Engineering Mathematics Test 1

Concept:

Determinant of a matrix is the product of its eigenvalues.

Determinant of a matrix is same as its transpose.

Determinant of a matrix is reciprocal to its inverse.

Calculation:

Eigenvalues of matrix A are 10 and y.

Determinant of matrix A = 10 × y

100 = 10 × y

∴ y = 10

 the other Eigen value is 10

Concept:

Methods to check Linearly dependent or Linearly Independent vectors:

Let x₁, x₂, x₃ ….. x_r are the n-vectors.

Consider A = [x₁, x₂, x₃…. x_r]_{n × r}

General Method:

1) of ρ(A) = number of vector (R), then Linearly Independent.

2) If ρ(A) < number of vector (r), then Linearly Dependent

Matrix method:

If A is a square matrix :

1) |A| ≠ 0, vectors are Linearly Independent.

2) If |A| = 0, vectors are Linearly Dependent.

Calculation:

For linearly dependent vectors, |A| = 0

⇒ \(A = \left[ {\begin{array}{*{20}{c}} 1&2&3\\ 4&\lambda &6\\ 3&4&5 \end{array}} \right]\)

⇒ |A| = 1(5λ - 24) – 2(20 - 18) + 3 (16 – 3λ)

⇒ |A| = 5λ – 24 – 4 + 48 – 9λ ⇒ λ = 5

Hence for the vectors to be linearly dependent λ = 5.

A square matrix ‘A’ such that A^k = 0 where ‘k’ is the least positive integer, is called the Nilpotent matrix of index ‘k’.

Calculation:

Here,\(A = \left[ {\begin{array}{*{20}{c}} 1&4\\ { - 1}&{ - 2} \end{array}} \right]\)

 \({A^2} = \left[ {\begin{array}{*{20}{c}} 1&4\\ { - 1}&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&4\\ { - 1}&{ - 2} \end{array}} \right]\) \( = \left[ {\begin{array}{*{20}{c}} 0&0\\ 0&0 \end{array}} \right] = 0\)

Hence the index of the given matrix is 2

Concept:

For real symmetric matrix, Eigen vector corresponding to different eigen values are Orthogonal.

Orthogonal vectors:

X₁ and X₂ are called Orthogonal if x₁ ⋅ x₂ = 0

Three or more vectors are Orthogonal if they are pair-wise orthogonal.

i.e. x₁ ⋅ x₂ = x₂ ⋅ x₃ = x₃ ⋅ x₁ = 0

Calculation:

For vectors \(\vec A = \left[ {\begin{array}{*{20}{c}} 2\\ 2\\ \lambda \end{array}} \right]\)and \(\vec B = \left[ {\begin{array}{*{20}{c}} 1\\ 0\\ 1 \end{array}} \right]\) to be orthogonal, \(\vec A \cdot \vec B = 0\) 

⇒ 2 × 1 + 2 × 0 + λ × 1 = 0 ⇒ λ = -2

Concept:

• The roots of characteristic equation |A - λI| = 0 are known as Eigenvalues of matrix A.
• To each Eigenvalue of λ if there exists a non-zero vector X such that AX = λX then X is called Eigenvector of matrix A corresponding to the Eigenvalue λ.
• Eigenvalue of adj (A) \(= \frac{{\left| A \right|}}{\lambda }\)

Calculation:

Characteristic Equation of A is |A - λI|  = 0

\(\left| {\begin{array}{*{20}{c}} {\left( {1 - \lambda } \right)}&2&2\\ 0&{\left( {2 - \lambda } \right)}&1\\ { - 1}&2&{\left( {2 - \lambda } \right)} \end{array}} \right| = 0\)

⇒ λ³ – 5λ² + 8λ - 4 = 0

⇒ (λ - 1) (λ - 2)² = 0

λ = 1, 2, 2 are the eigen values.

Eigenvalue of adj (A) \(= \frac{{\left| A \right|}}{\lambda }\)

\(\left| A \right| = \left| {\begin{array}{*{20}{c}} 1&2&2\\ 0&2&1\\ { - 1}&2&2 \end{array}} \right| = 1\left( {4 - 2} \right) + 2\left( { - 1} \right) + 2\left( 2 \right) = 4\)

Eigenvalues of adj(A) are \(\frac{4}{1},\frac{4}{2},\frac{4}{2} = 4,\;2,\;2\)

Concept:

Diagonalization of matrix:

If a square matrix Q of order n has n linearly independent Eigen vectors, then matrix P can be found such that \({P^{ - 1}}QP\) is a diagonal matrix.

Let Q be a square matrix of order 3.

Let λ1, λ2, and λ3 be Eigen values of matrix Q and \({X_1} = \left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{y_1}}\\ {{z_1}} \end{array}} \right],\;{X_2} = \left[ {\begin{array}{*{20}{c}} {{x_2}}\\ {{y_2}}\\ {{z_2}} \end{array}} \right],\;{X_3} = \left[ {\begin{array}{*{20}{c}} {{x_3}}\\ {{y_3}}\\ {{z_3}} \end{array}} \right]\) be the corresponding Eigen vectors.

Let denote the square matrix \(\left[ {\begin{array}{*{20}{c}} {{X_1}}&{{X_2}}&{{X_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{x_1}}&{{x_2}}&{{x_3}}\\ {{y_1}}&{{y_2}}&{{y_3}}\\ {{z_1}}&{{z_2}}&{{z_3}} \end{array}} \right]\) by P.

Now, the given matrix A can be diagonalized by \(D = {P^{ - 1}}QP\)

Or the matrix A can be represented by \(Q = PD{P^{ - 1}}\)

Where D is the diagonal matrix and it is represented by \(D = \left[ {\begin{array}{*{20}{c}} {{\lambda _1}}&0&0\\ 0&{{\lambda _2}}&0\\ 0&0&{{\lambda _3}} \end{array}} \right]\)

Properties of Eigen values:

The sum of Eigen values of a matrix A is equal to the trace of that matrix A

The product of Eigen values of a matrix A is equal to the determinant of that matrix A

Calculation:

\(P = \left[ {\begin{array}{*{20}{c}} 1&3\\ 0&1 \end{array}} \right]\) 

\({P^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}\\ 0&1 \end{array}} \right]\) 

The Eigen values of A = 2, -2

The Eigen vectors of \(A = \left[ {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right],\;\left[ {\begin{array}{*{20}{c}} 3\\ 1 \end{array}} \right]\) 

The Eigen values of A⁵ = 2⁵, -2⁵ = 32, -32

The Eigen vectors of \({A^5} = \left[ {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right],\;\left[ {\begin{array}{*{20}{c}} 3\\ 1 \end{array}} \right]\) 

A⁵ = P D P^-1

\( = \left[ {\begin{array}{*{20}{c}} 1&3\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {32}&0\\ 0&{ - 32} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 3}\\ 0&1 \end{array}} \right]\) 

\( = \left[ {\begin{array}{*{20}{c}} {32}&{ - 96}\\ 0&{ - 32} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 3}\\ 0&1 \end{array}} \right]\) 

\( = \left[ {\begin{array}{*{20}{c}} {32}&{ - 192}\\ 0&{ - 32} \end{array}} \right]\) 

Concept:

The number of solutions can be determined by finding out the rank of the Augmented matrix and rank of the Coefficient matrix.

• If rank(Augmented matrix) = rank(Coefficient matrix) = no. of variables then no of solutions = 1.
• If rank(Augmented matrix)  ≠ rank(Coefficient matrix) then no of solutions = 0.
• If rank(Augmented matrix) = rank(Coefficient matrix) < no. of variables, no of solutions = infinite.

Calculation:

Augmented matrix is \(A:B = \left[ {\begin{array}{*{20}{c}} 2&4&{5\;:6}\\ 2&5&{6\;:7}\\ 2&5&{\left( {a + 2} \right)\;:\left( {b + 2} \right)} \end{array}} \right]\)

Applying Elementary operations: R₂ → R₂ – R₁, R₃ →  R₃ – R₂

\(A\;:B = \left[ {\begin{array}{*{20}{c}} 2&4&{5\;:6}\\ 0&1&{1\;:1}\\ 0&0&{a - 4\;:b - 5} \end{array}} \right]\)

For unique solution, ρ[A] = ρ[A, B] = Number of variables

When, a – 4 ≠ 0 ⇒ a ≠ 4 and b can take any real value,

 ρ[A] = 3 and ρ[A, B] = 3 = Number of variables

Hence the system of solutions have unique solution when a ≠ 4 and b can take any real value.

\(adj\left( {adj\left( A \right)} \right)\; = \;{\left| A \right|^{n - 2}A}\)

n = 4 (since A_4x4)

A is upper triangular matrix

∴ product of diagonal elements = \(\left| A \right|\)

\(\left| A \right| = \left| {\begin{array}{*{20}{c}} 1&2&3&4\\ 0&5&6&7\\ 0&0&8&9\\ 0&0&0&{10} \end{array}} \right| = 1 \times 5 \times 8 \times 10 = 400\)

\({\left| A \right|^{n - 2}} = \;{\left| A \right|^{4 - 2}} = {400^2} = 160000\)

\(adj\left( {adj\left( A \right)} \right)\; = \;{\left| A \right|^2}A\)

\( = 160000 \times \left[ {\begin{array}{*{20}{c}} 1&2&3&4\\ 0&5&6&7\\ 0&0&8&9\\ 0&0&0&{10} \end{array}} \right]\)

\(= 16000 \times \left[ {\begin{array}{*{20}{c}} {10}&{20}&{30}&{40}\\ 0&50&{60}&{70}\\ 0&0&{80}&{90}\\ 0&0&0&{100} \end{array}} \right]\)

Note:

\(\left| A \right|\) is a determinant of matrix A

Characteristic equation of given matrix

\(\left| {\begin{array}{*{20}{c}} {3 - \lambda }&1&4\\ 0&{2\; - \;\lambda }&6\\ 0&0&{5 - \;\lambda } \end{array}} \right| = 0\)

\(\left( {3\; - \;\lambda } \right) \times \left( {2\; - \lambda } \right) \times \left( {5\; - \;\lambda } \right)\; = 0\)

Eigenvalues: 2, 3 and 5

Eigenvector corresponds to: λ = 2

\(\left[ {\begin{array}{*{20}{c}} 1&1&4\\ 0&0&6\\ 0&0&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right]\) = 0

\(x + y + 4z = 0\)

\(6z = 0\)

\(5z = 0\)

\(\therefore z = \;0\)

\(\therefore x = \; - y\)

if y = -2 then x = 2

Eigenvector: (2, -2, 0)

Eigenvector corresponds to: λ = 3

\(\left[ {\begin{array}{*{20}{c}} 0&1&4\\ 0&{ - 1}&6\\ 0&0&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right]\) = 0

\(y + 4z = 0\)

\(- y + \;6z = 0\)

\(2z = 0\)

\(\therefore z = \;0\)

\(\therefore y = \;0\)

x can take any value

Eigenvector: (5, 0, 0)

Eigen vector corresponds to: λ = 5

\(\left[ {\begin{array}{*{20}{c}} { - 2}&1&4\\ 0&{ - 3}&6\\ 0&0&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right]\) = 0

\(0 \times z = 0\)

∴ z can take any value

Let z = 1

\(- 3y + 6z = 0\)

\(\therefore y = \;2\)

\(- 2x + y + 4z = 0\)

\(\therefore x = \;3\)

Concept:

Gauss Seidel Method:

In Gauss Seidel method, the value of x calculated is used in next calculation putting other variable as 0.

x₁ + 2x₂ + 3x₃ = 5

Putting x₂ = 0, x₃ = 0 ⇒ x₁ = 5

2x₁ + 3x₂ + x₃ = 1

Putting x₁ = 5, x₃ = 0 ⇒ x₂ = -3

3x₁ + 2x₂ + x₃ = 3

Putting x₁ = 5, x₂ = -3 ⇒ x₃ = 3 – 3(5) – 2 (-3)

x₃ = 3 – 15 + 6

x₃ = -6