Engineering Mathematics Test 1

Concept:

A matrix is said to be symmetric matrix if A^T = A

A matrix is said to be orthogonal matrix if AA^T = I i.e. A^T = A^-1

Calculation:

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

So, A is a symmetric matrix.

It is given that. A is orthogonal matrix.

⇒ A^T = A^-1

⇒ A = A^-1

AX = B

⇒ X = A^-1B = AB

\( \Rightarrow X = \left[ {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 2 }}}&0&{\frac{1}{{\sqrt 2 }}}\\0&1&0\\{\frac{1}{{\sqrt 2 }}}&0&{\frac{{ - 1}}{{\sqrt 2 }}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0\\1\\{ - \sqrt 2 }\end{array}} \right]\) 

\( \Rightarrow \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}\\1\\{ + 1}\end{array}} \right]\) 

x + y + z = -1 + 1 + 1 = 1

\(A = \left( {\begin{array}{*{20}{c}} 1&1&1&1\\ 1&2&{ - 1}&\lambda \\ 5&{ 7}&1&{{\lambda ^2}} \end{array}} \right)\) 

R₂ → R₂ – R₁,

R₃ → R₃ – 5R₁

\(A = \left[ {\begin{array}{*{20}{c}} 1&1&1&1\\ 0&1&{ - 2}&{\lambda - 1}\\ 0&2&{ - 4}&{{\lambda ^2} - 5} \end{array}} \right]\) 

R₃ → R₃ – 2R₂,

\(A = \left[ {\begin{array}{*{20}{c}} 1&1&1&1\\ 0&1&{ - 2}&{\lambda - 1}\\ 0&0&0&{{\lambda ^2} - 5 - 2\lambda + 2} \end{array}} \right]\) 

Rank of the matrix A = number of non-zero rows

Given that Rank of A = 2.

So, the last row must be zero.

⇒ λ² – 2λ – 3 = 0

⇒ λ² – 3λ + λ – 3 = 0

⇒ λ(λ - 3) + 1(λ - 3) = 0

⇒ λ = -1, 3   

Concept:

Eigen vector (X) that corresponding to Eigen value (λ) satisfies the equation AX = λX.

Calculation:

\(A = \left[ {\begin{array}{*{20}{c}} 3&4\\ 2&t \end{array}} \right]\)

\(X = \left[ {\begin{array}{*{20}{c}} 4\\ { - 1} \end{array}} \right]\)

Let λ is an Eigen value,

AX = λX

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 3&4\\ 2&t \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4\\ { - 1} \end{array}} \right] = \lambda \left[ {\begin{array}{*{20}{c}} 4\\ { - 1} \end{array}} \right]\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 8\\ {8 - t} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {4\lambda }\\ { - \lambda } \end{array}} \right]\)

By comparing both the sides,

8 = 4λ ⇒ λ = 2

8 – t = -λ ⇒ 8 – t = -2 ⇒ t = 10 

Concept:

Consider the system of m linear equations

a11 x1 + a12 x2 + … + a1n xn = b1

a21 x1 + a22 x2 + … + a2n xn = b2

…

am1 x1 + am2 x2 + … + amn xn = bm

The above equations containing the n unknowns x1, x2, …, xn. To determine whether the above system of equations is consistent or not, we need to find the rank of following matrices.

\(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}\\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}\\ \ldots & \ldots & \ldots & \ldots \\ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}} \end{array}} \right]\) and \(\left[ {A{\rm{|}}B} \right] = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}&{{b_1}}\\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}&{{b_2}}\\ \ldots & \ldots & \ldots & \ldots & \ldots \\ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}}&{{b_m}} \end{array}} \right]\)

A is the coefficient matrix and [A|B] is called as augmented matrix of the given system of equations.

We can find the consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to rank of an augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution.

The rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to rank of an augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.

The rank of A = Rank of augmented matrix < n

(iii) If the rank of matrix A is not equal to rank of the augmented matrix, then the system is inconsistent, and it has no solution.

The rank of A ≠ Rank of an augmented matrix

Calculation:

The given system of equations can be represented in a matrix form as shown below.

\(A = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}\\ 0&5&3\\ 1&{ - 2}&{ - 5} \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right],B = \left[ {\begin{array}{*{20}{c}} { - 1}\\ { - 8}\\ 7 \end{array}} \right]\)

The Augmented matrix can be written by

\([A|B] = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}\\ 0&5&3\\ 1&{ - 2}&{ - 5} \end{array}{\rm{|}}\begin{array}{*{20}{c}} { - 1}\\ { - 8}\\ 7 \end{array}} \right]\)

R₃ → R₃ – R₁

\(= \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}\\ 0&5&3\\ 0&{ - 5}&{ - 3} \end{array}{\rm{|}}\begin{array}{*{20}{c}} { - 1}\\ { - 8}\\ 8 \end{array}} \right]\)

R₃ → R₃ + R₂

\(= \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}\\ 0&5&3\\ 0&0&0 \end{array}{\rm{|}}\begin{array}{*{20}{c}} { - 1}\\ { - 8}\\ 0 \end{array}} \right]\)

The rank of matrix A = 2

The rank of Augmented matrix = 2

Rank of A = Rank of augmented matrix = 2 < n = 3

\(A = \left[ {\begin{array}{*{20}{c}} {\cos x}&{ - \sin x}&1\\ {\sin x}&{\cos x}&1\\ {\cos \left( {x + y} \right)}&{ - \sin \left( {x + y} \right)}&0 \end{array}} \right]\)

Determinant value of A is

\(\left| A \right| = \left| {\begin{array}{*{20}{c}} {\cos x}&{ - \sin x}&1\\ {\sin x}&{\cos x}&1\\ {\cos \left( {x + y} \right)}&{ - \sin \left( {x + y} \right)}&0 \end{array}} \right|\)

= cos x (sin (x + y)) + sin x (- cos (x + y)) + (- sin x sin (x + y) – cos x cos (x + y))

= cos x sin (x + y) – sin x cos (x + y) – sin x sin (x + y) – cos x cos (x + y)

= cos x (sin (x + y) – cos (x + y)) – sin x (cos (x + y) + sin (x + y))

= cos x (sin x cos y + cos x sin y – cos x cos y + sin x sin y) – sin x (cos x cos y – sin x sin y + sin x cos y + cos x sin y)

= sin x cos x cos y + cos² x sin y – cos² x cos y + sin x sin y cos x – sin x cos x cos y + sin² x sin y – sin² x cos y – sin x cos x sin y

\( = \sin y - \cos y = \sqrt 2 \left( {\frac{1}{{\sqrt 2 }}\sin y - \frac{1}{{\sqrt 2 }}\cos y} \right) = \sqrt 2 \sin \left( {y - \frac{\pi }{4}} \right)\)

The value of sin x lies in the interval [-1, 1]

Therefore, determinant lies in the interval \(\left[ { - \sqrt 2 ,\;\sqrt 2 } \right]\)

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:

Consider the system of m linear equations

a11 x1 + a12 x2 + … + a1n xn = b1

a21 x1 + a22 x2 + … + a2n xn = b2

…

am1 x1 + am2 x2 + … + amn xn = bm

The above equations containing the n unknowns x1, x2, …, xn. To determine whether the above system of equations is consistent or not, we need to find the rank of following matrices.

\(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}\\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}\\ \ldots & \ldots & \ldots & \ldots \\ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}} \end{array}} \right]\) and \(\left[ {A{\rm{|}}B} \right] = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}&{{b_1}}\\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}&{{b_2}}\\ \ldots & \ldots & \ldots & \ldots & \ldots \\ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}}&{{b_m}} \end{array}} \right]\)

A is the coefficient matrix and [A|B] is called as augmented matrix of the given system of equations.

We can find the consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to rank of an augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution.

The rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to rank of an augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.

The rank of A = Rank of augmented matrix < n

(iii) If the rank of matrix A is not equal to rank of the augmented matrix, then the system is inconsistent, and it has no solution.

The rank of A ≠ Rank of an augmented matrix

Calculation:

The given system of equations can be represented in a matrix form as shown below.

\(A = \left[ {\begin{array}{*{20}{c}} 1&2&2\\ 1&a&3\\ 1&{11}&a \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right],B = \left[ {\begin{array}{*{20}{c}} 1\\ 3\\ b \end{array}} \right]\)

The Augmented matrix can be written by

\([A|B] = \left[ {\begin{array}{*{20}{c}} 1&2&2\\ 1&a&3\\ 1&{11}&a \end{array}{\rm{|}}\begin{array}{*{20}{c}} 1\\ 3\\ b \end{array}} \right]\)

R₃ → R₃ – R₁

R₂ → R₂ – R₁

\(= \left[ {\begin{array}{*{20}{c}} 1&2&2\\ 0&{a - 2}&1\\ 0&9&{a - 2} \end{array}{\rm{|}}\begin{array}{*{20}{c}} 1\\ 2\\ {b - 1} \end{array}} \right]\)

\({R_3} \to {R_3} - \left( {\frac{9}{{a - 2}}} \right){R_2}\)

\(= \left[ {\begin{array}{*{20}{c}} 1&2&2\\ 0&{a - 2}&1\\ 0&0&{\frac{{{{\left( {a - 2} \right)}^2} - 9}}{{a - 2}}} \end{array}{\rm{|}}\begin{array}{*{20}{c}} 1\\ 2\\ {\frac{{\left( {a - 2} \right)\left( {b - 1} \right) - 18}}{{a - 2}}} \end{array}} \right]\)

The system ho have a unique solution, \(\frac{{{{\left( {a - 2} \right)}^2} - 9}}{{a - 2}} \ne 0\)

⇒ (a - 2)² – 9 ≠ 0

⇒ a – 2 ≠ ±3

⇒ a ≠ -1 or 5

\(A = \left[ {\begin{array}{*{20}{c}}2&3\\x&4\end{array}} \right]\)

\(\left| {A - \lambda I} \right| = 0\)

\( \Rightarrow \left| {\begin{array}{*{20}{c}}{2 - \lambda }&3\\x&{4 - \lambda }\end{array}} \right| = 0\)

⇒ (2 – λ) (4 – λ) – 3x = 0

⇒ λ² – 6λ +8 – 3x = 0

Roots of the above equation are Eigen values and let the Eigen values be λ₁ and λ₂

From the properties of Eigen values,

Sum of the Eigen values, \({\lambda _1} + {\lambda _2} = - \frac{{\left( { - 6} \right)}}{1} = 6\)

Product of Eigen values, \({\lambda _1}{\lambda _2} = \frac{{\left( {8 - 3x} \right)}}{1} = 8 - 3x\)

For the Eigen values to be positive, λ₁λ₂ > 0

⇒ 8 – 3x > 0

⇒ x < 8/3

For roots to be real, \(\sqrt {{6^2} - 4\left( 1 \right)\left( {8 - 3x} \right)} \ge 0\)

⇒ 36 – 32 + 12x > 0

⇒ x > -1/3

The range of for which Eigen values of matrix are real and positive is: \( - \frac{1}{3} < x < \frac{8}{3}\)

Given that, \(A = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ i&{\frac{{ - 1 + i\sqrt 3 }}{2}}&0\\ 0&{1 + 2i}&{\frac{{ - 1 - i\sqrt 3 }}{2}} \end{array}} \right]\)

Let \(ω = \frac{{ - 1 -i\sqrt 3 }}{2}\)

\(\Rightarrow {ω ^2} = \frac{{ - 1 + i\sqrt 3 }}{2}\)

Now, \(A = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ i&ω &0\\ 0&{1 + 2i}&{{ω ^2}} \end{array}} \right]\)

Given matrix A is a lower triangular matrix. 

We know that the n values of a triangular matrix are diagonal elements.

1, ω, ω2   are cube roots of unity

1³ = 1

ω³ = 1

(ω²)³ =1

Hence the  Eigenvalues of A = 1, ω, ω²

If λ is an Eigenvalue of a matrix A, then λⁿ is the Eigenvalue of the matrix Aⁿ

Eigenvalue of A¹⁰² = (1)¹⁰², (ω)¹⁰², (ω²)¹⁰²

= 1, ω¹⁰², ω²⁰⁴

= 1, (ω³)³⁴, (ω³)⁶⁸

= 1, 1, 1.

We know that, a trace of a matrix = sum of eigenvalues.

Characteristic equation: |A – λI| = 0

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

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

⇒ (1 – λ) (-6 + 3λ – 2λ + λ² + 6) – 1(-15 – 5λ + 12) + 3(-5 + 4 – 2λ) = 0

⇒ λ³ = 0

According to Cayley Hamilton theorem every square matrix satisfies its characteristic polynomial.

⇒ A³ = 0

A¹⁴ + 3A – 2I = A¹² A² + 3A – 2I = 3A – 2I

\(= 3\left[ {\begin{array}{*{20}{c}}1&1&3\\5&2&6\\{ - 2}&{ - 1}&{ - 3}\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3&9\\{15}&4&{18}\\{ - 6}&{ - 3}&{ - 11}\end{array}} \right]\)