Engineering Mathematics Test 1

Concept:

det (AB) = det (A) × det (B)

If A is any square matrix of order n, we can form the matrix [A – λI], where I is the nth order unit matrix. The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.

The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

Properties of Eigenvalues:

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

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

Calculation:

\(M = \left[ {\begin{array}{*{20}{c}} 9&2&7&1\\ 0&7&2&1\\ 0&0&{11}&6\\ 0&0&{ - 5}&0 \end{array}} \right]\)

Characteristic equation: |M – λI| = 0

\( \Rightarrow \left| {\begin{array}{*{20}{c}} {9 - \lambda }&2&7&1\\ 0&{7 - \lambda }&2&1\\ 0&0&{11 - \lambda }&6\\ 0&0&{ - 5}&\lambda \end{array}} \right| = 0\)

⇒ (9 – λ) (7 – λ) (11λ – λ² + 30) = 0

⇒ λ = 5, 6, 7, 9

Eigen values of matrix M = 5, 6, 7, 9

Eigen values of (8I – M) = 3, 2, 1, -1

Determinant of (8I – M) = -6

det((8I – M)³) = -216

Concept:

Directional derivative of a function f along the vector \(\hat u \) is given by:

\(DD = \nabla f.\frac{{\vec u}}{{\left| u \right|}}\)

where \(\frac{{\vec u}}{{\left| u \right|}}={\hat{u}}\)

grad f or ∇ f is defined by the equation,

\(grad\;f = \nabla f = i\frac{{\partial f}}{{\partial x}} + j\frac{{\partial f}}{{\partial y}} + k\frac{{\partial f}}{{\partial z}}\)

Calculation:

z = y²e^2x

\(gradz = \nabla z = 2{y^2}{e^{2x}}\hat i + 2y{e^{2x}}\hat j\)

At (2, -1), \(\nabla z = 2{e^4}\hat i - 2{e^4}\hat j\)

Unit vector \(\vec b = \alpha \hat i + \beta \hat j\)

Directional derivative \( = \nabla z.\vec b = \left( {2{e^4}\hat i - 2{e^4}\hat j} \right).\alpha \hat i + \beta \hat j\)

= 2e⁴α – 2e⁴β

Given that, the directional derivative is zero.

⇒ 2e⁴α – 2e⁴β = 0

⇒ α = β

As b is a unit vector, \(\sqrt {{\alpha ^2} + {\beta ^2}} = 1\)

\( \Rightarrow \alpha = \beta = \frac{1}{{\sqrt 2 }}\)

\(\left| {\alpha + \beta } \right| = \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = \sqrt 2 \)

Concept:

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Calculation:

The given differential equation is, \(\frac{{{d^2}y}}{{d{x^2}}} + 5\frac{{dy}}{{dx}} + 6y = 0\)

D² + 5D + 6 = 0

Auxiliary equation: m² + 5m + 6

⇒ m = -2, -3

The solution is, y = C₁ e^-2x + C₂ e^-3x

y(0) = 2

⇒ C₁ + C₂ = 2

\(\frac{{dy}}{{dx}} = - 2{C_1}{e^{ - 2x}} - 3{C_2}{e^{ - 3x}}\)

\({\left( {\frac{{dy}}{{dx}}} \right)_{x = 0}} = 0\)

⇒ -2C₁ – 3C₂ = 0

By solving the above two equations, C₁ = 6 and C₂ = -4

Now, the solution is, y(x) = 6 e^-2x – 4 e^-3x

At x = 1, \(y = \frac{6}{{{e^2}}} - \frac{4}{{{e^3}}} = 0.61\)

Concept:

The probability density function of Poisson’s distribution is

\(p\left( x \right)=\frac{{{e}^{-\lambda }}\cdot {{\lambda }^{x}}}{x!}\)

Where

λ = mean = np

n = number of total outcomes

p = probability of success

Note: This distribution uses where the probability of success i.e. p is very small.

Calculation:

Poisson (2) represents that the distribution is Poisson with λ = 2.

Now, the distribution function \(= \frac{{{e^{ - 2}}{2^x}}}{{x!}}\)

\(E\left( {\frac{1}{{1 + x}}} \right) = \mathop \sum \limits_{x = 0}^\infty \frac{1}{{1 + x}}\left( {\frac{{{e^{ - 2}}{2^x}}}{{x!}}} \right)\)

\(= \mathop \sum \limits_{x = 0}^\infty \frac{{{e^{ - 2}}{2^x}}}{{\left( {x + 1} \right)!}}\)

\(= \frac{{{e^{ - 2}}}}{2}\mathop \sum \limits_{x = 0}^\infty \frac{{{2^{\left( {x + 1} \right)}}}}{{\left( {x + 1} \right)!}}\)

\(= \frac{{{e^{ - 2}}}}{2}\left( {\frac{2}{{1!}} + \frac{{{2^2}}}{{2!}} + \frac{{{2^3}}}{{3!}} + \ldots } \right)\)

\( = \frac{{{e^{ - 2}}}}{2}\left( {{e^2} - 1} \right)\)

\(= \frac{1}{2}\left( {1 - {e^{ - 2}}} \right)\)

1 – P(E^c) – P(F^c) = P(E) + P(F) – 1

= P(E) + P(F) – P (E ⋂ F) + P (E ⋂ F) – 1

= P (E U F) + P (E ⋂ F) – 1

Now, consider the option 1,

P (E ⋂ F) ≤ 1 – P(E^c) – P(F^c)

⇒ P (E ⋂ F) ≤ P (E U F) + P (E ⋂ F) – 1

⇒ P (E U F) ≥ 1

P (E U F) can not be greater than 1 and hence option 1 is not always true.

Number of distinguishable firecrackers = 10

Defective firecrackers = 4

Given that three firecrackers are drawn at random (without replacement) from the packet.

The probability that the first firecracker is defective = 4/10

The probability that the second firecracker is defective = 3/9

The probability that the third firecracker is defective = 2/8

Now, the probability that all three firecrackers are defective \(= \frac{4}{{10}} \times \frac{3}{9} \times \frac{2}{8} = \frac{1}{{30}}\)

The given distribution is uniform distribution i.e. the variable uniformly distributed between 0 to 10.

Y = X – [X]

For X = U (0, 1), Y > 0.25 when 1 > X > 0.25

For X = U (1, 2), Y > 0.25 when 2 > X > 1.25

Similarly, Y > 0.25 for 75% of area.

Hence the required probability = 0.75

Straight line from (0, 0) to (2, 4)

\(\frac{{x - 0}}{{2 - 0}} = \frac{{y - 0}}{{4 - 0}} = t\)

⇒ x = 2t, y = 4t

⇒ dx = 2dt, dy = 4dt

The limits of t are: 0 to 1

Now, the line integral is \(\mathop \smallint \limits_0^1 \left[ {2\left( {4t} \right)\hat i + \left( {2t} \right)\hat j} \right].\left( {\hat idx + \hat jdy} \right)\)

\(= \mathop \smallint \limits_0^1 8tdx + 2tdy\)

\(= \mathop \smallint \limits_0^1 16tdt + 8tdt\)

\(= \mathop \smallint \limits_0^1 24tdt = 12\)

\(\mathop \smallint \limits_{ - \pi }^\pi \left| x \right|\cos nxdx\)

The given function f(x) = |x| cos nx is even function. So, the given integral can be reduced to

\( = 2\mathop \smallint \limits_0^\pi \left| x \right|\cos nxdx\)

\( = 2\mathop \smallint \limits_0^\pi x\cos nxdx\)

\(\smallint x\cos nxdx = x\smallint \cos nxdx - \smallint \frac{{\sin nx}}{n}dx\)

\( = x\left( {\frac{{\sin nx}}{n}} \right) + \frac{1}{{{n^2}}}\left( {\cos nx} \right)\)

\( = 2\left[ {\frac{{x\sin nx}}{n} + \frac{{\cos nx}}{{{n^2}}}} \right]_0^\pi \)

\( = 2\left[ {\frac{{\cos n\pi - 1}}{{{n^2}}}} \right]\)

\( = \frac{{ - 2}}{{{n^2}}}\left( {1 - {{\left( { - 1} \right)}^n}} \right)\)

= 0 when n is even

\( = - \frac{4}{{{n^2}}}\) when n is odd

Concept:

A square matrix ‘A’ is said to be orthogonal if it follows the following condition.

AAT = ATA = I

Calculation:

\(M = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ a&b \end{array}} \right]\)

\({M^T} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&a\\ {\frac{1}{{\sqrt 2 }}}&b \end{array}} \right]\)

\(M{M^T} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ a&b \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&a\\ {\frac{1}{{\sqrt 2 }}}&b \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }}}\\ {\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }}}&{{a^2} + {b^2}} \end{array}} \right]\)

For orthogonal matrix, MM^T = I

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }}}\\ {\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }}}&{{a^2} + {b^2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

By comparing both the sides, we get

\(\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }} = 0 \Rightarrow a + b = 0\)

a² + b² = 1

\(\Rightarrow b = \pm \sqrt {1 - {a^2}}\)

Therefore, \(b = \sqrt {1 - {a^2}}\) is not always true.

⇒ (a + b)² – 2ab = 1

\(\Rightarrow ab = - \frac{1}{2}\)

By using the relation, a + b = 0 i.e. b = -a, we get

\( \Rightarrow a\left( { - a} \right) = - \frac{1}{2} \Rightarrow a = \pm \frac{1}{{\sqrt 2 }}\)

For \(a = \frac{1}{{\sqrt 2 }},b = - \frac{1}{{\sqrt 2 }}\), \(M = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{1}{{\sqrt 2 }}}&{\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right]\)

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

\( = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = {I_2}\)

For \(a = \frac{{ - 1}}{{\sqrt 2 }},b = \frac{1}{{\sqrt 2 }}\), \(M = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \right]\)

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

\( = \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&0 \end{array}} \right] \ne {I_2}\)

Therefore, M² = I₂ is not always true.

Concept:

Cauchy Integral formula

\({{f}^{n}}\left( a \right)=\frac{n!}{2\pi i}\underset{c}{\overset{{}}{\mathop \oint }}\,\frac{f\left( z \right)}{{{\left( z-a \right)}^{n+1}}}dz\)

Comparing we get

f(z) = e^2z

a = -1

n = 3

Calculations:

n = 3

f'''(z) = 8e^2z

f'''(z = -1) = 8e^-2

The given integral can be written as

\(8{{e}^{-2}}=\frac{3!}{2\pi i}\oint \frac{{{e}^{2z}}}{{{\left( z+1 \right)}^{4}}}dz\)

Concept:

If W = f(x, y, z) is a continuous function of n variables x, y, z, …, with continuous partial derivatives ∂W/∂x, ∂W/∂y, ∂W/∂z, … and if x, y, z, … are differentiable functions x = x(t), y = y(t), z = z(t), etc. of a variable t, then the total derivative of w with respect to t is given by

\(\frac{{dW}}{{dt}} = \frac{{\partial W}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial W}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial W}}{{\partial z}}\frac{{dz}}{{dt}}\)

Calculation:

f(x, y) = e^x sin y

\(\frac{{\partial f}}{{\partial x}} = {e^x}\sin y\)

\(\frac{{\partial f}}{{\partial y}} = {e^x}\cos y\)

At t = 0, \(\frac{{\partial f}}{{\partial x}} = e\sin 0 = 0\)

\(\frac{{\partial f}}{{\partial y}} = {e^1}\cos 0 = e\)

x = t³ + 1 and y = t⁴ + t

\(\frac{{dx}}{{dt}} = 3{t^2},\frac{{dy}}{{dt}} = 4{t^3} + 1\)

At t = 0, x = 1 and y = 0

At t = 0, \(\frac{{dx}}{{dt}} = 0,\frac{{dy}}{{dt}} = 1\)

\(\frac{{df}}{{dt}} = \frac{{\partial f}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dt}}\)

= 0 + e = e = 2.718

Concept:

The standard form of a first-order linear differential equation is,

Form 1: \(\frac{{dy}}{{dx}} + Py = Q\)

Where P and Q are the functions of x.

Integrating factor, \(IF = {e^{\smallint Pdx}}\)

Now, the solution for the above differential equation is,

\(y\left( {IF} \right) = \smallint IF.Qdx+C\)

Form 2: \(\frac{{dx}}{{dy}} + Px = Q\)

Where P and Q are the functions of y.

Integrating factor, \(IF = {e^{\smallint Pdy}}\)

Now, the solution for the above differential equation is,

\(x\left( {IF} \right) = \smallint IF.Qdy+C\)

Calculation:

\(\frac{{dy}}{{dx}} + 10y = f\left( x \right),x > 0\)

It is the first order linear differential equation.

Integrating factor, \(I.F. = {e^{\smallint 10dx}} = {e^{10x}}\)

The solution of the differential equation is,

\(y\left( x \right){e^{10x}} = \smallint {e^{10x}}f\left( x \right)dx + C\)

\( \Rightarrow y\left( x \right) = {e^{ - 10x}}\smallint {e^{10x}}f\left( x \right)dx + C{e^{ - 10x}}\)

\(\mathop {\lim }\limits_{x \to \infty } y\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{\smallint {e^{10x}}f\left( x \right)dx}}{{{e^{10x}}}} + \mathop {\lim }\limits_{x \to \infty } C{e^{ - 10x}}\)

\( = \mathop {\lim }\limits_{x \to \infty } \frac{{\smallint {e^{10x}}f\left( x \right)dx}}{{{e^{10x}}}}\)

By applying the L-Hospitals rule,

\( = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^{10x}}f\left( x \right)}}{{10{e^{10x}}}}\)

\( = \frac{1}{{10}}\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \frac{1}{{10}} = 0.1\)

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 an 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 the rank of an augmented matrix and it is equal to the number of unknowns, then 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 the 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 the rank of the augmented matrix, then the system is inconsistent, and it has no solution.

Rank of A ≠ Rank of augmented matrix

Calculation:

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

\(A = \left[ {\begin{array}{*{20}{c}} 1&1&5\\ 1&2&m\\ 1&2&4 \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right],B = \left[ {\begin{array}{*{20}{c}} 3\\ 5\\ k \end{array}} \right]\)

The Augmented matrix can be written by

\([A|B] = \left[ {\begin{array}{*{20}{c}} 1&1&5\\ 1&2&m\\ 1&2&4 \end{array}{\rm{|}}\begin{array}{*{20}{c}} 3\\ 5\\ k \end{array}} \right]\)

R₃ → R₃ – R₁, R₂ → R₂ – R₁

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

R₃ → R₃ – R₂

\(= \left[ {\begin{array}{*{20}{c}} 1&1&5\\ 0&1&{m - 5}\\ 0&0&{4 - m} \end{array}{\rm{|}}\begin{array}{*{20}{c}} 3\\ 2\\ {k - 5} \end{array}} \right]\)

The system is consistent if, the rank of matrix = rank of an augmented matrix.

Option 1: m ≠ 4

For any value of k, the rank (A) = rank (A|B) = 3

The system is consistent.

Option 2: k ≠ 5

For m = 4, the rank (A) ≠ rank (A|B)

The system is inconsistent.

Option 3: m = 4

For k ≠ 5, the rank (A) ≠ rank (A|B)

The system is inconsistent.

Option 4: k = 5

For m = 4, the rank (A) = rank (A|B) = 2

The system is consistent.

For m ≠ 4, the rank (A) = rank (A|B) = 3

The system is consistent.

Concept:

Binomial distribution:

Let p is the probability that an event will happen in a single trail (called the probability of success) and

q = 1 – p is the probability that an event will fail to happen (probability of failure)

The probability that the event will happen exactly r times in n trails (i.e. x successes and n – r failures will occur) is given by the probability function

\(f\left( x \right) = P\left( {X = r} \right) = {n_{{C_r}}}{p^r}{q^{n - r}}\)

where the random variable X denotes the number of successes in n trials and r = 0, 1, 2, … n

For Binomial distribution,

Mean = μ = np

Variance = σ2 = npq

Standard deviation = σ = √(npq)

Calculation:

Let V be the event that occurs in a trial with probability p. Mathematical expectation E of the number of trials to first occurrence of V in a sequence of trials is E = 1/p.

Similarly, the expected number of trials to get n^th success = n/p

Given that, the probability (p) = 1/5 = 0.2

The expected number of trials to get first success = 1/0.2 = 5

So, the expected number of failures preceding the first success is 4

The expected number of trials to get second success = 2/0.2 = 10

Expected number of successes in first 50 trials = np = 50 × 0.2 = 10

\(f\left( x \right) = \frac{{z + 1}}{{\left( {z - 3} \right)\left( {z - 4} \right)}}\)

At z = 2

Let z – 2 = t ⇒ z = t + 2

Now, the function becomes

\(f\left( t \right) = \frac{{t + 3}}{{\left( {t - 1} \right)\left( {t - 2} \right)}}\)

By using partial fractions, f(t) can be written as

\(\Rightarrow f\left( t \right) = \frac{{ - 4}}{{\left( {t - 1} \right)}} + \frac{5}{{\left( {t - 2} \right)}}\)

\(= \frac{{ - 4}}{{ - \left( {1 - t} \right)}} + \left( {\frac{5}{{ - 2\left( {1 - \frac{t}{2}} \right)\;}}} \right)\)

\(= 4{\left( {1 - t} \right)^{ - 1}} - \frac{5}{2}\;{\left( {1 - \frac{t}{2}} \right)^{ - 1}}\)

By using Binomial expansion,

\(\Rightarrow f\left( t \right) = 4\left[ {1 + t + {t^2} + {t^3} + \ldots } \right] - \frac{5}{2}\left[ {1 + \frac{t}{2} + {{\left( {\frac{t}{2}} \right)}^2} + {{\left( {\frac{t}{2}} \right)}^3} + \ldots } \right]\)

\(= 4\left[ {1 + t + {t^2} + {t^3} + \ldots } \right] - \frac{5}{2}\left[ {1 + \frac{t}{2} + \frac{{{t^2}}}{4} + \frac{{{t^3}}}{8} \ldots } \right]\)

\(= \left( {4 - \frac{5}{2}} \right) + t\left( {4 - \frac{5}{4}} \right)+ {t^2}\left( {4 - \frac{5}{8}} \right) + {t^3}\left( {4 - \frac{5}{{16}}} \right) + \ldots\)

\(\Rightarrow f\left( t \right) = \frac{3}{2} + \frac{{11}}{4}t + \frac{{27}}{8}{t^2} + \frac{{59}}{{16}}{t^3} + \ldots\)

Now put t = z – 2

\(\Rightarrow f\left( z \right) = \frac{3}{2} + \frac{{11}}{4}\left( {z - 2} \right) + \frac{{27}}{8}{\left( {z - 2} \right)^2} + \frac{{59}}{{16}}{\left( {z - 2} \right)^3} + \ldots \)

u(x, y) = x³ + ax²y + bxy² + 2y³

The given function is harmonic function.

From CR equations,

u_x = v_y , u_y = -v_x

u_x = 3x² + 2axy + by², u_y = ax² + 2bxy + 6y²

⇒ v_y = 3x² + 2axy + by²

By integrating on both sides with respect to y.

\(\Rightarrow \smallint {v_y}dy = \smallint \left( {3{x^2} + 2axy + b{y^2}} \right)dy\)

⇒ v = 3x²y + axy² + \(\frac{{b{y^3}}}{3} + \phi \left( x \right)\)

⇒ v_x = 6xy + ay² + ϕ’(x) = -u_y

⇒ 6xy + ay² + ϕ’(x) = -ax² – 2bxy – 6y²

By comparing on both sides,

-2b = 6 ⇒ b = -3

a = -6

ϕ’(x) = -ax²

\(\Rightarrow \phi \left( x \right) = - \frac{{a{x^3}}}{3} + C = \frac{{6{x^3}}}{3} + C = 2{x^3} + C\)

Now, v = 3x²y – 6xy² – y³ + 2x³ + C

Given that, v (0, 0) = 1

⇒ C = 1

Now, V = 3x²y – 6xy² – y³ + 2x³ + 1

v(1, 1) = 3 – 6 – 1 + 2 – 1 = -1

Concept:

Properties of Rank:

Rank of a matrix is the number of independent rows in the given matrix. Given two square matrices A and B of order n × n, we have following properties:

1. Rank of product of A and B i.e. Rank (AB) ≥ Rank (A) + Rank (B) – order of square matrix

2. Rank of sum of A and B i.e. Rank (A + B) ≤ Rank (A) + Rank (B).

Properties of Determinant:

Given two square matrices A and B of order n × n and their determinants Det(A) and Det(B) respectively, determinant of their product i.e Det( AB) = Det(A) * Det(B). However, the same does not hold for the addition of the given matrices.

Example:

Consider two square matrices A and B each of order 2×2.

\(A = \left[ {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right]\;\;\;\;\;\;Rank\left( A \right) = 2,\;\;Det\left( A \right) = 5\)

\(B = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\;\;\;\;\;Rank\left( B \right) = 2,\;\;\;\;\;\;\;\;Det\left( B \right) = 1\)

\(AB = \left[ {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right]\;\;\;\;\;\;\;\;\;\;\;\;\;\;Rank\left( {AB} \right) = 2,\;\;\;\;\;Det\left( {AB} \right) = 5\)

\(A + B = \;\left[ {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}}3&1\\3&5\end{array}} \right]\;\;\;\;\;\;\;Rank\left( {A + B} \right) = 2,\;Det\left( {A + B} \right) = 12\)

Statement I is FALSE.

Rank of product matrix AB is 2. Product of Rank(A) and Rank(B) is: 2*2 = 4. Therefore, rank of product matrix is not equal to the product of the rank of individual matrices.

Statement II is TRUE. 

Det(AB)= 5 = Det(A) * Det(B)

Statement III is TRUE.  

Rank(A + B) = 2. Sum of rank of A and B is: 2 + 2 = 4. Therefore, the relation: Rank (A + B) ≤ Rank (A) + Rank (B) holds true

Therefore, rank of the addition matrix is less than or equal to the sum of rank of the individual matrices.

Statement IV is FALSE.  

Try to find out the eigen values by taking the example of matrix with rank 2. Then check if given statement follows from that or not.

Calculation:

A be (n x n) real valued square symmetric matrix of rank 2.

\(\mathop \sum \limits_{i = 1}^n \mathop \sum \limits_{j = 1}^n A_{ij}^2 = 50\),

which means, sum of square of all elements of A = 50.

Also, rank of A = 2 i.e. we have (n – 2) eigen values are 0.

So eigen values are in the form of a₁, a_2,0, 0, ……

Let us consider a matrix A = \(\left[ {\begin{array}{*{20}{c}}{ - 5}&0\\0&5\end{array}} \right]\)

Here, for this eigen values are [-5, 5].

Take another example, B = \(\left[ {\begin{array}{*{20}{c}}5&0&0\\0&5&0\\0&0&0\end{array}} \right]\) and C = \(\left[ {\begin{array}{*{20}{c}}6&0&0\\0&{\surd 14}&0\\0&0&0\end{array}} \right]\)

Matrix B and C are of rank 2 and symmetric.

Eigen value for A = 5, 5, 0

And eigen value for B = \(6,\;\sqrt {14} ,\;0\)

As 0 and 5 are in the range of [-5, 5], So statement 1 is correct.

But eigen value with largest magnitude which is 6 for C must be strictly greater than 5 and in case of B largest value is 5 which is not strictly greater than 5, So, second statement is incorrect here.

Concept:

Xp + Yq = Z

Where \(p\to \frac{dz}{dx}\)

\(q\to \frac{dz}{dy}\) & X, Y and Z are function of x, y & z

Then it’s auxiliary equation will be

\(\frac{dx}{X}=\frac{dy}{Y}=\frac{dz}{Z}\) & on solving above we get u = c₁ & v = c₂

Where c₁ & c₂ = constants u & v are functions of x, y, z

Then solution of the above equation will be

f(u, v) = 0 or u = ϕ (v)

Calculation:

We are given \(\frac{{{x}^{2}}dz}{dx}+\frac{yd{{z}^{2}}}{dy}=\left( x+y \right)z\)

Then its auxiliary equation will be,

\(\frac{dx}{{{x}^{2}}}=\frac{dy}{{{y}^{2}}}=\frac{dz}{\left( x+y \right)z}\) ---(1)

On solving first two parts of the above equation we

\(\int \frac{dx}{{{x}^{2}}}=\int \frac{dy}{y}\Rightarrow \frac{1}{x}=\frac{1}{y}+{{c}_{1}}\Rightarrow {{c}_{1}}=\frac{1}{x}-\frac{1}{y}\)

We can write equation (1) as expressed below

\(\frac{\frac{dx}{x}}{x}=\frac{\frac{dy}{y}}{y}=\frac{\frac{dz}{z}}{x+y}\)

\(\Rightarrow \frac{dx}{x}+\frac{dy}{y}=\frac{dz}{z}\)

By integrating the above equation,

In x + In y = In z + In c₂

⇒ In c₂ = In x + In y – In z

\(\Rightarrow {{c}_{2}}=\frac{xy}{z}\)

Hence solution of given equation will be

f(c₁, c₂) = 0

Explanation:

Given:

Differential equation: (2D³ – 7D² + 7D – 2) y = e^-8x

Particular integral, \(PI=\frac{1}{f\left( D \right)}\phi \left( x \right)\)

\(=\frac{1}{2{{D}^{3}}-7{{D}^{2}}+7D-2}{{e}^{-8x}}\)

\(=\frac{1}{2{{\left( -8 \right)}^{3}}-7{{\left( -8 \right)}^{2}}+7\left( -8 \right)-2}{{e}^{-8x}}\)

Given lines are:

4x + 3y + 7 = 0

3x + 4y + 8 = 0

Let  4x + 3y + 7 = 0 be the line of x on y and 3x + 4y + 8 = 0 be the line of y on x.

\(4x + 3y + 7 = 0 \)

\(x = - \frac{{7}}{4} - \frac{3}{4}y\)

\(3x + 4y + 8 = 0 \)

\(y = - \frac{8}{4} - \frac{3}{4}x\)

The regression coefficient of x on y is:

\({b_{xy}} = - \frac{3}{4}\)

The regression coefficient of y on x is:

\({b_{yx}} = - \frac{3}{4}\)

The correlation coefficient will be:

\(r = \sqrt {{b_{xy}}.{b_{yx}}} = \sqrt {\left( { - \frac{3}{4} \times - \frac{3}{4}} \right)}\)

\(\Rightarrow r = \frac{3}{4} = 0.75\)

Since \({b_{xy}},\;{b_{yx}}\) are both negative, r is also negative. ⇒ r = -0.75