drmgriyearbtech.com · Web viewProve that a square matrix and its transpose has the same eigen values. Solution: Let A be a square matrix of order n. The characteristic equation of

UNIT-IIMATRICES Introduction In mathematics, a matrix is a rectangular array numbers.Matrices consisting of only one column or row are called vrctors,while higher –dimensional, arrays of numbers are called tensors.Matrix can also keep track of the coefficients in a system of linear equations.For a square matrix, the determinant and inverse matrix govern the behavior of solutions to the corresponding system of linear equations, and eigen values and eigen vectors provide insight into the geometry of the associated linear transformation. Applications1.Physics makes use of them in various domains, for example in geometrical optics and matrix mechanics.2.Matrices encoding distances of knot points in a graph, such as cities connected by roads, are used in graph theory, and computer graphics use matrices to encode projections of three-dimensional spact onto a two-dimensional screen.3. Serialism and dodecaphonism are musical movements of the 20th century that utilize a square mathematical matrix to determine the pattern of music intervals.Characteristic EquationFor a Linear transformation the characteristic equation [Latent equation] can be defined as |A−λI |=0 where A is the given matrix from the linear equation and is the eigen constant or characteristic constant and I is the unit matrix with respect to the order of A.

45

Problems Find the characteristic equation of ( 2 1

−1 2).Solution:Let A=( 2 1

−1 2) The characteristic equation of A is λ2−s1 λ+s2=0 s1=¿sum of the main diagonal elements=2+2=4s2=|A| =| 2 1

−1 2|=4+1=5Hence the required characteristic equation is λ2−4 λ+5=0 Find the characteristic equation of (1 2

0 2)Solution:Let A=(1 2

0 2) The characteristic equation of A is λ2−s1 λ+s2=0s1=¿Sum of the main diagonal elements=1+2=3s2=|A| =|1 2

0 2|=2−0=2Hence the required characteristic equation is λ2−3 λ+2=0

46

Find the characteristic equation of (1 1 11 2 21 2 3) .Solution:Given matrix is a 3x3 matrixThe characteristic equation of A is λ3−s1 λ2+s2 λ−s3=0

s1=¿sum of the main diagonal elements=1+2+3=6s2=¿sum of the minors of main diagonal elements ¿|2 2

2 3|+|1 11 3|+|1 1

1 2| ¿ (6−4 )+(3−1¿+(2−1)=2+2+1=5

s3=|A|

¿|1 1 11 2 21 2 3|

¿1 (6−4 )−1 (3−2 )+1 (2−2 )

¿2−1=1 Hence the required characteristic equation is λ3−6 λ2+5 λ−1=0 Find the characteristic equation of [2 0 1

0 2 01 0 2 ].Solution:Given matrix is a 3x3 matrixThe characteristic equation of A is λ3−s1 λ2+s2 λ−s3=0

s1=¿Sum of the main diagonal elements=2+2+2=6s2=¿Sum of the minors of main diagonal element

47

¿|2 00 2|+|2 1

1 2|+|2 00 2|

=4+ (4−1) +4=4+3+4=11 s3=|A|

¿|2 0 10 2 01 0 2| ¿2 (4−0 )−0+1 (0−2 )

¿8−2=6 Hence the required characteristic equation isλ3−6 λ2+11 λ−6=0

EIGEN VALUES : Let A be a square matrix .The characteristic equation of A is |A−λI|=0.The roots of the characteristic equation are called Eigen values of A.EIGEN VECTOR: Let A be a square matrix .If there exists a non –zero column vector X such that AX=λX, then the vector X is called an Eigen vector of A corresponding to the Eigen value of λ .Properties of Eigen values.Solution:i)The sum of the eigen values of a matrix is equal to the trace of the matrix and product of the eigen values is equal to the determinant of the matrix.ii)A square matrix A and its transpose AT have the same eigen values.48

Problems Prove that a square matrix and its transpose has the same eigen values.Solution:Let A be a square matrix of order n.The characteristic equation of A and AT are |A−λI|=0∧¿

|AT−λI|=0Since the determinant value is unaltered by the interchange of rows and columns.(i.e) |A|=|AT|Hence characteristic equation of A and AT are identicalTherefore the eigen values of A and ATare the same. If A=(4 1

3 2), find the eigen value of A3.Solution:The characteristic equation of A is λ2−s1 λ+s2=0

s1=4+2=6

s2=|4 13 2|

¿8−3=5The characteristic equation of A is λ2−6 λ+5=0 (λ−5¿ ( λ−1 )=0

λ=1,5Eigen values of the given matrix A are 1,5Eigen values of the matrixA3 are 1,12549

Find the eigen values of A=( 1 1−1 1).Solution:The characteristic equation of A is λ2−s1 λ+s2=0

s1=1+1=2

s2=| 1 1−1 1|

¿1+1=2The characteristic equation of A is λ2−2 λ+2=0λ=2±√4−8

2

λ=2±2i2

λ=1±i

∴The eigen values are 1+i and 1−i Find the sum and product of the eigen values of (−15 4 3

10 −12 620 −4 2)

without finding the eigen values.Solution:Sum of the eigen values of A = trace of he matrix A

¿−15−12+2

¿−25 Product of the Eigen Values of A=|A| =−15(−24+24)−4 (20−120)+3¿+240)¿−4 ¿)+3(200)

50

=400+600 =1000 Find the sum and product of the eigen values of (−2 2 −3

2 1 −6−1 −2 0 )

without finding the eigen values.Solution:Sum of the eigen values of A = trace of he matrix A¿−2+1+0=−1Product of the Eigen Values of A=|A| ¿−2(0−12)−2(0−6)−3(−4+1) =24+12+9 =45

Find the sum and product of the eigen values of (−10 −2 −52 2 3

−5 3 5 )without finding the eigen values.Solution: Sum of the eigen values of A = trace of the matrix A ¿10+2+5=17Product of the Eigen Values of A=|A| ¿10(10−9)+2(10+15)−5¿6+10) =10+2(25 ¿−5(16) =10+50−80=−20

51

If the matrix A=(1 1 31 5 13 1 1), find the eigen values of A−1.

Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0s1=1+5+1=7

s2=|1 11 5|+|5 1

1 1|+|1 33 1|

=(5−1¿+(5−1)+¿) ¿4+4−8=0 s3=|1 1 3

1 5 13 1 1|

=1(5−1¿−1(1−3)+3¿) =4+2−¿42=−¿36 The characteristic equation of A isλ3−7 λ2+36=0If λ=−2 , then (−2 )3−¿7(−2 )2+36=0

∴ λ=−¿2 is a root.Using synthetic division, -2 1 -7 0 36 0 -2 18 -36

1 -9 18 0∴ λ=−2∧λ2−9 λ+18=0

52

( λ−3¿ ( λ−6 )=0

λ=3,6Hence, The Eigen values of A are −¿2,3,6 The Eigen values of A−1 are −12 , 13 , 16 Find the eigen values of 2 A2 ifA=(4 1

3 2).Solution:The characteristic equation of A is λ2−s1 λ+s2=0s1=4+2=6

s2=|4 13 2|

=8−¿3=5The characteristic equation of A is λ2−6 λ+5=0λ=1,5Eigen values of A are 1,5Eigen values of A2are 1,25Eigen values of 2A2 are 2,50

Two eigen values of the matrix A=(2 2 11 3 11 2 2)are equal to 1(one)

each. Find the eigen values ofA−1.Solution:Let λ1 , λ2 , λ3be the Eigenvalues of A . Given that λ1=λ2=1 .We know that,

53

Sum of the eigen values of A = trace of the matrix ATherefore , λ1+λ2+λ3=2+3+2 ⟹1+ 1+λ3=7

λ3=5

∴ λ1=1 , λ2=1 , λ3=5 Therefore ,Eigen values of A−1are 1,1, 15 If 3 and 15 are two eigen values ofA=( 8 −6 2

−6 7 −42 −4 3 ), find |A|.

Solution:Let λ1 , λ2 , λ3be the Eigenvalues of A .Given λ1=3 , λ2=15 , λ3=?We know that,Sum of the eigen values =sum of the main diagonal elements

⟹ λ1+λ2+ λ3=8+7+3 ⟹ 3+15+λ3=18⟹ λ3=0 ∴ The eigen values are 3,15,0. |A|=¿Product of the eigen values =(3)(15)(0) ⟹|A|=0

Find the eigen values of the inverse of the matrixA=(3 0 08 4 02 2 5).Solution:Given matrix A is a lower triangular matrix.

54

The eigen values of a triangular matrix are just the diagonal elements of the matrixHence eigen values of A are 3,4,5∴ Eigen values of A−1 are 13 , 14 , 15

If 3 and 6 are the eigen values of A=(1 1 31 5 13 1 1). Write down the

eigen values of A−1 and 3A.Solution:Let λ1 , λ2 , λ3be the Eigenvaluesof AGiven that λ1=3∧ λ2=6We know that,Sum of the eigen values =sum of the main diagonal elements

⟹ λ1+λ2+ λ3=1+5+1 ⟹ 3+6+λ3=7⟹ λ3=7−9

⟹ λ3=−2 Hence the eigen values of A are −¿2,3,6Eigen values of 3A are −¿6,9,18Eigen values of A−1are −12 , 13 , 16

Find the eigen values of A3given A=(1 2 30 2 −70 0 3 ).

Solution:55

For a Triangular matrix, the diagonal elements are its Eigen values. Therefore, The eigen values of A are 1,2,3 The eigen values of A3 are 1,8,27 Two eigen values of A=( 6 −2 2

−2 3 −12 −1 3 )are 2 and 8. Find the

third eigen values.Solution:Let λ1 , λ2 , λ3be the Eigenvalues of AGiven λ1=2 , λ2=8 , λ3=?We know that,Sum of the eigen values =sum of the main diagonal elements

⟹ λ1+λ2+ λ3=6+3+3 ⟹ 2+8+λ3=12⟹ λ3=12−10⟹ λ3=2

The product of the two eigen values of A=( 6 −2 2−2 3 −12 −1 3 )is 16.

Find the third eigen value.Solution:Let the eigen values of A be λ1 , λ2 , λ3Given λ1 λ2=16. By the property, |A|= product of the eigen values⟹|A|=λ1 λ2 λ3

56

λ1 λ2 λ3=| 6 −2 2−2 3 −12 −1 3 |

16λ3=6 (9−1 )+2(−6+2)+2(2−6)

=6(8¿+2(−4)+2(−4) =48−8−8=32

⟹ λ3=3216 ⟹ λ3=2 Find the sum of the squares of the eigen values of

A=(3 1 40 2 60 0 5) .Solution:

Given matrix A is a upper triangular matrix.For a Triangular matrix, the diagonal elements are its Eigen

values.

∴ Eigen values of A are 3,2,5Sum of the squares of the eigen values of A=9+4+25=38 If the sum of two eigen values and trace of the matrix A are equal, find the value of |A|.Solution:

Let λ1 , λ2 , λ3be the Eigenvalues of A .

57

By the property, Sum of the eigen values of A =Trace of the matrix A( i .e ) λ1+λ2+λ3=λ1+λ2

⟹ λ3=0By the property, |A|= product of the eigen values( i .e ) |A|=λ1 λ2 λ3=0 Prove that the eigen values of −3 A−1 are the same as those of

A=(1 22 1).Solution: The characteristic equation of A is λ2−s1 λ+s2=0

s1=1+1=2

s2=|1 22 1|

=1−¿4¿−3

The characteristic equation of A is λ2— 2λ−3=0⇒ ( λ+1 ) ( λ−3 )=0⇒ λ=−1,3Eigen values of A are −¿1,3

Eigen values of A−1are −¿1,13 Eigen values of −3 A−1 are 3,−¿1∴ Eigen values of A =Eigen values of −3 A−1

58

Two eigen values of A=( 4 6 61 3 2

−1 −5 −2) are equal and they are double the third. Find the eigen value of A2.Solution:

Let λ1 , λ2 , λ3be the Eigenvaluesof A .Given λ1=λ2=2λ3Sum of the eigen values =sum of the main diagonal elements ⟹ λ1+ λ2+ λ3=4+3−¿2 ⟹ λ1+ λ1+ λ12 =5

⟹ 2 λ1+ λ12 =5, where λ3= λ12⟹

5 λ12

=5

⟹λ12

=1

λ1=2=λ22 λ3=2⇒ λ3=1Eigen values of A are 2,2,1Eigen values of A2 are 4,4,1.

Problems Find the eigen values and eigen vectors of A=( 6 −6 5

14 −13 107 −6 4 )

59

Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0s1=6−13+4

=10−13=−3

s2=| 6 −614 −13|+|−13 10

−6 4 |+|6 57 4|

¿(−78+84)+(−52+60)+(24−35)

=6+8−11=3

s3=|A|

=6(8¿+6 (56−70)+5 (−84+91)

¿48−84+35=−1

The characteristic equation of A isλ3+3 λ2+3 λ+1=0If λ=−1 , then (−1 )3+3−3+1=0

∴ λ=−1 is a root.Using synthetic division, -1 1 3 3 1 0 -1 -2 -1

1 2 1 0∴ λ=−1∧λ2+2 λ+1=0

⟹ λ=−1 ,−1 ,−1

60

To find eigen vectors solve (A−λI )X=0

⇒ (6−λ −6 514 −13−λ 107 −6 4−λ)(

x1x2x3)=(000)−−−−−−(1)

Case (i) : When λ=−1∈(1 ) ,

(6+1 −6 514 −13+1 107 −6 4+1)(

x1x2x3)=(000)

( 7 −6 514 −12 107 −6 5 )( x1x2x3)=(000)

⇒ 7x1−6 x2+5x3=0 14x1−12 x2+10 x3=0 7x1−6 x2+5x3=0The above equations represents the same equation 7x1−6 x2+5x3=0Choosing arbitrary values for x1 , let x1=0 6x2=5 x3

x25

=x36

X1=(056)Choosing arbitrary values for x2 , let x2=0 7x1=−5 x3

61

x1−5

=x37

X2=(−507 )Choosing arbitrary values for x3 , let x3=0 7x1=6 x2

x16

=x27

X3=(670)∴Eigen vectors of A are

X1=(056) , X2=(−507 ) , X3=(670) Find the eigen values and eigen vectors of A=( 6 −2 2

−2 3 −12 −1 3 ) Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0

s1=6+3+3=12

s2=| 6 −2−2 3 |+| 3 −1

−1 3 |+|6 22 3|

=(18-4)+(9-1)+(18-4)62

=14+8+14=36s3=|A|

=6(9−1¿+2(−6+2)+2(2−6)

¿48−8−8=32

The characteristic equation of A isλ3−12 λ2+36 λ−32=0If λ=2 ,then (2 )3−12 (2 )2+36 (2 )−32=0

∴ λ=2 is a root.Using synthetic division, 2 1 -12 36 -32 0 2 -20 32

1 -10 16 0∴ λ=2∧λ2−10 λ+16=0

⟹ λ=2,2,8


⇒ (6−λ −2 2−2 3− λ −12 −1 3−λ)(

x1x2x3)=(000)−−−−−−(1)

Case (i): When λ=2 in (1), we get,63

(6−2 −2 2−2 3−2 −12 −1 3−2)(

x1x2x3)=(000)

( 4 −2 2−2 1 −12 1 1 )(x1x2x3)=(000)

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

The above equations represents the same equation 2 x1−x2+x3=0

Choosing arbitrary values for x1 , let x1=0x2=x3

X1=(011)Choosing arbitrary values for x2, let x2=0 2x1=−x3

x1−1

=x32

X2=(−102 )Case (ii): When λ=8 in (1),

64

(6−8 −2 2−2 3−8 −12 −1 3−8)(

x1x2x3)=(000)

(−2 −2 2−2 −5 −12 −1 −5)(

x1x2x3)=(000)

−¿2x1−2 x2+2x3=0−¿2x1−5 x2−x3=0 2x1−x2−5 x3=0Solving first two equations using cross rule method

x12+10

=x2

−4−2=

x310−4

x112

=x2

−6=x36

x12

=x2−1

=x31

X3=( 2−11 )∴Eigen vectors of A are

X1=(011) , X2=(−102 ) , X3=( 2−11 ) Find the eigen values and eigen vectors of A=(2 −2 2

1 1 11 3 −1) Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0

s1=2+1−1=2

65

s2=|2 −21 1 |+|1 1

3 −1|+|2 21 −1|

=(2+2)+(−1−3)+(−2−2)

¿4−4−4=−4

s3=|A|

¿2(−1−3)+2 (−1−1)+2(3−1)¿−8−4+4=−8

The characteristic equation of A isλ3−2 λ2−4 λ+8=0If λ=2 ,then (2 )3−2 (2 )2−4 (2 )+8=0

∴ λ=2 is a root.Using synthetic division, 2 1 -2 -4 8 0 2 0 -8

1 0 -4 0∴ λ=2∧λ2−4=0

⟹ λ=2,2,−2


⇒ (2−λ −2 21 1−λ 11 3 −1−λ)(

x1x2x3)=(000)−−−−−−(1)

Case (i): When λ=2∈(1 ) .

66

(2−2 −2 21 1−2 11 3 −1−2)(

x1x2x3)=(000)

(0 −2 21 −1 11 3 −3)(

x1x2x3)=(000)

0x1−2 x2+2x3=0x1−x2+x3=0

x1+3x2−3 x3=0Solving first two equations using cross rule methodx1

−2+2=x22−0

=x30+2

x10

=x22

=x32

X1=(011)X2=(011) as an eigen vector corresponding to λ=2

Case (ii): When λ=−2∈(1)

(2+2 −2 21 1+2 11 3 −1+2)(

x1x2x3)=(000)

(4 −2 21 3 11 3 1)(

x1x2x3)=(000)

4x1−2 x2+2x3=067

x1+3 x2+x3=0

x1+3 x2+x3=0Solving first two equations using cross rule methodx1

−2−6=

x22−4

=x312+2

x1−8

=x2−2

=x314

x1−4

=x2

−1=x37

X3=(−4−17 )

∴Eigen vectors of A are X1=(011) , X2=(011) , X3=(−4−1

7 ) Find the eigen values and eigen vectors of A=(1 0 0

0 3 −10 −1 3 ).


s2=|1 00 3|+| 3 −1

−1 3 |+|1 00 3|

=3+(9−¿1)+3=14 s3=|A|

=1(9−1¿−0+0=868

The characteristic equation of A isλ3−7 λ2+14 λ−8=0If λ=1 ,then 1-7+14-8=0∴ λ=1 is a root.Using synthetic division, 1 1 -7 14 -8

0 1 -6 8 1 -6 8 0

∴ λ=1∧ λ2−6 λ+8=0

⟹ λ=1,2,4


⇒ (1−λ 0 00 3−λ −10 −1 3−λ)(

x1x2x3)=(000)−−−−−−(1)

Case (i): When λ=1(1−1 0 00 3−1 −10 −1 3−1)(

x1x2x3)=(000)

(0 0 00 2 −10 −1 2 )(x1x2x3)=(000)

0x1+0x2+0x3=069

0 x1+2x2−x3=0

0 x1−x2+2 x3=0Solving last two equations using cross rule methodx14−1

=−x20−0

=x30−0

x13

=x20

=x30

X1=(100)Case (ii) : when λ=2∈(1)

(1−2 0 00 3−2 −10 −1 3−2)(

x1x2x3)=(000)

(−1 0 00 1 −10 −1 1 )(x1x2x3)=(000)

−x1+0 x2+0x3=00 x1+x2−x3=0

0 x1−x2+x3=0Solving first two equations using cross rule methodx10−0

=−x21−0

=x3

−1−0

x10

=x2−1

=x3

−1

X2=( 0−1−1)Case (iii) : when λ=4∈(1)

70

(1−4 0 00 3−4 −10 −1 3−4)(

x1x2x3)=(000)

(−3 0 00 −1 −10 −1 −1)(

x1x2x3)=(000)

−3 x1+0x2+0x3=00 x1−x2−x3=0

0 x1−x2−x3=0Solving first two equations using cross rule methodx10−0

=−x23−0

=x33−0

x10

=x2−3

=x33

x10

=x2−1

=x31

X3=( 0−11 ) ∴Eigen vectors of A are

X1=(100) , X2=( 0−1−1) , X3=( 0−11 ) Find the eigen values and eigen vectors of (−2 2 −3

2 1 −6−1 −2 0 ).

Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0s1=−2+1+0=−1

71

s2=| 1 −6−2 0 |+|−2 −3

−1 0 |+|−2 22 1|

=−12−3−6=−21

s3=|A| =(−2)(0−12)−2(0−6)+(−3)¿) =24+12+9 =45The characteristic equation of A isλ3+ λ2−21 λ−45=0If λ=−3 ,then (−3 )3+(−3 )2−21 (−3 )−45=0

∴ λ=−3 is a root.Using synthetic division, -3 1 1 -21 -45 0 -3 6 45

1 -2 -15 0λ=−3 , λ2−2 λ−15=0

⟹ λ=−3 ,−3,5

To find eigen vectors solve(A−λI )X=0

72

⇒ (−2−λ 2 −32 1−λ −6

−1 −2 0−λ)(x1x2x3)=(000)−−−−−−(1)

Case (i): When λ=−3 in (1),(−2+3 2 −32 1+3 −6

−1 −2 0+3) (x1x2x3)=(000)

( 1 2 −32 4 −6

−1 −2 3 ) (x1x2x3)=(000)x1+2x2−¿3x3=02 x1+4 x2−6 x3=0

−x1−2 x2+3 x3=0

The above equations represents the same equation x1+2x2−3 x3=0

Choosing arbitrary values for x1 , let x1=02 x2=3 x3

x23

=x32

X1=(032)Choosing arbitrary values for x2 , let x2=0

x1=3 x3

73

x13

=x31

X2=(301)Case(ii) : When λ=5 in (1),

(−2−5 2 −32 1−5 −6

−1 −2 0−5) (x1x2x3)=(000)

(−7 2 −32 −4 −6

−1 −2 −5) (x1x2x3)=(000)

−7 x1+2x2−¿3x3=02 x1−4 x2−6 x3=0

−x1−2 x2−5 x3=0Solving first two equations using cross rule methodx1

−12−12=

−x242+6

=x3

28−4

x1−24

=x2

−48=x324

x11

=x22

=x3−1

X3=( 12−1)

74

∴Eigen vectors of A are X1=(032) , X2=(301) , X3=( 12−1)

Find the eigen values and eigen vectors of A =( 7 −2 −2−2 1 4−2 4 1 ). Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0

s1=7+1+1=9

s2=| 7 −2−2 1 |+|1 4

4 1|+| 7 −2−2 1 |

¿(7−4)+(1−16)+(7−4 )

¿3−15+3=−9

s3=|A|

¿7(−15)+2(−2+8)−2(−8+2)¿−105+12+12=−81

The characteristic equation of A isλ3−9 λ2−9 λ+81=0If λ=3 ,then (3 )3−9 (3 )2−9 (3 )+81=0

∴ λ=3 is a root.Using synthetic division, 3 1 -9 -9 8175

0 3 -18 -81 1 -6 -27 0

λ=3 , λ2−6 λ−27=0

⟹ λ=3,9 ,−3To find eigen vectors solve (A−λI )X=0

⇒ (7−λ −2 −2−2 1−λ 4−2 4 1− λ)(

x1x2x3)=(000)−−−−−−(1)

Case (i): When λ=¿3 in (1),(7−3 −2 −2−2 1−3 4−2 4 1−3)(

x1x2x3)=(000)

( 4 −2 −2−2 −2 4−2 4 −2)(

x1x2x3)=(000)

4 x1−2x2−2 x3=0

−2 x1−2x2+4 x3=0

−2 x1+4 x2−2x3=0Solving first two equations using cross rule methodx1

−8−4=

−x216−4

=x3

−8−4

x1−12

=x2

−12=x3

−12

x11

=x21

=x31

76

X1=(111)Case (ii): When λ=¿9 in (1),

(7−9 −2 −2−2 1−9 4−2 4 1−9)(

x1x2x3)=(000)

(−2 −2 −2−2 −8 4−2 4 −8)(

x1x2x3)=(000)

−2 x1−2x2−2 x3=0

−2 x1−8 x2+4 x3=0

−2 x1+4 x2−8 x3=0

Solving first two equations using cross rule methodx1

−8−16=

−x2−8−4

=x3

16−4

x1−2

=x21

=x31

X2=(−211 )Case (iii): When λ=−¿3 in (1)

(7+3 −2 −2−2 1+3 4−2 4 1+3)(

x1x2x3)=(000)

( 10 −2 −2−2 4 4−2 4 4 )(x1x2x3)=(000)

77

10 x1−2x2−2 x3=0

−2 x1+4 x2+4 x3=0

−2 x1+4 x2+4 x3=0


−8+8=

−x240−4

=x3

40−4

x10

=x2

−36=x336

X3=( 0−11 )

∴Eigen vectors of A are X1=(111) , X2=(−211 ) , X3=( 0−11 )

Find the eigen values and eigen vectors of (0 1 11 0 11 1 0). Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0

s1=0+0+0=0

s2=|0 11 0|+|0 1

1 0|+|0 11 0|

¿−1−1−1=−3

78

s3=|A|

¿0−1(0−1)+1(1)=2

The characteristic equation of A isλ3−0 λ2−3 λ−2=0If λ=2 ,then (2 )3−0 (2 )2−3 (2 )−2=0

∴ λ=2 is a root.Using synthetic division, 2 1 0 -3 -2 0 2 4 2

1 2 1 0λ=2 , λ2+2 λ+1=0

⟹ λ=2 ,−1 ,−1


⇒ (0−λ 1 11 0−λ 11 1 0−λ)(

x1x2x3)=(000)−−−−−−(1)

Case (i): When λ=2 in (1)(0−2 1 11 0−2 11 1 0−2)(

x1x2x3)=(000)

(−2 1 11 −2 11 1 −2)(

x1x2x3)=(000)

−2 x1+x2+x3=0

79

x1−2 x2+x3=0

x1+ x2−2x3=0

Solving first two equations using cross rule methodx11+2

=−x2

−2−1=

x34−1

x13

=x23

=x33

x11

=x21

=x31

X1=(111)Case (ii): whenλ=−1∈(1)

(0+1 1 11 0+1 11 1 0+1)(

x1x2x3)=(000)

(1 1 11 1 11 1 1)(

x1x2x3)=(000)

x1+ x2+x3=0

x1+ x2+x3=0

x1+ x2+x3=0

The above equations represents the same equation x1+ x2+x3=0

Choosing arbitrary values for x1 , let x1=0x2=−x3

80

x21

=x3−1

X2=( 01−1)Choosing arbitrary values for x2, let x2=0 x1=−x3

x1−1

=x31

X3=(−101 )∴Eigen vectors of A are

X1=(111) , X2=( 01−1) , X3=(−101 )

Cayley-Hamilton theorem.Every square matrix satisfies its own characteristic equation.Problems Verify Cayley Hamilton theorem for the matrix A=(5 3

1 3). Solution:The characteristic equation of A is λ2−s1 λ+s2=0By Cayley Hamilton theorem,A2−s1 A+s2 I=0Here, s1=5+3=881

s2=|5 31 3|

=15−¿3=12To prove A2−8 A+12 I=0

A2=(5 31 3)(5 3

1 3) =(25+3 15+9

5+3 3+9 )¿(28 248 12)

A2−8 A+12 I=(25+3 15+95+3 3+9 )+(−40 −24

−8 −24)+(12 00 12)

=(0 00 0)

∴ Cayley Hamilton theorem is verified. Use Cayley Hamilton theorem to find of the given matrix A=(5 3

1 3)Solution:The characteristic equation of A is λ2−s1 λ+s2=0s1=5+3

¿8

s2=|5 31 3| =15−¿3 =12By Cayley Hamilton theorem,we have A2−8 A+12 I=0Premultiply by A on both sides , we get A3−8 A2+12 A=0

82

⇒ A3=8 A2−12 AA2=(5 3

1 3)(5 31 3)=(28 24

8 12)∴ A3=8(28 24

8 12)−12(5 31 3)

=(224 19264 96 )−(60 36

12 36) =(164 156

52 60 ) Use Cayley Hamilton theorem to find of the given matrix A=(−1 3

2 4).Solution:The characteristic equation of A is λ2−s1 λ+s2=0

s1=−1+4

=3s2=|−1 3

2 4|=−4−6=−10

By Cayley Hamilton theorem, we have A2−3 A−10 I=0Premultiply by A on both sides , we get A3−3 A2−10 A =0⇒ A3=3 A2+10A

A2=(−1 32 4)(−1 3

2 4 ) =( 1+6 −3+12

−2+8 6+16 )

83

=(7 96 22)

∴ A3=3(7 96 22)+10(−1 3

2 4) =(21 27

18 66)+(−10 3020 40)

=(11 5738 106)

Verify Cayley –Hamilton therorem . Also find A−1∧A4 ,ifA=[1 0 3

2 1 −11 −1 1 ].


s2=|1 02 1|+| 1 −1

−1 1 |+|1 31 1|

=1+(1−1¿+¿) =1+0−2=−1

s3=|A|

¿1(1−1)−0(2+1)+3 (−2−1)

¿0−0+3(−3)=−9

The characteristic equation of A isλ3−3 λ2− λ+9=0 By Cayley Hamilton Theorem,A3−3 A2−A+9 I=0−−−−−−−−(1)

84

Verification: A2=[1 0 3

2 1 −11 −1 1 ][1 0 3

2 1 −11 −1 1 ]

=[ 1+0+3 0+0−3 3+0+32+2−1 0+1+1 6−1−11−2+1 0−1−1 3+1+1 ]

=[4 −3 63 2 40 −2 5 ]

A3=A . A2

¿ [1 0 32 1 −11 −1 1 ] [4 −3 6

3 2 40 −2 5]

¿ [ 4+0+0 −3+0−6 6+0+158+3+0 −6+2+2 12+4−54−3+0 −3−2−2 6−4+5 ]

¿ [ 4 −9 2111 −2 111 −7 7 ]

A3−3 A2−A+9 I=[ 4 −9 2111 −2 111 −7 7 ]−3[4 −3 6

3 2 40 −2 5 ]−[1 0 3

2 1 −11 −1 1 ]+[9 0 0

0 9 00 0 9]

=[0 0 00 0 00 0 0 ]

To find A−1: Multiply both sides by A−1 in (1), we get A2−3 A−I+9 A−1=0

⟹ A−1=−19

[ A2−3 A−I ]

85

A−1=−19 [[4 −3 6

3 2 40 −2 5]+[−3 0 −9

−6 −3 3−3 3 −3]+[−1 0 0

0 −1 00 0 −1] ]

=−19 [ 0 −3 −3

−3 −2 7−3 1 1 ]

=19 [0 3 33 2 −73 −1 −1]

To Find A4:Multiply both sides by A in (1), we getA4−3 A3−A2+9 A=0

⟹ A4=3 A3+A2−9 A

¿3[ 4 −9 2111 −2 111 −7 7 ]+[4 −3 6

3 2 40 −2 5 ]−9[

1 0 32 1 −11 −1 1 ]

¿ [12 −27 6333 −6 333 −21 21]+[4 −3 6

3 2 40 −2 5 ]+[ −9 0 −27

−18 −9 9−9 9 −9 ]

¿ [ 12+4−9 −27−3+0 63+6−2733+3−18 −6+2−9 33+4+93+0−9 −21−2+9 21+5−9 ]

¿ [ 7 −30 4218 −13 46−6 −14 17 ]

Verify Cayley –Hamilton therorem . Also find A−1∧A4 ,ifA=[1 2 −3

2 5 −43 7 −51].Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0

86

s1=1+5−5=1

s2=|1 22 5|+|5 −4

7 −5|+|1 −23 −5|

¿(5−4)+(−25+28)+(−5+6)

=1+3+1=5s3=|A|

¿1(3)−2(−10+12)−2(14−15)

¿3−4+2=1

The characteristic equation of A isλ3−λ2+5 λ−1=0 By Cayley Hamilton Theorem,A3−A2+5 A−I=0−−−−−−(1)Verification:

A2=[1 2 −22 5 −43 7 −5] [

1 2 −22 5 −43 7 −5 ]

=[−1 −2 00 1 −42 6 −9 ]

A3=A2. A

¿ [−1 −2 00 1 −42 6 −9] [

1 2 −22 5 −43 7 −5 ]

¿ [ −5 −12 10−10 −23 16−13 −29 17]

87

A3−A2+5 A−I=[ −5 −12 10−10 −23 16−13 −29 17]−[−1 −2 0

0 1 −42 6 −9]+5 [

1 2 −22 5 −43 7 −5 ]−[1 0 0

0 1 00 0 1 ]

=[0 0 00 0 00 0 0 ]To find A−1: Multiply both sides by A−1 in (1), we get

A−1=A2−A+5 I

A2=[1 2 −22 5 −43 7 −5] [

1 2 −22 5 −43 7 −5 ]

=[−1 −2 00 1 −42 6 −9 ]

A−1=[−1 −2 00 1 −42 6 −9]−[1 2 −2

2 5 −43 7 −5]+[5 0 0

0 5 00 0 5 ]

=[ 3 −4 2−2 1 0−1 −1 1]

To Find A4: Multiply both sides by A in (1), we getA4−A3+5 A2−A=0

A4=A3−5 A2+A

A4=[ −5 −12 10−10 −23 16−13 −29 17]−5[

−1 −2 00 1 −42 6 −9 ]+[1 2 −2

2 5 −43 7 −5 ]

¿ [ −5 −12 10−10 −23 16−13 −29 17]+[ 5 10 0

0 −5 20−10 −30 45]+[1 2 −2

2 5 −43 7 −5 ]

88

¿ [ 1 0 8−8 −23 32−20 −52 57]

Verify Cayley –Hamilton therorem . Also find A−1∧A4 , ifA=[1 2 3

2 −1 43 1 −1].Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0

s1=1−1−1=−1

s2=|1 22 −1|+|−1 4

1 −1|+|1 33 −1|

¿(−1−4 )+(1−4)+(−1−9)

¿−5−3−10=−18

s3=|A|

¿1(−3)−2(−2−12)+3(2+3)

¿−3+28+15=40

The characteristic equation of A isλ3+ λ2−18 λ−40=0 By Cayley Hamilton Theorem,A3+A2−18 A−40 I=0−−−−−−(1)Verification:

A2=[1 2 32 −1 43 1 −1][

1 2 32 −1 43 1 −1]

89

=[14 3 812 9 −22 4 14 ]

A3=[14 3 812 9 −22 4 14 ] [1 2 3

2 −1 43 1 −1]=[44 33 46

24 13 7452 14 8 ]

A3+A2−18 A−40 I=[44 33 4624 13 7452 14 8 ]+[14 3 8

12 9 −22 4 14 ]−18 [1 2 3

2 −1 43 1 −1]−40[

1 0 00 1 00 0 1]

=[0 0 00 0 00 0 0 ]

To find A−1: Multiply both sides by A−1 in (1), we getA−1= 1

40[ A2+A−18 I ]

A−1= 140 [[14 3 8

12 9 −22 4 14 ]+[1 2 3

2 −1 43 1 −1]+[−18 0 0

0 −18 00 0 −18]]

= 140 [−3 5 11

14 −10 25 5 −5]

To Find A4: Multiply both sides by A in (1), we getA4+A3−18 A2−40 A=0

A4=−A3+18 A2+40 A

A4=−[44 33 4624 13 7452 −14 8 ]+18[14 3 8

12 9 −22 4 14 ]+40[1 2 3

2 −1 43 1 −1]

¿ [−44 −33 −46−24 −13 −74−52 14 −8 ]+[252 54 144

216 162 −3636 72 252 ]+[ 40 80 120

80 −40 160120 40 −40]

90

¿ [248 101 218272 109 50104 126 204]

Verify Cayley –Hamilton therorem . Also find A−1∧A4 , ifA=[ 2 −1 2

−1 2 −11 −1 2 ].


s2=| 2 −1−1 2 |+| 2 −1

−1 2 |+|2 21 2|

¿(4−1)+(4−1)+(4−2)

=3+3+2=8s3=|A|

¿2(4−1)+1(−2+1)+2(1−2)

¿6−1−2=3The characteristic equation of A isλ3−6 λ2+8 λ−3=0 By Cayley Hamilton Theorem,A3−6 A2+8 A−3 I=0−−−−−−(1)Verification:

A2=[ 2 −1 2−1 2 −11 −1 2 ][ 2 −1 2

−1 2 −11 −1 2 ]

91

=[ 7 −6 9−5 6 −65 −5 7 ]

A3=[ 2 −1 2−1 2 −11 −1 2 ][ 7 −6 9

−5 6 −65 −5 7 ]

=[ 14+5+10 −12−6−10 18+6+14−7−10−5 6+12+5 −9−12−77+5+10 −6−6−10 9+6+14 ]

=[ 29 −28 38−22 23 −2822 −22 29 ]

A3−6 A2+8 A−3 I=[ 29 −28 38−22 23 −2822 −22 29 ]−6[ 7 −6 9

−5 6 −65 −5 7 ]+8 [ 2 −1 2

−1 2 −11 −1 2 ]−3 [1 0 0

0 1 00 0 1]

¿(0 0 00 0 00 0 0)

To find A−1: Multiply both sides by A−1 in (1), we getA−1=1

3[A2−6 A+8 I ]

A−1=13 [ [ 7 −6 9

−5 6 −65 −5 7 ]−6 [ 2 −1 2

−1 2 −11 −1 2 ]+8[1 0 0

0 1 00 0 1 ]]

=13 [ 3 0 −31 2 0

−1 1 3 ]To find A4: Multiply both sides by A in (1), we get

A4−6 A3+8 A2−3 A=0

92

A4=6 A3−8 A2+3 A ¿6 [6 A2−8 A+3 I ]-8A2+3 A ¿36 A2−48 A+18 I−8 A2+3 A ¿28 A2−45 A+18 I

A4=28 [ 7 −6 9−5 6 −65 −5 7 ]-45[ 2 −1 2

−1 2 −11 −1 2 ]+[18 0 0

0 18 00 0 18]

¿ [ 196 −168 252−140 168 −168140 −140 196 ]−[ 90 −45 90

−45 90 −4545 −45 90 ]+[18 0 0

0 18 00 0 18]

¿ [ 124 −123 162−95 96 −12395 −95 124 ]

Verify Cayley –Hamilton therorem . Also find A−1∧A4 , ifA=[1 0 −2

2 2 40 0 2 ].


s2=|1 02 2|+|2 4

0 2|+|1 −20 2 |

=(2-0)+(4-0)+(2-0)=2+4+2=8

s3=|A|

=1(4-0)+0-2(0-0) =493

The characteristic equation of A isλ3−5 λ2+8 λ−4=0 By Cayley Hamilton Theorem,A3−5 A2+8 A−4 I=0−−−−−−(1)Verification:

A2=[1 0 −22 2 40 0 2 ] [1 0 −2

2 2 40 0 2 ]

=[1+0+0 0+0+0 −2+0−42+4+0 0+4+0 −4+8+80+0+0 0+0+0 0+0+4 ]

=[1 0 −66 4 120 0 4 ]

A3=A2. A

=[1 0 −66 4 120 0 4 ] [1 0 −2

2 2 40 0 2 ]

=[1+0+0 0+0+0 −2+0−126+8+0 0+8+0 −12+16+240+0+0 0+0+0 0+0+8 ]

=[ 1 0 −1414 8 280 0 8 ]

A3−5 A2+8 A−4 I=[ 1 0 −1414 8 280 0 8 ]−5[1 0 −6

6 4 120 0 4 ]+8 [1 0 −2

2 2 40 0 2 ]−4 [1 0 0

0 1 00 0 1]

=[0 0 00 0 00 0 0 ]

To find A−1: Multiply both sides by A−1 in (1), we get94

A−1=14

[A3−5 A2+8 A ]

A−1= 14 [[ 1 0 −1414 8 280 0 8 ]−5[1 0 −6

6 4 120 0 4 ]+8 [1 0 −2

2 2 40 0 2 ]]

=14 [ [ 1 0 −1414 8 280 0 8 ]+[ −5 0 30

−30 −20 −600 0 −20]+[ 8 0 −16

16 16 320 0 16 ]]

=14 [ 4 0 00 4 00 0 4 ]

=ITo find A4: Multiply both sides by A in (1), we getA4−5 A3+8 A2−4 A=0

A4=5 A3−8 A2+4 A

A4=[ 5 0 −7070 40 1400 0 40 ]+[ −8 0 48

−48 −32 −960 0 −32]+[4 0 −8

8 8 160 0 8 ]

¿ [ 1 0 −3030 16 600 0 16 ]

Verify Cayley –Hamilton theorem . Also find A−1∧A4 , ifA=[−1 0 3

8 1 −7−3 0 8 ].

Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0s1=−1+1+8=8

95

s2=|−1 08 1|+|1 −7

0 8 |+|−1 3−3 8|

=−1+8+(−8+9)

=8+1−1=8 s3=|A|

¿−1(8−0)+0+3(0+3)=1The characteristic equation of A isλ3−8 λ2+8 λ−1=0 By Cayley Hamilton Theorem,

A3−8 A2+8 A−I=0−−−−−−(1)Verification:A2=[−1 0 3

8 1 −7−3 0 8 ][−1 0 3

8 1 −7−3 0 8 ]

=[ 1+0−9 0+0+0 −3+0+24−8+8+21 0+1+0 24−7−563+0−24 0+0+0 −9+0+64 ]

=[ −8 0 2121 1 −39

−21 0 55 ]A3=[−1 0 3

8 1 −7−3 0 8 ][ −8 0 21

21 1 −39−21 0 55 ]

=[ 8+0−63 0+0+0 −21+0+165−64+21+147 0+1+0 168−39−38524+0−168 0+0+0 −63+0+440 ]

=[ −55 0 144104 1 −256

−144 0 377 ]96

A3−8 A2+8 A−I=[ −55 0 144104 1 −256

−144 0 377 ]-8[ −8 0 2121 1 −39

−21 0 55 ]+8[−1 0 38 1 −7

−3 0 8 ]-[1 0 00 1 00 0 1 ]

=[0 0 00 0 00 0 0 ]

To find A−1: Multiply both sides by A−1 in (1), we getA−1=A2−8 A+8 I

A−1=[ −8 0 2121 1 −39

−21 0 55 ]+[ 8 0 24−64 −8 5624 0 −64 ]+[8 0 0

0 8 00 0 8]

=[ 8 0 −3−43 10 173 0 −1]

To find A4: Multiply both sides by A in (1), we getA4−8 A3+8 A2−A=0

A4=8 A3−8 A2+A

¿8 [8 A2−8 A+ I ]-8A2+A ¿56 A2−63 A+8 I

¿56[ −8 0 2121 1 −39

−21 0 55 ]−63 [−1 0 38 1 −7

−3 0 8 ]+[8 0 00 8 00 0 8 ]

¿56[ −8 0 2121 1 −39

−21 0 55 ]+[ 63 0 −189−504 −63 441189 0 −504 ]+[8 0 0

0 8 00 0 8 ]

=[−377 0 987672 1 2625

−987 0 2584 ]97

Verify Cayley –Hamilton therorem . Also find A−1∧A4 ,ifA=[ 7 2 −2

−6 −1 26 2 −1].

The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0s1=7−1−1=5

s2=| 7 2−6 −1|+|−1 2

2 −1|+|7 −26 −1|

¿(−7+12)+(1−4)+(−7+12)

=5−3+5=7 s3=|A|

=7(−3)−2(6−12)−2(−12+6) ¿−21+12+12=3

The characteristic equation of A isλ3−5 λ2+7 λ−3=0 By Cayley Hamilton Theorem,A3−5 A2+7 A−3 I=0−−−−−−(1)Verification:

A2=[ 7 2 −2−6 −1 26 2 −1] [

7 2 −2−6 −1 26 2 −1]

=[49−12−12 14−2−4 −14+4+2−42+6+12 −12+1+4 12−2−242−12−6 12−2−2 −12+4+1]

98

=[ 25 8 −8−24 −7 824 8 −7]

A3=A . A2

=[ 7 2 −2−6 −1 26 2 −1] [

25 8 −8−24 −7 824 8 −7 ]

=[ 175−48−48 56−14−16 −56+16+14−150+24+48 −48+7+16 48−8−14150−48−24 48−14−8 −48+16+7 ]

=[ 79 26 −26−78 −25 2678 26 −25 ]

A3−5 A2+7 A−3 I=[ 79 26 −26−78 −25 2678 26 −25]−¿ 5[ 25 8 −8

−24 −7 824 8 −7]+7[ 7 2 −2

−6 −1 26 2 −1]-3

[1 0 00 1 00 0 1 ] =[0 0 0

0 0 00 0 0 ]

To find A−1: Multiply both sides by A−1 in (1), we getA−1=1

3[A2−5 A+7 I ]

A−1=13 [ [ 25 8 −8

−24 −7 824 8 −7 ]+[−35 −10 10

30 5 −10−30 −10 5 ]+[7 0 0

0 7 00 0 7]]

A−1=13 [−3 −2 26 5 −2

−6 −2 5 ] To Find A4: Multiply both sides by A in (1), we get

99

A4−5 A3+7 A2−3 A=0 A4=5 A3−7 A2+3 A

A4=5[ 79 26 −26−78 −25 2678 26 −25]−7[

25 8 −8−24 −7 824 8 −7]+3 [

7 2 −2−6 −1 26 2 −1]

¿ [ 395 130 −130−390 −125 130390 130 −125]+[−175 −56 56

168 49 −56−168 −56 49 ]+[ 21 6 −6

−18 −3 618 6 −3 ]

=[ 241 80 −80

−240 −79 80240 80 −79]

Orthogonal reduction of a symmetric matrix to Diagonal form. Orthogonal matrix

A square matrix A is said to be an orthogonal matrix, if AT A=A AT=I

Diagonalisation of a matrixThe process of finding a matrix M such that M−1 AM=D, where D is the diagonal matrix, is called as Diagonalisation of A.

Problems100

Show that A =( cosθ sinθ−sinθ cosθ) is orthogonal.

Solution:( i .e ) AT A=A AT=I

A AT=( cosθ sinθ−sinθ cosθ)(cosθ −sinθ

sinθ cosθ )

=( cos2θ+sin 2θ −cosθsinθ+sinθcosθ−sinθco sθ+cosθsinθ sin2θ+cos2θ )

=(1 00 1)

=ISimilarly, AT A=I

∴ AT A=A AT=IHence A is orthogonal matrix. Construct a Diagonalised matrix by an orthogonal transformation of

A=[ 6 −2 2−2 3 −12 −1 3 ].Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0

s1=6+3+3=12

s2=| 6 −2−2 3 |+| 3 −1

−1 3 |+|6 22 3|

¿(18−4)+(9−1)+(18−4)

=14+8+14=36101

s3=|A|

¿6(9−1)+2(−6+2)+2(2−6)

¿48−8−8=32The characteristic equation of A isλ3−12 λ2+36 λ−32=0If λ=2 ,then (2 )3−12 (2 )2+36 (2 )−32=0

∴ λ=2 is a root.Using synthetic division, 2 1 -12 36 -32 0 2 -20 32

1 -10 16 0λ=2∧ λ2−10 λ+16=0

λ=2,2,8


⇒ (6−λ −2 2−2 3− λ −12 −1 3−λ)(

x1x2x3)=(000)−−−−−−(1)

Case (i): When λ=8 in (1), (6−8 −2 2

−2 3−8 −12 −1 3−8)(

x1x2x3)=(000)

102

(−2 −2 2−2 −5 −12 −1 −5)(

x1x2x3)=(000)

−¿2x1−2 x2+2x3=0−¿2x1−5 x2−x3=0 2x1−x2−5 x3=0Solving first two equations using cross rule method

x12+10

=x2

−4−2=

x310−4

x112

=x2

−6=x36

x12

=x2−1

=x31

X1=( 2−11 )Case(ii): When λ=2 in (1),

(6−2 −2 2−2 3−2 −12 −1 3−2)(

x1x2x3)=(000)

( 4 −2 2−2 1 −12 1 1 )(x1x2x3)=(000)

4 x1−2x2+2x3=0−2 x1+x2−x3=0

2 x1−x2+x3=0

The above equations represents the same equation 2 x1−x2+x3=0Choosing arbitrary values for x1 , let x1=0

103

x2=x3

X2=(011)Let X3=( lmn )

X1T X 3=0⟹2l−m+n=0 ………………..(1)

X2T X3=0⟹0 l+m+n=0 …………………(2)

Solving (1) and (2) we getl

−1−1= −m2−0

= n2−0

l−2

=−m2

=n2

X3=(−2−22 )=( 11−1)

Now clearly anytwo eigen vectorsare pairwise orthogonal.i.e X1 X2T=X2 X3T=X , 3X1T=0∴ The Modal Matrix M=( 2 0 1

−1 1 11 1 −1)

To Prove : NTAN=D(8,2,2)To find Normalised matrix

104

N=[2√6

0 1√3

−1√6

1√2

1√3

1√6

1√2

−1√3

]NT=[

2√6

−1√6

1√6

0 1√2

1√2

1√3

1√3

1√3

]To find AN

AN=[ 6 −2 2−2 3 −12 −1 3 ][

2√6

0 1√3

−1√6

1√2

1√3

1√6

1√2

−1√3

] ¿ [

12+2+2√6

0−2+2√2

6−2−2√3

−4−3−1√6

0+3−1√2

−2+3+1√3

4+1+3√6

0−1+3√2

2−1−3√3

]¿ [16√6

0 2√3

−8√6

2√2

2√3

8√6

2√2

−2√3

]Calculate D=N T AN

105

D=[2√6

−1√6

1√6

0 1√2

1√2

1√3

1√3

1√3

] [16√6

0 2√3

−8√6

2√2

2√3

8√6

2√2

−2√3

] =[

32+8+86

0−2+2√12

4−2−2√18

0−8+8√12

0+2+22

0+2−2√6

16−8−8√18

0+2−2√6

2+2+23

] =[8 0 0

0 2 00 0 2 ]=D(8,2,2)

Construct a Diagonalised matrix by an orthogonal transformation of A=[ 10 −2 −5

−2 2 3−5 3 5 ].

Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0s1=10+2+5=¿17s2=|10 −2

−2 2 |+|2 33 5|+|10 −5

−5 5 | =(20−4¿+(10−9)+¿25) =16+1+25=42 s3=|A|

106

=10(10−9¿+2(−10+15)−5¿+10) =10(1)+2(5¿−5¿) =10+10−¿20=0

The characteristic equation of A isλ3−17 λ2+42 λ=0λ(λ2−17 λ+42¿=0

λ ( λ−3 ) ( λ+14 )=0

λ=0,3,14


⇒ (10−λ −2 −5−2 2−λ 3−5 3 5− λ)(

x1x2x3)=(000)−−−−−−(1)

Case (i): When λ=0∈(1 ) ,

(10−0 −2 −5−2 2−0 3−5 3 5−0)(

x1x2x3)=(000)

( 10 −2 −5−2 2 3−5 3 5 )( x1x2x3)=(000)10 x1−2x2−5 x3=0

−2 x1+2 x2+3 x3=0

−5 x1+3−2 x3=0


— 6+10=

−x230−10

=x3

20−4

107

x14

=x2

−20=x316

x11

=x2−5

=x34

X1=( 1−54 )Case (ii): Whenλ=3∈(1 ) ,

(10−3 −2 −5−2 2−3 3−5 3 5−3)(

x1x2x3)=(000)

( 7 −2 −5−2 −1 3−5 3 2 )( x1x2x3)=(000)7 x1−2 x2−5x3=0

−2 x1−x2+3 x3=0

−5 x1+3+2x3=0


— 6−5=

x221−10

=x3

−7−4

x1−11

=x2

−11=

x3−11

x11

=x21

=x31

X2=(111)Case (iii): Whenλ=14∈(1 ) ,

108

(10−14 −2 −5−2 2−14 3−5 3 5−14)(

x1x2x3)=(000)

(−4 −2 −5−2 −12 3−5 3 −9)(

x1x2x3)=(000)

−4 x1−2 x2−5 x3=0

−2 x1−12x2+3 x3=0

−5 x1+3 x2−9 x3=0


— 6−60=

−x2−12−10

=x3

48−4

x1−66

=x222

=x344

x1−3

=x21

=x32

X3=(−312 )Now clearly anytwo eigen vectorsare pairwise orthogonal. i.e X1 X2T=X2 X3T=X3 X1T=0 ∴ The Modal Matrix M=( 1 1 −3

−5 1 14 1 2 )

To Prove : NT AN=D(0,3,14)To find Normalised matrix

109

N=[1

√421√3

−3√14

−5√42

1√3

1√14

4√42

1√3

2√14

]NT=[

1√42

−5√42

4√42

1√3

1√3

1√3

−3√14

1√14

2√14

]To find AN

AN=( 10 −2 −5−2 2 3−5 3 5 ) [

1√42

1√3

−3√14

−5√42

1√3

1√14

4√42

1√3

2√14

] =[

10+10−20√42

10−2−5√3

−30−2−10√14

−2−10+12√42

−2+2+3√3

6+2+6√14

−5−15+20√42

−5+3+5√3

15+3+10√14

]=[03√3

−42√14

0 3√3

14√14

0 3√3

28√14

]Calculate D=N T AN

D=[1

√42−5√42

4√42

1√3

1√3

1√3

−3√14

1√14

2√14

][03√3

−42√14

0 3√3

14√14

0 3√3

28√14

]110

=[03−5+12

√126−42−70+112

√5880 3+3+3

3−42+14+28

√420 −9+3+6

√42126+14+56

14]

=[0 0 00 3 00 0 14 ]=D(0,3,14)

Construct a Diagonalised matrix by an orthogonal transformation of A=[ 8 −6 2

−6 7 −42 −4 3 ].

Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0s1=8+7+3=¿18s2=| 8 −6

−6 7 |+| 7 −4−4 3 |+|8 2

2 3| =(56−36¿+(21−16)+(24−4)

=20+5+20=45s3=|A|

=8(21−16¿+6(−18+8)+2 (24−14) =8(5¿+6 (−10)+2¿) =40−60+20=0

111

The characteristic equation of A isλ3−18 λ2+45 λ=0⟹ λ(λ2−18 λ+45¿=0

⟹ λ ( λ−15 ) ( λ−3 )=0

⟹λ=0,3,15


⇒ (8−λ −6 2−6 7−λ −42 −4 3−λ)(

x1x2x3)=(000)−−−−−−(1)


(8−0 −6 2−6 7−0 −42 −4 3−0)(

x1x2x3)=(000)

( 8 −6 2−6 7 ¿

−4¿3¿4)( x1x2x3)=(000)8 x1−6 x2+2x3=0

−6 x1+7 x2−4 x3=0

2 x1+3 x2+3x3=0


24−14=

−x2−32+12

=x3

56−36

x110

=x220

=x320

x11

=x22

=x32

112

X1=(122)Case (ii): Whenλ=3∈(1 ) ,

(8−3 −6 2−6 7−3 ¿

−4¿3−3¿4)(x1x2x3)=(000)( 5 −6 2−6 4 ¿

−4¿0¿4)( x1x2x3)=(000)5 x1−6 x2+2 x3=0

−6 x1+4 x2−4 x3=0

2 x1+3 x2+0 x3=0Solving last two equations using cross rule methodx1

24−8=

−x2−20+12

=x3

20−36

x116

=x28

=x3

−16

x12

=x21

=x3−2

X2=( 21−2) Case (iii): Whenλ=15 ,

(8−15 −6 2−6 7−15 ¿

−4¿3−15¿4 )(x1x2x3)=(000)(−7 −6 2−6 −8 ¿

−4¿−12¿4)(x1x2x3)=(000)−7 x1−6 x2+2x3=0

113

−6 x1−8 x2−4 x3=0

2 x1−4 x2−12 x3=0solve last two equations using cross rule methodx1

24+16=

−x228+12

=x3

56−36

x140

=x2

−40=x320

x12

=x2−2

=x31

X3=( 2−21 )Now clearly any two eigen vectorsare pairwise orthogonal. i.e X1 X2T=X2 X3T=X3 X1T=0 ∴ The Modal Matrix M=(1 2 2

2 1 22 −2 1)


N=[13

23

23

23

13

−23

23

−23

13

]=13 [1 2 22 1 −22 −2 1 ]

NT=13 [1 2 22 1 −22 −2 1 ]114

To find ANAN=( 8 −6 2

−6 7 ¿−4¿3¿4) 13 [1 2 2

2 1 −22 −2 1 ]

=13 [ 8−12+4 16−6−4 16+12+12−6+14−8 −12+7+8 −12−14−42−8+6 4−4−6 4+8+3 ]

¿13 [0 6 300 3 −300 −6 15 ]

Calculate D=N T AND=

13 [1 2 22 1 −22 −2 1 ] 13 [0 6 30

0 3 −300 −6 15 ]

D=19 [0 0 00 27 00 0 135 ]=[0 0 0

0 3 00 0 15]=D(0,3,15)

Construct a Diagonalised matrix by an orthogonal transformation of A=[ 1 −1 −1

−1 1 −1−1 −1 1 ].

Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0s1=1+1+1=¿3

s2=| 1 −1−1 1 |+| 1 −1

−1 1 |+| 1 −1−1 1 |

115

=(1−1¿+(1−1)+¿)=0s3=|A|

=1(1−1¿+1(−1−1)−1(1+1)=−4

The characteristic equation of A isλ3−3 λ2+4=0If λ=−1 ,then −¿1−¿3+4=0 ∴ λ=−1 is a root.Using synthetic division, -1 1 -3 0 4 0 -1 4 -4

1 -4 4 0λ=−1∧ λ2−4 λ+4=0

λ=−1 , ( λ−2 )2=0

λ=−1,2,2 ,


⇒ (2−λ −2 21 1−λ 11 3 −1−λ)(

x1x2x3)=(000)−−−−−−(1)

Case (i): When λ=−1∈(1 ) ,

(1+1 −1 −1−1 1+1 −1−1 −1 1+1)(

x1x2x3)=(000)

116

( 2 −1 −1−1 2 −1−1 −1 2 )(x1x2x3)=(000)2 x1−x2−x3=0

−x1+2 x2−x3=0

−x1−x2+2x3=0

Solving first two equations using cross rule methodx11+2

=−x2

−2−1=

x34−1

x13

=x23

=x33

x11

=x21

=x31

X1=(111)Case (ii): Whenλ=2∈(1 ) ,

(1−2 −1 −1−1 1−2 −1−1 −1 1−2)(

x1x2x3)=(000)

(−1 −1 −1−1 −1 −1−1 −1 −1)(

x1x2x3)=(000)

−x1−x2−x3=0

−x1−x2−x3=0

−x1−x2−x3=0

The above equations represents the same equation −x1−x2−x3=0

117

Choosing arbitrary values for x1 , let x1=0x2=−x3

x21

=x3−1

X2=( 01−1)To find the third eigen vector orthogonal to X1and X2,since the matrix A is symmetric.Let X3=( lmn )

X1T X 3=0⟹ l+m+n=0 ………………..(1)

X2T X3=0⟹0 l+m−n=0 …………………(2)

Solving (1) and (2) we getx1

−1−1=

−x2−1−0

=x31−0

x1−2

=x21

=x31

X3=(−211 )Now clearly anytwo eigen vectorsare pairwise orthogonal. i.e X1 X2T=X2 X3T=X3 X1T=0

118

∴ The Modal Matrix M=(1 0 −21 1 11 −1 1 )

To Prove : NT AN=D(−¿1,2,2) To find Normalised matrix

N=[1√3

0 −2√6

1√3

1√2

1√6

1√3

−1√2

1√6

]NT=[

1√3

1√3

1√3

0 1√2

−1√2

−2√6

1√6

1√6

]To find AN

AN=[ 1 −1 −1−1 1 −1−1 −1 1 ][

1√3

0 −2√6

1√3

1√2

1√6

1√3

−1√2

1√6

]=[

−1√3

0 −4√6

−1√3

2√2

2√6

−1√3

−2√2

2√6

]119

Calculate D=N T AND=[

1√3

1√3

1√3

0 1√2

−1√2

−2√6

1√6

1√6

] [−1√3

0 −4√6

−1√3

2√2

2√6

−1√3

−2√2

2√6

] =[−1 0 0

0 2 00 0 2]=D(−1,2,2)

Construct a Diagonalised matrix by an orthogonal transformation of A=[1 0 0

0 3 −10 −1 3 ].


s2=| 3 −1−1 3 |+|1 0

0 3|+|1 00 3|

=9−1+3+3=14s3=|A|

¿1(9−1)−0+0=8The characteristic equation of A isλ3−7 λ2+14 λ−8=0If λ=1 ,then1-7+14-8=0

∴ λ=1 is a root.Using synthetic division120

1 1 -7 14 -8 0 1 -6 8

1 -6 8 0λ=1 , λ2−6 λ+8=0

λ=1 , λ=2,4

λ=1,2,4


⇒ (1−λ 0 00 3−λ −10 −1 3−λ)(

x1x2x3)=(000)−−−−−−(1)


(1−1 0 00 3−1 −10 −1 3−1)(

x1x2x3)=(000)

(0 0 00 2 −10 −1 2 )(x1x2x3)=(000)0x1+0x2+0 x3=00 x1+2x2−x3=0

0 x1−x2+2x3=0

Solving last two equations using cross rule methodx14−1

=−x20−0

=x30−0

121

x13

=x20

=x30

x11

=x20

=x30

X1=(100)Case (ii): Whenλ=2∈(1 ) ,

(1−2 0 00 3−2 −10 −1 3−2)(

x1x2x3)=(000)

(−1 0 00 1 −10 −1 1 )(x1x2x3)=(000)−x1+0 x2+0 x3=0

0 x1+x2−x3=0

0 x1−x2+x3=0

Solving first two equations using cross rule methodx10−0

=−x21−0

=x3

−1−0

x10

=x2−1

=x3

−1

x10

=x21

=x31

X2=(011)Case (iii) : Whenλ=4∈(1 ) ,

122

(1−4 0 00 3−4 −10 −1 3−4)(

x1x2x3)=(000)

(−3 0 00 −1 −10 −1 −1)(

x1x2x3)=(000)

−3 x1+0x2+0 x3=0

0 x1−x2−x3=0

0 x1−x2−x3=0


=−x23−0

=x33−0

x10

=x2−3

=x33

x10

=x2−1

=x31

X3=( 0−11 ) Now clearly any two eigen vectors are pairwise orthogonal. i.e X1 X2T=X2 X3T=X3 X1T=0 ∴ The Modal Matrix M=(1 0 0

0 1 −10 1 1 )


123

N=[1 0 0

0 1√2

−1√2

0 1√2

1√2

]NT=[

1 0 0

0 1√2

1√2

0 −1√2

1√2

] To Find ANAN=[1 0 0

0 3 −10 −1 3 ][

1 0 0

0 1√2

−1√2

0 1√2

1√2

] =[

1+0+0 0+0+0 0+0+0

0+0+0 0+ 3√2

− 1√2

0− 3√2

− 1√2

0+0+0 0− 1√2

+ 3√2

0+ 1√2

+ 3√2

] =[1 0 0

0 √2 −2√20 √2 2√2 ]

Calculate D=N T AND=[

1 0 0

0 1√2

1√2

0 −1√2

1√2

][1 0 00 √2 −2√20 √2 2√2 ]=[1 0 0

0 2 00 0 4]=D(1,2,4)

124

Construct a Diagonalised matrix by an orthogonal transformation of A=[ 7 −2 0

−2 6 −20 −2 5 ].


s2=| 7 −2−2 6 |+| 6 −2

−2 5 |+|7 00 5|

=(42−4 ¿+(30−4 )+(35−0)

=38+26+35=99s3=|A|

=7(30−4¿+2(−10−0)+0 =7(26)+2¿)=162

Thecharacteristic equationof A is λ3−18 λ2+99 λ−162=0If λ=3 ,then(3¿¿3−18 (3 )2+99 (3 )−162=0

∴ λ=3 is a root.Using synthetic division 3 1 -18 99 -162 0 3 -45 162

125

1 -15 54 0λ=3 , λ2−15 λ+54=0

λ=3 , λ=9,6

λ=3,9,6


⇒ (7−λ −2 0−2 6−λ −20 −2 5−λ)(

x1x2x3)=(000)−−−−−−(1)


(7−3 −2 0−2 6−3 −20 −2 5−3)(

x1x2x3)=(000)

( 4 −2 0−2 3 −20 −2 2 )(x1x2x3)=(000)4x1−2 x2+0 x3=0

−2 x1+3 x2−2 x3=0

0 x1−2 x2+2 x3=0


=−x2

−8−0=

x312−4

x14

=x28

=x38

x11

=x22

=x32

126

X1=(122)Case (ii): Whenλ=6∈(1 ) ,

(7−6 −2 0−2 6−6 −20 −2 5−6)(

x1x2x3)=(000)

( 1 −2 0−2 0 −20 −2 −1)(

x1x2x3)=(000)

x1−2 x2+0 x3=0

−2 x1+0 x2−2 x3=0

0 x1−2 x2−x3=0


=−x2

−2−0=

x30−4

x14

=x22

=x3

−4

x14

=x22

=x3

−4

X2=( 21−2)Case (iii): Whenλ=9∈(1 ) ,

(7−9 −2 0−2 6−9 −20 −2 5−9)(

x1x2x3)=(000)

(−2 −2 0−2 −3 −20 −2 −4)(

x1x2x3)=(000)

127

−¿2x1−2 x2+0 x3=0−2 x1−3x2−2x3=0

0 x1−2 x2−4 x3=0


=−x24−0

=x36−4

x14

=x2

−4=x32

x12

=x2−2

=x31

X3=( 2−21 ) Now clearly any two eigen vectorsare pairwise orthogonal. i.e X1 X2T=X2 X3T=X3 X1T=0 ∴ The Modal Matrix M=(1 2 2

2 1 −22 −2 1 )


N=13 [1 2 22 1 −22 −2 1 ]

NT=13 [1 2 22 1 −22 −2 1 ]128

To Find ANAN=[ 7 −2 0

−2 6 −20 −2 5 ]13 [1 2 2

2 1 −22 −2 1 ]

=13 [ 7−4+0 14−2+0 14+4+0

−2+12−4 −4+6+4 −4−12−20−4+10 0−2−10 0+4+5 ]

=13 [3 12 186 6 −186 −12 9 ]=[1 4 6

2 2 −62 −4 3 ]

Calculate D=N T AND=

13 [1 4 62 2 −62 −4 3 ] [1 2 2

2 1 −22 −2 1 ]

¿13 [1+4+4 4+4−8 6−12+62+2−4 8+2+8 12−6−62−4+2 8−4−4 12+12+3]

= 13 [9 0 00 18 00 0 27]=[3 0 0

0 6 00 0 9 ]=D(3,6,9)

Verify that the eigenvectors of the real symmetric matrix A=[ 2 0 −10 2 0

−1 0 2 ] are orthogonal in pairs.Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0

s1=2+2+2=6

129

s2=|2 00 2|+| 2 −1

−1 2 |+|2 00 2|

=4+3+4=11s3=|A|

=2(4−0¿+0(1)−1¿) =8−2=6

The characteristic equation of A isλ3−6 λ2+11 λ−6=0If λ=1 ,then 1-6+11-6=0∴ λ=1 is a root.Using synthetic division 1 1 -6 11 -6

0 1 -5 6 1 -5 6 0

λ=1 , λ2−5 λ+6=0

λ=1 , λ=2,3

λ=1,2,3


⇒ (2−λ 0 −10 2−λ 0

−1 0 2−λ)(x1x2x3)=(000)−−−−−−(1)


130

(2−1 0 −10 2−1 0

−1 0 2−1)(x1x2x3)=(000)

( 1 0 −10 1 0

−1 0 1 )(x1x2x3)=(000)x1+0x2−x3=0

0 x1+x2+0 x3=0

−x1+0 x2+x3=0

Solve first two equations using cross rule methodx10+1

=−x20−0

=x31−0

x11

=x20

=x31

X1=(101) Case(ii): Whenλ=2∈(1 ) ,

(2−2 0 −10 2−2 0

−1 0 2−2)(x1x2x3)=(000)

( 0 0 −10 0 0

−1 0 0 )(x1x2x3)=(000)0 x1+0 x2−x3=0

0 x1+0x2+0 x3=0

−x1+0 x2+0 x3=0

131

Solving first and last equations using cross rule methodx10+0

=−x20−1

=x30+0

x10

=x21

=x30

X2=(010)Case (iii) : Whenλ=3∈(1 ) ,

(2−3 0 −10 2−3 0

−1 0 2−3)(x1x2x3)=(000)

(−1 0 −10 −1 0

−1 0 −1)(x1x2x3)=(000)

−x1+0 x2− x3=0

0 x1−x2+0x3=0

−x1+0 x2− x3=0


=−x20+0

=x31−0

x1−1

=x20

=x31

X3=(−101 )

132

∴Eigen vectors of A are X1=(101) , X2=(010) , X3=(−101 )

To prove the orthogonality condition :X1 X2

T=0, X2 X3T=0, X3 X1T=0 i.e X1 X2T=(101) (0 1 0 )=0X2 X3

T=(010) (−1 0 1 )=0X3 X1

T=(−101 ) (1 0 1 )=0

i.e X1 X2T=X2 X3T=X3 X1T=0∴The eigenvectors are pairwise orthogonal.

Verify that the eigenvectors of the real symmetric matrix A=[ 3 −1 1−1 5 −11 −1 3 ] are orthogonal in pairs.


s2=| 5 −1−1 3 |+|3 1

1 3|+| 3 −1−1 5 |

133

=14+8+14=36s3=|A|

=3(14¿+1 (−3+1)+1(1−5) =42−2−4=36

The characteristic equation of A isλ3−11 λ2+36 λ−36=0If λ=2 ,then(2¿¿3−11 (2 )2+36 (2 )−36=0

∴ λ=2 is a root.Using synthetic division 2 1 -18 36 -36 0 2 -18 36

1 -9 18 0λ=2 , λ2−9 λ+18=0

λ=2 , λ=3,6

λ=2,3,6


⇒ (3−λ −1 1−1 5−λ −11 −1 3−λ)(

x1x2x3)=(000)−−−−−−(1)


134

(3−2 −1 1−1 5−2 −11 −1 3−2)(

x1x2x3)=(000)

( 1 −1 1−1 3 −11 −1 1 )(x1x2x3)=(000)x1−x2+x3=0

−x1+3x2−x3=0

x1−x2+x3=0


=− x2

−1+1=x33−1

x1−2

=x20

=x32

X1=(−202 )=( 10−1) Case (ii): whenλ=3∈(1 ) ,

(3−3 −1 1−1 5−3 −11 −1 3−3)(

x1x2x3)=(000)

( 0 −1 1−1 2 −11 −1 0 )(x1x2x3)=(000)0 x1−x2+x3=0

−x1+2 x2−x3=0

x1−x2+0x3=0Solving first two equations using cross rule method135

x11−2

=−x20+1

=x30−1

x1−1

=x2−1

=x3−1

x11

=x21

=x31

X2=(111)Case (iii): Whenλ=6∈(1 ) ,

(3−6 −1 1−1 5−6 −11 −1 3−6)(

x1x2x3)=(000)

(−3 −1 1−1 −1 −11 −1 −3)(

x1x2x3)=(000)

−3 x1−x2+x3=0

−x1−x2−x3=0

x1−x2−3 x3=0Solving first two equations using cross rule methodx11+1

=−x23+1

=x33−1

x12

=x2

−4=x32

x11

=x2−2

=x31

X3=( 1−21 )

136

∴Eigen vectors of A are X1=( 10−1) , X2=(111) , X3=( 1−21 )

To prove the orthogonality condition :X1 X2

T=0, X2 X3T=0, X3 X1T=0 i.e X1 X2T=( 10−1) (1 1 1 )=0X2 X3

T=(111) (1 −2 1 )=0 X3 X1T=( 1−21 ) (1 0 −1 )=0


Verify that the eigenvectors of the real symmetric matrix A=( 1 −1 0−1 2 10 1 1) are orthogonal in pairs. Solution:The characteristic equation of A isλ3−s1 λ2+s2 λ−s3=0s1=1+2+1=4

s2=|2 11 1|+|1 0

0 1|+| 1 −1−1 2 |

137

=(2−1¿+(1−0)+¿1) =1+1+1=3s3=|A|

=1(2−1¿+1 ¿)+0=0The characteristic equation of A isλ3−4 λ2+3 λ−0=0

λ (λ2−4 λ+3)=0

λ=0 , λ2−4 λ+3=0λ=0,1,3


⇒ (1−λ −1 0−1 2−λ 10 1 1−λ)(

x1x2x3)=(000)−−−−−−(1)


(1−0 −1 0−1 2−0 10 1 1−0)(

x1x2x3)=(000)

( 1 −1 0−1 2 10 1 1)(

x1x2x3)=(000)

x1−x2+0x3=0

−x1+2x2+x3=0

0 x1+x2+x3=0

Solving first two equations using cross rule method138

x1−1−0

=−x21−0

=x32−1

x1−1

=x2−1

=x31

x11

=x21

=x3−1

X1=( 11−1)Case (ii): Whenλ=1∈(1 ) ,

(1−1 −1 0−1 2−1 10 1 1−1)(

x1x2x3)=(000)

( 0 −1 0−1 1 10 1 0)(

x1x2x3)=(000)

0 x1−x2+0x3=0

−x1+ x2+x3=0

0 x1+x2+0 x3=0


−1−0=

−x20−0

=x30−1

x1−1

=x20

=x3−1

x11

=x20

=x31

X2=(101)Case (iii): Whenλ=3∈(1 ) ,

139

(1−3 −1 0−1 2−3 10 1 1−3)(

x1x2x3)=(000)

(−2 −1 0−1 −1 10 1 −2)(

x1x2x3)=(000)

−2 x1−x2+0 x3=0

−x1−x2+ x3=0

0 x1+x2−2x3=0


−1−0=

−x2−2−0

=x32−1

x1−1

=x22

=x31

X3=(−121 )∴Eigen vectors of A are

X1=( 11−1) , X2=(101) , X3=(−121 )To prove the orthogonality condition :X1 X2

T=0, X2 X3T=0, X3 X1T=0 i.e X1 X2T=( 11−1) (1 0 1 )=0X2 X3

T=(101) (−1 2 1 )=0140

X3 X1T=(−121 ) (1 1 −1 )=0


141

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