 69
Eigen vector X1 corresponding to λ1= 3 is given by
0x0,x
02x
06xx
04xx
0
0
0
x
x
x
200
61-0
410
x
x
x
Xwhere0,)X3(A
23
3
32
32
3
2
1
3
2
1
11
==⇒
=
=+−
=+⇒










=










⇒










==− I

70. The characteristic vector corresponding to eigen
value λ1 = 3 is given by
When λ2 = 2, let X2 be the eigen vector then
(A – 2I) X2 = 0 where X2 = [x1 x2 x3]’
0k,kXwhere
0
0
k
x
x
x
X 111
1
3
2
1
1 ≠=










=










=










=




















⇒
0
0
0
300
600
411
3
2
1
x
x
x

71. 71










−=










−=∴
=
−=
=∴
≠==
−
=
−==+∴
=⇒=
=
=++⇒
0
1
1
k
0
k
k
XisvectoreigenRequired
0x
kx
kx
0kwhere(say)k
0
x
1
x
1
x
xxor0xx
0x03x
06x
04xxx
22
2
2
3
22
21
22
321
2121
33
3
321

72. 72
When λ3 = 5, let X3 be the eigen vector then
(A – 5I) X3 = 0 where X3 = [x1 x2 x3]’
Solving above equations by cross – multiplication,
we get
06x3x
04xx2x
0
0
0
x
x
x
000
630
412
32
321
3
2
1
=+−
=++−⇒










=




















−
−
⇒

73. 73
Required eigen vector is
333231
33
321
321
kx,2kx,3kx
0kwhere(say)k
1
x
2
x
3
x
6
x
12
x
126
x
===⇒
≠===⇒
==
+










=










=
1
2
3
k
k
2k
3k
X 3
3
3
3
3
∴

74. 1.6 CAYLEY HAMILTON
THEOREM
Every square matrix satisfies its own
characteristic equation.
Let A = [aij]n×n be a square matrix
then,
nnnn2n1n
n22221
n11211
a...aa
................
a...aa
a...aa
A
×












=

75. Let the characteristic polynomial of A be φ (λ)
Then,
The characteristic equation is
 
 
 
 
 
 
11 12 1n
21 22 2n
n1 n2 nn
φ(λ) = A - λI
a -λ a ... a
a a -λ ... a
=
... ... ... ...
a a ... a -λ
| A -λI|= 0

76. Note 1:- Premultiplying equation (1) by A-1
, we
have
⇒ n n-1 n-2
0 1 2 n
n n-1 n-2
0 1 2 n
We are to prove that
pλ +p λ +p λ +...+p = 0
p A +p A +p A +...+p I= 0 ...(1)
I
⇒
n-1 n-2 n-3 -1
0 1 2 n-1 n
-1 n-1 n-2 n-3
0 1 2 n-1
n
0 =p A +p A +p A +...+p +p A
1
A =- [p A +p A +p A +...+p I]
p

77. This result gives the inverse of A in terms of
(n-1) powers of A and is considered as a practical
method for the computation of the inverse of the
large matrices.
Note 2:- If m is a positive integer such that m > n
then any positive integral power Am
of A is linearly
expressible in terms of those of lower degree.

78. Verify Cayley – Hamilton theorem for the matrix
A = . Hence compute A-1
.
Solution:- The characteristic equation of A is










−
−−
−
211
121
112
tion)simplifica(on049λ6λλor
0
λ211
1λ21
11λ2
i.e.,0λIA
23
=−+−
=
−−
−−−
−−
=−
Example 1:-

79. To verify Cayley – Hamilton theorem, we have to
show that A3
– 6A2
+9A – 4I = 0 … (1)
Now,










−
−−
−−
=










−
−−
−










−
−−
−
=×=










−
−−
−
=










−
−−
−










−
−−
−
=
222121
212221
212222
211
121
112
655
565
556
655
565
556
211
121
112
211
121
112
23
2
AAA
A

80. A3
-6A2
+9A – 4I = 0
= - 6 + 9
-4
=
This verifies Cayley – Hamilton theorem.
∴










−
−−
−−
222121
212221
212222










−
−−
−
655
565
556










−
−−
−
211
121
112










100
010
001
0
000
000
000
=











81. 81
Now, pre – multiplying both sides of (1) by A-1
, we
have
A2
– 6A +9I – 4 A-1
= 0
=> 4 A-1
= A2
– 6 A +9I










−
−
=∴










−
−
=










+










−
−−
−
−










−
−−
−
=⇒
−
−
311
131
113
4
1
311
131
113
100
010
001
9
211
121
112
6
655
565
556
4
1
1
A
A

82. 82
Given find Adj A by using Cayley –
Hamilton theorem.
Solution:- The characteristic equation of the given
matrix A is










−
−
−
=
113
110
121
A
tion)simplifica(on035λ3λλor
0
λ113
1λ10
1-2λ1
i.e.,0λIA
23
=++−
=
−−
−−
−
=−
Example 2:-

83. 83
By Cayley – Hamilton theorem, A should satisfy
A3
– 3A2
+ 5A + 3I = 0
Pre – multiplying by A-1
, we get
A2
– 3A +5I +3A-1
= 0










−
−
−
=










−
−−
−−
=










−
−
−










−
−
−
==
+−−=⇒
339
330
363
3A
146
223
452
113
110
121
113
110
121
A.AANow,
(1)...5I)3A(A
3
1
A
2
21-

84. 84
AAAAdj.
A
AAdj.
Athat,knowWe
173
143
110
3
1
500
050
005
339
330
363
146
223
452
3
1
AFrom(1),
1
1
1
−
−
−
=∴
=










−
−
−−
−=




















+










−
−
−
−










−
−−
−−
−=∴

85. 85










−
−
−−
=










−
−
−−






−−=∴
−=
−
−
−
=
173
143
110
AAdj.
173
143
110
3
1
3)(AAdj.
3
113
110
121
ANow,

86. 86
1.7 DIAGONALISATION OF A
MATRIX
Diagonalisation of a matrix A is the
process of reduction A to a diagonal form.
If A is related to D by a similarity transformation,
such that D = M-1
AM then A is reduced to the
diagonal matrix D through modal matrix M. D is
also called spectral matrix of A.

87. 87
1.8 REDUCTION OF A MATRIX
TO DIAGONAL FORM
If a square matrix A of order n has n linearly
independent eigen vectors then a matrix B can
be found such that B-1
AB is a diagonal matrix.
Note:- The matrix B which diagonalises A is called
the modal matrix of A and is obtained by
grouping the eigen vectors of A into a square
matrix.

88. 88
Similarity of matrices:-
A square matrix B of order n is said to be a
similar to a square matrix A of order n if
B = M-1
AM for some non singular
matrix M.
This transformation of a matrix A by a non –
singular matrix M to B is called a similarity
transformation.
Note:- If the matrix B is similar to matrix A, then B
has the same eigen values as A.

89. 89
Reduce the matrix A = to diagonal form by
similarity transformation. Hence find A3
.
Solution:- Characteristic equation is
=> λ = 1, 2, 3
Hence eigen values of A are 1, 2, 3.










−
−
300
120
211
0=










−
−
λ-300
1λ-20
21λ1-
Example:-

90. 90
Corresponding to λ = 1, let X1 = be the eigen
vector then










3
2
1
x
x
x










=∴
===∴
=
=−
=+−⇒










=




















−
−
=−
0
0
1
kX
x0x,kx
02x
0xx
02xx
0
0
0
x
x
x
200
110
210
0X)I(A
11
3211
3
32
32
3
2
1
1

91. 91
Corresponding to λ = 2, let X2 = be the eigen
vector then,










3
2
1
x
x
x










=∴
===∴
=
=−
=+−−⇒










=




















−
−
=−
0
1-
1
kX
x-kx,kx
0x
0x
02xxx
0
0
0
x
x
x
100
100
211-
0X)(A
22
32221
3
3
321
3
2
1
2
0,
I2

92. 92
Corresponding to λ = 3, let X3 = be the eigen
vector then, 









3
2
1
x
x
x










=∴
−
===∴
=−−
=+−−⇒










=




















−
−
=−
2
2-
3
kX
xk-x,kx
0x
02xxx
0
0
0
x
x
x
000
11-0
212-
0X)(A
33
13332
3
321
3
2
1
3
3
2
2
3
,
2
I3
k
x

93. 93
Hence modal matrix is














−==∴









 −
=
−=










−=
−
2
1
00
11-0
2
1-
11
M
MAdj.
M
1-00
220
122-
MAdj.
2M
200
21-0
311
M
1

94. 94
Now, since D = M-1
AM
=> A = MDM-1
A2
= (MDM-1
) (MDM-1
)
= MD2
M-1
[since M-1
M = I]










=










−










−
−














−
−
=−
300
020
001
200
21-0
311
300
120
211
2
1
00
11-0
2
1
11
AMM 1

95. 95
Similarly, A3
= MD3
M-1
=
A3
=
























−
−




















−
2700
19-80
327-1
2
1
00
11-0
2
1
11
2700
080
001
200
21-0
311

96. 96
1.9 ORTHOGONAL
TRANSFORMATION OF A SYMMETRIC
MATRIX TO DIAGONAL FORM
A square matrix A with real elements is said to
be orthogonal if AA’ = I = A’A.
But AA-1
= I = A-1
A, it follows that A is orthogonal if
A’ = A-1
.
Diagonalisation by orthogonal transformation is
possible only for a real symmetric matrix.

97. 97
If A is a real symmetric matrix then eigen
vectors of A will be not only linearly independent but
also pairwise orthogonal.
If we normalise each eigen vector and use
them to form the normalised modal matrix N then it
can be proved that N is an orthogonal matrix.

98. 98
The similarity transformation M-1
AM = D takes
the form N’AN = D since N-1
= N’ by a property of
orthogonal matrix.
Transforming A into D by means of the
transformation N’AN = D is called as orthogonal
reduction or othogonal transformation.
Note:- To normalise eigen vector Xr, divide each
element of Xr, by the square root of the sum of the
squares of all the elements of Xr.

99. 99
Diagonalise the matrix A = by means of an
orthogonal transformation.
Solution:-
Characteristic equation of A is
204
060
402
66,2,λ
0λ)16(6λ)λ)(2λ)(6(2
0
λ204
0λ60
40λ2
−=⇒
=−−−−−⇒
=
−
−
−
Example :-

100. 100
I
 
 
 
  
     
     
     
          
⇒
∴
 
 ∴  
  
1
1 2
3
1
1
2
3
1 3
2
1 3
1 1 2 3 1
1 1
x
whenλ = -2,let X = x be theeigenvector
x
then (A + 2 )X = 0
4 0 4 x 0
0 8 0 x = 0
4 0 4 x 0
4x + 4x = 0 ...(1)
8x = 0 ...(2)
4x + 4x = 0 ...(3)
x = k ,x = 0,x = -k
1
X = k 0
-1

101. 101
2
2I
0
 
 
 
  
     
     
     
          
⇒ −
∴
1
2
3
1
2
3
1 3
1 3
1 3 2
2 2 3
x
whenλ = 6,let X = x betheeigenvector
x
then (A -6 )X = 0
-4 0 4 x 0
0 0 x = 0
4 0 -4 x 0
4x +4x = 0
4x - 4x = 0
x = x and x isarbitrary
x must be so chosen that X and X are orthogonal among th
.1
emselves
and also each is orthogonal with X

102. 102
   
   
   
      
∴
∴
 
 
 
  
∴
2 3
3 1
3 2
3
1α
Let X = 0 and let X =β
1γ
Since X is orthogonal to X
α - γ = 0 ...(4)
X is orthogonal to X
α + γ = 0 ...(5)
Solving (4)and(5), we getα = γ = 0 and β is arbitra ry.
0
Takingβ =1, X = 1
0
1 1 0
Modal matrix is M = 0 0 1
-1 1
 
 
 
  0

103. 103
 
 
 
 
 
 
  
 
  
    
    
    
        
    
 
 
 
  
The normalised modal matrix is
1 1
0
2 2
N = 0 0 1
1 1
- 0
2 2
1 1
0 - 1 1
02 2 2 0 4 2 2
1 1
D =N'AN = 0 0 6 0 0 0 1
2 2
4 0 2 1 1
- 00 1 0
2 2
-2 0 0
D = 0 6 0 which is the required diagonal matrix
0 0 6
.

104. 104
1.10 QUADRATIC FORMS
DEFINITION:-DEFINITION:-
A homogeneous polynomial of second degree
in any number of variables is called a quadratic
form.
For example,
ax2
+ 2hxy +by2
ax2
+ by2
+ cz2
+ 2hxy + 2gyz + 2fzx and
ax2
+ by2
+ cz2
+ dw2
+2hxy +2gyz + 2fzx + 2lxw +
2myw + 2nzw
are quadratic forms in two, three and four variables.

105. 105
In n – variables x1,x2,…,xn, the general quadratic form
is
In the expansion, the co-efficient of xixj = (bij + bji).
∑∑= =
≠
n
1j
n
1i
jiijjiij bbwhere,xxb
).b(b
2
1
awherexxaxxb
baandaawherebb2aSuppose
jiijijji
n
1j
n
1i
ijji
n
1j
n
1i
ij
iiiijiijijijij
+==
==+=
∑∑∑∑ = == =

106. 106
Hence every quadratic form can be written as
( ) ( )
getweform,matrixin
formsquadraticofexamplessaidabovethewritingNow
.x,...,x,xXandaAwhere
symmetric,alwaysisAmatrixthethatso
AX,X'xxa
n21ij
ji
n
1j
n
1i
ij
==
=∑∑= =












=++
y
x
bh
ha
y][xby2hxyax(i) 22

107. 107
[ ]
[ ]
























=
+++++++++




















=+++++
w
z
y
x
dnml
ncgf
mgbh
lfha
wzyx
2nzw2myw2lxwzx2f2gyz2hxydw2czbyax(iii)
z
y
x
cgf
gbh
fha
zyx2fzx2gyz2hxyczbyax(ii)
222
222

108. 108
1.11 NATURE OF A QUADRATIC
FORM
A real quadratic form X’AX in n variables is said
to be
i. Positive definite if all the eigen values of A > 0.
ii. Negative definite if all the eigen values of A < 0.
iii. Positive semidefinite if all the eigen values of A 0
and at least one eigen value = 0.
iv. Negative semidefinite if all the eigen values of
A 0 and at least one eigen value = 0.
v. Indefinite if some of the eigen values of A are + ve
and others – ve.
≥
≤

109. 109
Find the nature of the following quadratic forms
i. x2
+ 5y2
+ z2
+ 2xy + 2yz + 6zx
ii. 3x2
+ 5y2
+ 3z2
– 2yz + 2zx – 2xy
Solution:-
i. The matrix of the quadratic form is










=
113
151
311
A
Example :-

110. 110
The eigen values of A are -2, 3, 6.
Two of these eigen values being positive and
one being negative, the given quadratric form is
indefinite.
ii. The matrix of the quadratic form is
The eigen values of A are 2, 3, 6. All these eigen
values being positive, the given quadratic form
is positive definite.










−
−−
−
=
311
151
113
A

111. 111
1.12 REDUCTION OF QUADRATIC
FORM TO CANONICAL FORM
A homogeneous expression of the second
degree in any number of variables is called a
quadratic form.
For instance, if
which is a quadratic form.
(i)....2hxy2gzx2fyzczbyaxAXX'then
],zyx[X'and
z
y
x
X,
cfg
fbh
gha
A
222
+++++=
=










=










=

112. 112
Let λ1, λ2, λ3 be the eigen values of the matrix A
and
be its corresponding eigen vectors in the
normalized form (i.e., each element is divided by
square root of sum of the squares of all the three
elements in the eigen vector).










=










=










=
3
3
3
3
2
2
2
2
1
1
1
1
z
y
x
X,
z
y
x
X,
z
y
x
X

113. 113
Then B-1
AB = D, a diagonal matrix.
Hence the quadratic form (i) is reduced to a sum of
squares (i.e., canonical form).
λ1x2
+ λ2y2
+ λ3z2
And B is the matrix of transformation which is an
orthogonal matrix.
Note:-
1. Here some of λi may be positive or negative or
zero
2. If ρ(A) = r, then the quadratic form X’AX will
contain only r terms.

114. 114
1.13 INDEX AND SIGNATURE OF
THE QUADRATIC FORM
The number p of positive terms in the
canonical form is called the index of the quadratic
form.
(The number of positive terms) – ( the number of
negative terms)
i.e., p – (r – p) = 2p – r is called signature of the
quadratic form, where ρ(A) = r.

115. 115
1.14 LINEAR TRANSFORMATION
OF A QUADRATIC FORM.
Let X’AX be a quadratic form in n- variables
and let X = PY ….. (1) where P is a non –
singular matrix, be the non – singular
transformation.
From (1), X’ = (PY)’ = Y’P’ and hence
X’AX = Y’P’APY = Y’(P’AP)Y
= Y’BY …. (2)
where B = P’AP.

116. 116
Therefore, Y’BY is also a quadratic form in n-
variables. Hence it is a linear transformation of
the quadratic form X’AX under the linear
transformation X = PY and B = P’AP.
Note. (i) Here B = (P’AP)’ = P’AP = B
(ii) ρ(B) = ρ(A)
Therefore, A and B are congruent matrices.

117. 117
Reduce 3x2
+ 3z2
+ 4xy + 8xz + 8yz into canonical
form.
Or
Diagonalise the quadratic form 3x2
+ 3z2
+ 4xy + 8xz
+ 8yz by linear transformations and write the
linear transformation.
Or
Reduce the quadratic form 3x2
+ 3z2
+ 4xy + 8xz +
8yz into the sum of squares.
Example:-

118. 118
Solution:- The given quadratic form can be
written as X’AX where X = [x, y, z]’ and the
symmetric matrix
A =
Let us reduce A into diagonal matrix. We know tat
A = I3AI3.










344
402
423






























=










100
010
001
344
402
423
100
010
001
344
402
423

119. 119
− −
−
   
   
    
    − = −     
      
   − −
   
21 31OperatingR ( 2 / 3),R ( 4 / 3)
(for A onL.H.S.andpre factor on R.H.S.),weget
3 2 4 1 0 0
1 0 0
4 4 2
0 1 0 A 0 1 0
3 3 3
0 0 1
4 7 4
0 0 1
3 3 3














−−
















−
−=
















−
−
−
−−
100
010
3
4
3
2
1
A
10
3
4
01
3
2
001
3
7
3
4
0
3
4
3
4
0
423
getweR.H.S),onfactorpostandL.H.S.onA(for
4/3)(C2/3),(COperating 3121

120. 120














−−










−
−=










−
−
100
010
3
4
3
2
1
A
112
01
3
2
001
100
3
4
3
4
0
003
getwe(1),ROperating 32
APP'1,
3
4
3,Diagor
100
110
2
3
2
1
A
112
01
3
2
001
100
0
3
4
0
003
getwe(1),COperating 32
=





−−














−−










−
−=










−
−

121. 121
The canonical form of the given quadratic form is
Here ρ(A) = 3, index = 1, signature = 1 – (2) = -1.
Note:- In this problem the non-singular
transformation which reduces the given quadratic
form into the canonical form is X = PY.
i.e.,




















−
−
=










3
2
1
112
01
3
2
001
y
y
y
z
y
x
[ ]
2
3
2
2
2
1
3
2
1
321
yy
3
4
3y
y
y
y
100
0
3
4
0
003
yyyAP)Y(P'Y'
−−=




















−
−
=

122. 122
1.15 REDUCTION OF QUADRATIC
FORM TO CANANONICAL FORM BY
ORTHOGONAL TRANSFORMATION
Let X’AX be a given quadratic form. The modal
matrix B of A is the matrix whose columns are
characteristic vectors of A. If B represents the
orthogonal matrix of A,

123. 123
then X = BY will reduce X’AX to Y’ diag(λ1, λ2,…, λn) Y,
where λ1, λ2,…, λn are characteristic values of A.
Note. This method works successfully if the
characteristic vectors of A are linearly independent
which are pairwise orthogonal.

124. 124
Reduce 8x2
+ 7y2
+ 3z2
– 12xy + 4xz – 8yz into
canonical form by orthogonal reduction.
Solution:- The matrix of the quadratic form is










−
−−
−
=
342
476
268
A
Example 1:-

125. 125
The characteristic roots of A are given by 0|| =− IλA
153,0,λ
015)3)(λλ(λor
0
λ342
4λ76
26λ8
=∴
=−−
=
−−
−−−
−−
.,.ei

126. 126
Characteristic vector for λ = 0 is given by
[A – (0)I] X1 = 0
'
11
321
321
321
321
2)2,(1,kXvectoreigenthegiving
2
x
2
x
1
x
getwe,twofirstSolving
03x4x2x
04x7x6x
02x6x8xi.e.,
=
==
=+−
=−+−
=+−

127. 127
When λ = 3, the corresponding characteristic
vector is given by [A – 3I] X2 = 0
i.e.,
Solving any two equations, we get X2 = k2 (2, 1, -2)’.
Similarly characteristic vector corresponding to
λ = 15 is X3 = k3 (2, -2, 1)’.
04x2x
04x4x6x
02x6x5x
21
321
321
=−
=−+−
=+−

128. 128
Now, X1, X2, X3 are pairwise orthogonal
i.e., X1 . X2 = X2 . X3 = X3 . X1 = 0.
The normalised modal matrix is
















−
−=
3
1
3
2
3
2
3
2
3
2
3
1
3
2
3
1
3
2
B
∴

129. 129
Now B is orthogonal matrix and 1B =










=
















−
−
















−
−
=== −−
1500
000
003
3
1
3
2
3
2
3
2
3
2
3
1
3
2
3
1
3
2
A
3
1
3
2
3
2
3
2
3
2
3
1
3
2
3
1
3
2
ie.,
15}0,{3,diagDABBandBBi.e., 1T1

130. 130
which is the required canonical form.
Note. Here the orthogonal transformation is X =BY,
rank of the quadratic form = 2; index = 2,
signature = 2. It is positive definite.
[ ]
2
3
2
2
2
1
3
2
1
321
1
15y0.y3y
y
y
y
1500
000
003
yyy
DYY'AB)Y(BY'AXX'
++=




















=
== −

131. 131
TEST YOUR KNOWLEDGE
1. If then find the eigen value of
2. Write the matrix of the Quadratic form
3. Obtain the characteristic equation of the matrix whose
eigen values are 1,-2 and 3.
4. A = , then find the eigen values of
3A3
+5A2
-6A+2I.
1 2 3
0 3 2
0 0 2
A
− 
 ÷
=  ÷
 ÷− 
( )
2
3A I−
1 2 3 1 2 32 2 2x x x x x x+ −
.










−
200
320
061

132. 132
5. Write the quadratic form corresponding to the
following symmetric matrix
6. Find the sum of the eigen values of the inverse of.
7. Obtain the latent roots of A4
where A =
8. If A is idempotent matrix then A2
=A. What will be
the eigen values of A.










−
−
342
411
210
1 0 0
2 3 0
0 5 2
 
 ÷
− ÷
 ÷
 
5 4
1 2
 
 ÷
 

133. 133
9. If are the eigen values of the matrix A,
whose characteristic equation is
Obtain using the property.
10. (i) Using Cayley-Hamilton Theorem find the
inverse of the matrix
(ii) Find the Characteristic roots and
Characteristic vectors of the matrix
1 2 2 3 3 1λ λ λ λ λ λ+ +
3 2
21 45 0λ λ λ+ − − =
1 2 3, andλ λ λ
1 0 3
2 1 1
1 1 1
A
 
 ÷
= − ÷
 ÷− 
1 1 3
1 5 1
3 1 1
A
 
 ÷
= ÷
 ÷
 

134. 134
11. Reduce the quadratic form
to the canonical form by orthogonal transformation.
Also specify the matrix of transformation. Obtain
its index, signature and nature of the quadratic
form.
12. (i) Find the eigen value and eigen vector of the
matrix
(ii) Using Cayley Hamiltonian find the inverse of
2 2 2
5 2 2 6x y z xy yz xz+ + + + +










−−
−−
−−
6410
527
7411










−
−
111
112
301

135. 135
13. Discuss the nature, index and signature of the
quadratic form
14. Diagonalise the matrix by
orthogonal reduction and provide the normalized
modal matrix.
15. Reduce the quadratic form 2x1x2+2x1x3-2x2x3 to the
canonical form by an Orthogonal transformation.
2 2 2
10 2 5 6 10 4x y z yz zx xy+ + + − −
8 6 2
6 7 4
2 4 3
A
− 
 ÷
= − − ÷
 ÷− 

136. 136
16. (i) Find the eigen value and eigen vector of the
matrix
(ii) For A = , compute the value of
, Using Cayley-
Hamilton theorem.
17. Reduce the quadratic form 3x1
2
+ 5x2
2
+ 3x3
2
-2x2x3+2x3x1-2x1x2 to a Canonical form by orthogonal
reduction.
2 1 0
0 2 1
0 0 2
 
 ÷
 ÷
 ÷
 










−
−
111
112
301
6 5 4 3 2
5 8 2 9 31 36A A A A A A I− + − − + −

137. THANK YOU