Unit 1: Part B Solutions (Matrices and Calculus)

Question 1

Find the eigenvalues and eigenvectors of a matrix

201
020
102

Solution

Step 1: Find the characteristic equation.
Let A be the given matrix. The characteristic equation is |A - λI| = 0.

|A - λI| =

2-λ01
02-λ0
102-λ
= 0

Expanding along the second row (R2):
(2-λ) * [(2-λ)(2-λ) - (1)(1)] = 0
(2-λ) * [(2-λ)2 - 1] = 0
(2-λ) * [λ2 - 4λ + 4 - 1] = 0
(2-λ) * (λ2 - 4λ + 3) = 0
(2-λ) * (λ - 1) * (λ - 3) = 0

The eigenvalues (λ) are 1, 2, 3.

Step 2: Find the eigenvectors for each eigenvalue.
The eigenvector X = [x1, x2, x3]T is found by solving (A - λI)X = 0.

Case 1: λ = 1
(A - 1I)X = 0

101
010
101
x1
x2
x3
 =
0
0
0

This gives the equations:
x1 + x3 = 0 → x1 = -x3
x2 = 0
Let x3 = k. Then x1 = -k, x2 = 0.
The eigenvector X1 is k * [-1, 0, 1]T.

Case 2: λ = 2
(A - 2I)X = 0

001
000
100
x1
x2
x3
 =
0
0
0

This gives the equations:
x3 = 0
x1 = 0
x2 is a free variable. Let x2 = k.
The eigenvector X2 is k * [0, 1, 0]T.

Case 3: λ = 3
(A - 3I)X = 0

-101
0-10
10-1
x1
x2
x3
 =
0
0
0

This gives the equations:
-x1 + x3 = 0 → x1 = x3
-x2 = 0 → x2 = 0
Let x3 = k. Then x1 = k, x2 = 0.
The eigenvector X3 is k * [1, 0, 1]T.

Eigenvalues: λ = 1, 2, 3
Eigenvectors: X1 = [-1, 0, 1]T, X2 = [0, 1, 0]T, X3 = [1, 0, 1]T

Question 2

Using Cayley-Hamilton theorem, find A4 if A =

123
2-14
311

Solution

Step 1: Find the characteristic equation of A.
The characteristic equation is λ3 - S1λ2 + S2λ - S3 = 0.

  • S1 = Sum of diagonal elements = 1 + (-1) + 1 = 1
  • S2 = Sum of minors of diagonal elements = ((-1)(1) - (4)(1)) + ((1)(1) - (3)(3)) + ((1)(-1) - (2)(2)) = (-1 - 4) + (1 - 9) + (-1 - 4) = -5 - 8 - 5 = -18
  • S3 = Determinant of A = |A|
    = 1((-1)(1) - (4)(1)) - 2((2)(1) - (4)(3)) + 3((2)(1) - (-1)(3))
    = 1(-1 - 4) - 2(2 - 12) + 3(2 + 3)
    = 1(-5) - 2(-10) + 3(5) = -5 + 20 + 15 = 30
The characteristic equation is: λ3 - λ2 - 18λ - 30 = 0.

Step 2: Apply the Cayley-Hamilton Theorem.
By the theorem, the matrix A satisfies its own characteristic equation:
A3 - A2 - 18A - 30I = 0
From this, we get: A3 = A2 + 18A + 30I.

Step 3: Express A4 in terms of lower powers.
Multiply the equation from Step 2 by A:
A4 = A(A3) = A(A2 + 18A + 30I)
A4 = A3 + 18A2 + 30A
Now, substitute the expression for A3 back into this equation:
A4 = (A2 + 18A + 30I) + 18A2 + 30A
A4 = 19A2 + 48A + 30I

Step 4: Calculate A2.
A2 = A × A =

123
2-14
311
123
2-14
311
 =
1+4+92-2+33+8+3
2-2+124+1+46-4+4
3+2+36-1+19+4+1
 =
14314
1296
8614

Step 5: Calculate A4.
A4 = 19 *

14314
1296
8614
+ 48 *
123
2-14
311
 + 30 *
100
010
001

A4 =

26657266
228171114
152114266
+
4896144
96-48192
1444848
 +
3000
0300
0030

A4 =

266+48+3057+96+0266+144+0
228+96+0171-48+30114+192+0
152+144+0114+48+0266+48+30

A4 =
344153410
324153306
296162344

Question 3

Reduce the quadratic form 3x12 + 3x22 + 3x32 + 2x1x2 + 2x1x3 - 2x2x3 to canonical form through an orthogonal transformation. Also find its nature, index, signature.

Solution

Step 1: Write the matrix of the quadratic form.
The matrix A is constructed with diagonal elements as coefficients of squared terms and off-diagonal elements as half the coefficients of cross-product terms.
A =

311
13-1
1-13

Step 2: Find the eigenvalues of A.
The characteristic equation is λ3 - S1λ2 + S2λ - S3 = 0.

  • S1 = 3 + 3 + 3 = 9
  • S2 = (9 - 1) + (9 - 1) + (9 - 1) = 8 + 8 + 8 = 24
  • S3 = |A| = 3(9 - 1) - 1(3 - (-1)) + 1(-1 - 3) = 3(8) - 1(4) - 1(4) = 24 - 4 - 4 = 16
Equation: λ3 - 9λ2 + 24λ - 16 = 0.
By inspection (testing factors of 16):
  • If λ = 1: 1 - 9 + 24 - 16 = 0. So (λ - 1) is a factor.
  • If λ = 4: (4)3 - 9(4)2 + 24(4) - 16 = 64 - 9(16) + 96 - 16 = 64 - 144 + 96 - 16 = 160 - 160 = 0. So (λ - 4) is a factor.
Since the sum of eigenvalues is S1 = 9, we have 1 + 4 + λ3 = 9, which gives λ3 = 4.
The eigenvalues are λ1 = 1, λ2 = 4, λ3 = 4.

Step 3: Write the canonical form.
The canonical form is C = λ1y12 + λ2y22 + λ3y32.

Canonical Form: y12 + 4y22 + 4y32

Step 4: Determine nature, index, and signature.

  • Nature: Since all eigenvalues (1, 4, 4) are positive, the nature is Positive Definite.
  • Index (p): The number of positive eigenvalues = 3.
  • Signature (s): (Number of positive eigenvalues) - (Number of negative eigenvalues) = 3 - 0 = 3.

Question 4

Reduce the quadratic form 6x2 + 3y2 + 3z2 - 4xy - 2yz + 4xz into a canonical form through an orthogonal reduction.

Solution

Step 1: Write the matrix of the quadratic form.
A =

6-22
-23-1
2-13

Step 2: Find the eigenvalues of A.
The characteristic equation is λ3 - S1λ2 + S2λ - S3 = 0.

  • S1 = 6 + 3 + 3 = 12
  • S2 = (9 - 1) + (18 - 4) + (18 - 4) = 8 + 14 + 14 = 36
  • S3 = |A| = 6(9 - 1) - (-2)(-6 - (-2)) + 2(2 - 6) = 6(8) + 2(-4) + 2(-4) = 48 - 8 - 8 = 32
Equation: λ3 - 12λ2 + 36λ - 32 = 0.
By inspection (testing factors of 32):
  • If λ = 2: (2)3 - 12(2)2 + 36(2) - 32 = 8 - 12(4) + 72 - 32 = 8 - 48 + 72 - 32 = 80 - 80 = 0. So (λ - 2) is a factor.
  • If λ = 8: (8)3 - 12(8)2 + 36(8) - 32 = 512 - 12(64) + 288 - 32 = 512 - 768 + 288 - 32 = 800 - 800 = 0. So (λ - 8) is a factor.
Sum of eigenvalues is S1 = 12. We have 2 + 8 + λ3 = 12, which gives λ3 = 2.
The eigenvalues are λ1 = 2, λ2 = 2, λ3 = 8.

Step 3: Write the canonical form.
The canonical form is C = λ1y12 + λ2y22 + λ3y32, where [x, y, z]T = NY and N is the normalized modal matrix.

Canonical Form: 2y12 + 2y22 + 8y32

Question 5

Find the eigenvalues and eigenvectors of a matrix

220
211
-72-3

Solution

Step 1: Find the characteristic equation.
Let A be the given matrix. The characteristic equation is λ3 - S1λ2 + S2λ - S3 = 0.

  • S1 = 2 + 1 + (-3) = 0
  • S2 = ((1)(-3) - (1)(2)) + ((2)(-3) - (0)(-7)) + ((2)(1) - (2)(2)) = (-3 - 2) + (-6 - 0) + (2 - 4) = -5 - 6 - 2 = -13
  • S3 = |A| = 2((1)(-3) - (1)(2)) - 2((2)(-3) - (1)(-7)) + 0
    = 2(-3 - 2) - 2(-6 + 7) = 2(-5) - 2(1) = -10 - 2 = -12
The characteristic equation is: λ3 - (0)λ2 + (-13)λ - (-12) = 0 → λ3 - 13λ + 12 = 0.

By inspection (testing factors of 12):

  • If λ = 1: 1 - 13 + 12 = 0. So (λ - 1) is a factor.
  • If λ = 3: (3)3 - 13(3) + 12 = 27 - 39 + 12 = 0. So (λ - 3) is a factor.
Sum of eigenvalues is S1 = 0. We have 1 + 3 + λ3 = 0, which gives λ3 = -4.
The eigenvalues (λ) are 1, 3, -4.

Step 2: Find the eigenvectors for each eigenvalue.
Solve (A - λI)X = 0.

Case 1: λ = 1
(A - 1I)X = 0

120
201
-72-4
x1
x2
x3
 = 0
From R1: x1 + 2x2 = 0 → x1 = -2x2
From R2: 2x1 + x3 = 0 → x3 = -2x1 = -2(-2x2) = 4x2
Let x2 = k. Then x1 = -2k, x3 = 4k.
The eigenvector X1 is k * [-2, 1, 4]T.

Case 2: λ = 3
(A - 3I)X = 0

-120
2-21
-72-6
x1
x2
x3
 = 0
From R1: -x1 + 2x2 = 0 → x1 = 2x2
From R2: 2x1 - 2x2 + x3 = 0 → 2(2x2) - 2x2 + x3 = 0 → 4x2 - 2x2 + x3 = 0 → 2x2 + x3 = 0 → x3 = -2x2
Let x2 = k. Then x1 = 2k, x3 = -2k.
The eigenvector X2 is k * [2, 1, -2]T.

Case 3: λ = -4
(A - (-4)I)X = 0

620
251
-721
x1
x2
x3
 = 0
From R1: 6x1 + 2x2 = 0 → x2 = -3x1
From R2: 2x1 + 5x2 + x3 = 0 → 2x1 + 5(-3x1) + x3 = 0 → 2x1 - 15x1 + x3 = 0 → -13x1 + x3 = 0 → x3 = 13x1
Let x1 = k. Then x2 = -3k, x3 = 13k.
The eigenvector X3 is k * [1, -3, 13]T.

Eigenvalues: λ = 1, 3, -4
Eigenvectors: X1 = [-2, 1, 4]T, X2 = [2, 1, -2]T, X3 = [1, -3, 13]T

Question 6

Using Cayley-Hamilton theorem find its inverse of the matrix A =

123
245
356

Solution

Step 1: Find the characteristic equation of A.
The characteristic equation is λ3 - S1λ2 + S2λ - S3 = 0.

  • S1 = 1 + 4 + 6 = 11
  • S2 = (24 - 25) + (6 - 9) + (4 - 4) = -1 - 3 + 0 = -4
  • S3 = |A| = 1(24 - 25) - 2(12 - 15) + 3(10 - 12) = 1(-1) - 2(-3) + 3(-2) = -1 + 6 - 6 = -1
The characteristic equation is: λ3 - 11λ2 - 4λ + 1 = 0.

Step 2: Apply the Cayley-Hamilton Theorem.
A3 - 11A2 - 4A + 1I = 0
(Note: Since S3 = |A| = -1 ≠ 0, the inverse exists.)

Step 3: Find A-1.
Multiply the equation by A-1:
A-1(A3 - 11A2 - 4A + I) = A-1(0)
A2 - 11A - 4I + A-1 = 0
A-1 = -A2 + 11A + 4I

Step 4: Calculate A2.
A2 =

123
245
356
123
245
356
 =
1+4+92+8+153+10+18
2+8+154+16+256+20+30
3+10+186+20+309+25+36
 =
142531
254556
315670

Step 5: Calculate A-1.
A-1 = -

142531
254556
315670
+ 11 *
123
245
356
 + 4 *
100
010
001

A-1 =

-14-25-31
-25-45-56
-31-56-70
+
112233
224455
335566
 +
400
040
004

A-1 =

-14+11+4-25+22+0-31+33+0
-25+22+0-45+44+4-56+55+0
-31+33+0-56+55+0-70+66+4

A-1 =
1-32
-33-1
2-10

Question 7

Diagonalise the matrix A =

8-62
-67-4
2-43
by Orthogonality transformation.

Solution

To diagonalize A, we find the diagonal matrix D = NTAN, where N is the normalized modal matrix (matrix of normalized eigenvectors).

Step 1: Find the eigenvalues of A.
λ3 - S1λ2 + S2λ - S3 = 0

  • S1 = 8 + 7 + 3 = 18
  • S2 = (21 - 16) + (24 - 4) + (56 - 36) = 5 + 20 + 20 = 45
  • S3 = |A| = 8(21 - 16) - (-6)(-18 - (-8)) + 2(24 - 14) = 8(5) + 6(-10) + 2(10) = 40 - 60 + 20 = 0
Equation: λ3 - 18λ2 + 45λ = 0 → λ(λ2 - 18λ + 45) = 0
λ(λ - 3)(λ - 15) = 0.
The eigenvalues are λ = 0, 3, 15.

Step 2: Find the eigenvectors.

Case 1: λ = 0
(A - 0I)X = 0

8-62
-67-4
2-43
x1
x2
x3
 = 0
Using cross-multiplication on R1 and R2: x1 / ((-6)(-4) - (2)(7)) = x2 / ((2)(-6) - (8)(-4)) = x3 / ((8)(7) - (-6)(-6))
x1 / (24 - 14) = x2 / (-12 + 32) = x3 / (56 - 36)
x1 / 10 = x2 / 20 = x3 / 20 → x1 / 1 = x2 / 2 = x3 / 2
X1 = [1, 2, 2]T.

Case 2: λ = 3
(A - 3I)X = 0

5-62
-64-4
2-40
x1
x2
x3
 = 0
From R3: 2x1 - 4x2 = 0 → x1 = 2x2
From R1: 5x1 - 6x2 + 2x3 = 0 → 5(2x2) - 6x2 + 2x3 = 0 → 10x2 - 6x2 + 2x3 = 0 → 4x2 + 2x3 = 0 → x3 = -2x2
Let x2 = 1. X2 = [2, 1, -2]T.

Case 3: λ = 15
(A - 15I)X = 0

-7-62
-6-8-4
2-4-12
x1
x2
x3
 = 0
Using R1 and a simplified R3 (x1 - 2x2 - 6x3 = 0): x1 / ((-6)(-6) - (2)(-2)) = x2 / ((2)(1) - (-7)(-6)) = x3 / ((-7)(-2) - (-6)(1))
x1 / (36 + 4) = x2 / (2 - 42) = x3 / (14 + 6)
x1 / 40 = x2 / -40 = x3 / 20 → x1 / 2 = x2 / -2 = x3 / 1
X3 = [2, -2, 1]T.

Step 3: Normalize the eigenvectors.
Length of X1 = √(12 + 22 + 22) = √9 = 3.
Length of X2 = √(22 + 12 + (-2)2) = √9 = 3.
Length of X3 = √(22 + (-2)2 + 12) = √9 = 3.

The normalized modal matrix N is: N =

1/32/32/3
2/31/3-2/3
2/3-2/31/3
= (1/3)
122
21-2
2-21

Step 4: Write the diagonalized matrix D.
The diagonalized matrix D is formed by the eigenvalues.

The diagonal matrix D = NTAN =
000
030
0015

Question 8

Reduce the quadratic form x12 + 2x22 + x32 - 2x1x2 + 2x2x3 to the canonical form through an orthogonal transformation and find its nature.

Solution

Step 1: Write the matrix of the quadratic form.
A =

1-10
-121
011

Step 2: Find the eigenvalues of A.
λ3 - S1λ2 + S2λ - S3 = 0

  • S1 = 1 + 2 + 1 = 4
  • S2 = (2 - 1) + (1 - 0) + (2 - 1) = 1 + 1 + 1 = 3
  • S3 = |A| = 1(2 - 1) - (-1)(-1 - 0) + 0 = 1(1) + 1(-1) = 0
Equation: λ3 - 4λ2 + 3λ = 0 → λ(λ2 - 4λ + 3) = 0
λ(λ - 1)(λ - 3) = 0.
The eigenvalues are λ = 0, 1, 3.

Step 3: Write the canonical form.
The canonical form is C = λ1y12 + λ2y22 + λ3y32.

Canonical Form: 0y12 + 1y22 + 3y32 = y22 + 3y32

Step 4: Determine the nature.
Since the eigenvalues are 0, 1, 3 (all non-negative, with one being zero), the nature is positive semi-definite.

Nature: Positive Semi-definite

Question 9

Obtain an orthogonal transformation which will transform the quadratic form Q = 2xy + 2yz + 2xz to canonical form.

Solution

This requires finding the normalized modal matrix N for the transformation X = NY.

Step 1: Write the matrix of the quadratic form.
A =

011
101
110

Step 2: Find the eigenvalues of A.
λ3 - S1λ2 + S2λ - S3 = 0

  • S1 = 0 + 0 + 0 = 0
  • S2 = (0 - 1) + (0 - 1) + (0 - 1) = -3
  • S3 = |A| = 0 - 1(0 - 1) + 1(1 - 0) = 1 + 1 = 2
Equation: λ3 - 3λ - 2 = 0.
By inspection:
  • If λ = 2: 8 - 6 - 2 = 0. So (λ - 2) is a factor.
  • If λ = -1: -1 + 3 - 2 = 0. So (λ + 1) is a factor.
Sum of eigenvalues is S1 = 0. We have 2 + (-1) + λ3 = 0, which gives λ3 = -1.
The eigenvalues are λ = 2, -1, -1.

Step 3: Find the eigenvectors.

Case 1: λ = 2
(A - 2I)X = 0 →

-211
1-21
11-2
x
y
z
 = 0
Cross-multiplying R1 and R2: x / (1 - (-2)) = y / (1 - (-2)) = z / (4 - 1) → x/3 = y/3 = z/3.
X1 = [1, 1, 1]T.

Case 2: λ = -1 (repeated root)
(A - (-1)I)X = 0 →

111
111
111
x
y
z
 = 0
This gives one equation: x + y + z = 0.
We must find two orthogonal vectors (X2, X3) that satisfy this.
Let X2 be a simple vector satisfying this: Let y = -1, x = 1, then z = 0.
X2 = [1, -1, 0]T.
Now find X3 = [l, m, n]T such that it is orthogonal to both X1 and X2.
  • X3 ⊥ X1 → l + m + n = 0
  • X3 ⊥ X2 → l - m + 0 = 0 → l = m
Substitute l = m into the first equation: l + l + n = 0 → 2l + n = 0 → n = -2l.
Let l = 1. Then m = 1 and n = -2.
X3 = [1, 1, -2]T.

Step 4: Normalize the eigenvectors.
l(X1) = √(12+12+12) = √3.
l(X2) = √(12+(-1)2+02) = √2.
l(X3) = √(12+12+(-2)2) = √6.

Step 5: Write the transformation.
The transformation is X = NY, where N is the normalized modal matrix.

The orthogonal transformation is X = NY:
x
y
z
 =
1/√31/√21/√6
1/√3-1/√21/√6
1/√30-2/√6
y1
y2
y3

This transforms Q into the canonical form 2y12 - y22 - y32.

Question 10

Diagonalise the symmetric matrix A =

204
060
402
by Orthogonality transformation.

Solution

Step 1: Find the eigenvalues of A.
The characteristic equation is |A - λI| = 0.

|A - λI| =

2-λ04
06-λ0
402-λ
= 0

Expanding along R2:
(6-λ) * [(2-λ)(2-λ) - (4)(4)] = 0
(6-λ) * [(2-λ)2 - 16] = 0
(6-λ) * [λ2 - 4λ + 4 - 16] = 0
(6-λ) * (λ2 - 4λ - 12) = 0
(6-λ) * (λ - 6) * (λ + 2) = 0

The eigenvalues (λ) are 6, 6, -2.

Step 2: Find the eigenvectors.

Case 1: λ = -2
(A - (-2)I)X = 0 →

404
080
404
x1
x2
x3
 = 0
From R1: 4x1 + 4x3 = 0 → x1 = -x3
From R2: 8x2 = 0 → x2 = 0
Let x3 = 1. X1 = [-1, 0, 1]T.

Case 2: λ = 6 (repeated root)
(A - 6I)X = 0 →

-404
000
40-4
x1
x2
x3
 = 0
This gives one equation: -4x1 + 4x3 = 0 → x1 = x3.
x2 is a free variable.
We need two orthogonal vectors (X2, X3) satisfying x1 = x3.
Let X2 be a simple vector: Let x2 = 1, x1 = 0. Then x3 = 0.
X2 = [0, 1, 0]T.
Let X3 be another vector: Let x2 = 0, x1 = 1. Then x3 = 1.
X3 = [1, 0, 1]T.
Check for orthogonality: X2 · X3 = 0(1) + 1(0) + 0(1) = 0. They are orthogonal.

Step 3: Normalize the eigenvectors.
l(X1) = √((-1)2 + 02 + 12) = √2.
l(X2) = √(02 + 12 + 02) = √1 = 1.
l(X3) = √(12 + 02 + 12) = √2.

Step 4: Write the normalized modal matrix N and diagonal matrix D.
N =

-1/√201/√2
010
1/√201/√2

The diagonal matrix D is NTAN, which is simply the matrix of eigenvalues.

The diagonal matrix D =
-200
060
006