Engineering Mathematics Test 1

3yy’ + 4x = 0

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

3ydy = - 4x dx

On integration both the sides:

\(\frac{3}{2}{y^2} = - 2{x^2} + c\)

Where c is the constant of integration.

\(\frac{3}{2}{y^2} + 2{x^2} = c\)

\(\frac{{{x^2}}}{{\left( {\frac{1}{2}} \right)c}} + \frac{{{y^2}}}{{\left( {\frac{2}{3}} \right)c}} = 1\)

For Trapezoidal Rule, error = \(\rm -\frac {(b-a)}{12}\times h^2.f^{ii}(\zeta)=O(n^2)\)

For Simpson's 1/3rd  Rule, error = \(\rm -\frac {(b-a)}{180}\times h^4.f^{iv}(\zeta)=O(n^4)\)

Concept:

If x is Eigen vector matrix corresponding to the matrix A & λ is the Eigen value then AX = λX

Calculation:

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

\(\lambda X = \lambda \left[ {\begin{array}{*{20}{c}}1\\1\\2\end{array}} \right]\) 

\(AX = \lambda X \Rightarrow 2\left[ {\begin{array}{*{20}{c}}1\\1\\2\end{array}} \right] = \lambda \left[ {\begin{array}{*{20}{c}}1\\1\\2\end{array}} \right] \Rightarrow \lambda = 2\) 

According to question, λ = x

Hence [λ = 2 or λ = x = 2]

Concept:

Concept:

The equation

\({\rm{A}}\frac{{{\partial ^2}{\rm{u}}}}{{\partial {{\rm{x}}^2}}} + {\rm{B}}\frac{{{\partial ^2}{\rm{u}}}}{{\partial {\rm{x\;}}\partial {\rm{y}}}} + {\rm{C}}\frac{{{\partial ^2}{\rm{u}}}}{{\partial {{\rm{y}}^2}}} + {\rm{D}}\frac{{\partial {\rm{u}}}}{{\partial {\rm{y}}}} + {\rm{E}}\frac{{\partial {\rm{u}}}}{{\partial {\rm{x}}}} + {\rm{Fu}} = {\rm{y\;is}}\)

Elliptic if B2 – 4AC < 0

Parabolic if B2 – 4AC = 0

Hyperbolic if B2 – 4AC > 0

Calculation:

Comparing the given equation with standard form,

A = 1; B = 1 / 2; C = 0

Now, \({B^2} - 4AC = {\left( {\frac{1}{2}} \right)^2} - 4\times1.0 = \frac{1}{4} > 0\) 

⇒ Hyperbolic

Explanation:

For continuous function:

\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = f\left( 2 \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)\)

\(\mathop {\lim }\limits_{x \to {2^ - }} \left( {4 - x} \right) = \mathop {\lim }\limits_{x \to 2} \left( {4 - x} \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {kx - 4} \right)\)

(4 - 2) = 2 k - 4

k = 3

A box contains 2 red and 3 black balls

Sample space = The number of possibilities to choose three balls from the box

\( = {5_{{C_3}}} = 10\)

The number of possibilities to choose 1 red and 2 black balls \( = {2_{{C_1}}}{3_{{C_2}}} = 6\)

f(x, y, z) = (x² + y² – 2z²)(y² + z²)

\(\frac{{\partial f}}{{\partial x}} = \left( {2x} \right)\left( {{y^2} + {z^2}} \right)\) 

At the point, x = 2, y = 1 and z = 3 is

\(\frac{{\partial f}}{{\partial x}} = 2\left( 2 \right)\left( {{1^2} + {3^2}} \right) = 40\) 

\(A=\left[ \begin{matrix} -1 & 2 & -1 & 0 \\ 2 & 4 & 4 & 2 \\ 0 & 0 & 1 & 5 \\ 1 & 6 & 3 & 2 \\\end{matrix} \right]\)

R₄ → R₄ – R₁

\(=\left[ \begin{matrix} -1 & 2 & -1 & 0 \\ 2 & 4 & 4 & 2 \\ 0 & 0 & 1 & 5 \\ 2 & 4 & 4 & 2 \\\end{matrix} \right]\)

R₄ → R₄ – R₂

R₂ → R₂ + 2R₁

\(=\left[ \begin{matrix} -1 & 2 & -1 & 0 \\ 0 & 8 & 2 & 2 \\ 0 & 0 & 1 & 5 \\ 0 & 0 & 0 & 0 \\\end{matrix} \right]\)

Now this matrix is in echelon form.

Explanation:

Given:

\({y^2} = 2C\left( {x + \sqrt C } \right)\)      ---(1)

⇒ 2y y¹ = 2C      ---(2)

Using (2), (1) becomes

\({y^2} = 2y{y^1}\left( {x + {{\left( {y{y^1}} \right)}^{1/2}}} \right)\) 

\( \Rightarrow y = 2{y^1}x + 2{y^{\frac{1}{2}}}{\left( {{y^1}} \right)^3}\) 

\( \Rightarrow {\left( {y - 2x{y^1}} \right)^2} = 2y{\left( {{y^1}} \right)^3}\) 

∴ order = 1 and degree = 3.

According to Simpson’s 1/3^rd rule

\(\mathop \smallint \limits_a^b f\left( x \right)dx = \frac{h}{3}\left[ {\left( {{y_o} + {y_n}} \right) + 2\left( {{y_2} + {y_4} + \ldots } \right) + 4\left( {{y_1} + {y_3} + \ldots } \right)} \right]\)

\(\therefore \mathop \smallint \limits_0^b \left( {\frac{1}{{1 + {x^2}}}} \right)dx = \frac{1}{3}\left[ {\left( {{y_0} + {y_6}} \right) + 2\left( {{y_2} + {y_4}} \right) + 4\left( {{y_1} + {y_3} + {y_5}} \right)} \right]\)

\( = \frac{1}{3}\left[ {\left( {1 + 0.027} \right) + 2\left( {0.2 + 0.0588} \right) + 4 + \left( {0.5 + 0.1 + 0.0385} \right)} \right]\)

\(= \frac{1}{3}\left[ {4.0986} \right]\)

= 1.3662

Concept:

Variance = npd

where n = total no. of events, p = probability of success, d = probability of failure.

Standard deviation = \(\sqrt {variance} \) 

Calculation:

Variance = npd

n = 8, p = probability success \( = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}\) 

d = probability of failure \( = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}\) 

∴ Variance npd \( = 8\times \frac{1}{3}\times \frac{2}{3} = \frac{{16}}{9}\) 

∴ Standard deviation = \(\sqrt {variance} = \sqrt {\frac{{16}}{9}} = \frac{4}{3} = 1.33\) 

f(y) = y3 - 7y2 + 5

The auxiliary function is f₁(y) = y3 - 7y2 = y2(y - 7)

f(0) = f(7)=5.

The function f(x) is everywhere continuous and differentiable because it is a cubic polynomial.

Consequently, it satisfies all the conditions of Rolle’s theorem on the interval [0,7].

Therefore the value q is 7.

The given differential equation is:

\(4\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} + 12\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + 9{\rm{y}} = 0\)

Characteristics equation will be:

4m² + 12m + 9 = 0

\({\rm{m}} = \frac{{ - 12 + \sqrt {144 - 144} }}{{2 \times 4}}\)

\({\rm{m}} = \frac{{ - 12 \pm 0}}{8}\)

\({{\rm{m}}_1} = - \frac{3}{2},{\rm{\;}}{{\rm{m}}_2} = - \frac{3}{2}\)

The roots of the equation are real & equal.

Thus, solution for this equation will be written as:

\({\rm{y}}\left( {\rm{x}} \right) = \left( {{{\rm{c}}_1} + {{\rm{c}}_2}{\rm{x}}} \right){{\rm{e}}^{ - \frac{3}{2}{\rm{x}}}}\)      ----(i)

Now given y (0) = 1

\(1 = \left( {{{\rm{c}}_1} + {{\rm{c}}_2} \times 0} \right){{\rm{e}}^{ - \frac{3}{2} \times 0}}\)

C₁ = 1

Now, On differentiating equation (i):

\({\rm{y'}}\left( {\rm{x}} \right) = \left( {{{\rm{c}}_1} + {{\rm{c}}_2}{\rm{x}}} \right){\rm{\;}}\left( { - \frac{3}{2}} \right){{\rm{e}}^{ - \frac{3}{2} \times {\rm{x}}}} + {{\rm{c}}_2}{\rm{\;}}{{\rm{e}}^{ - \frac{3}{2} \times {\rm{x}}}}\)

Given \({\rm{y'}}\left( 0 \right) = - 4\)

Thus, \({\rm{y'}}\left( 0 \right) = \left( {{{\rm{c}}_1} + {{\rm{c}}_2} \times 0} \right)\left( { - \frac{3}{2}} \right){{\rm{e}}^{ - \frac{3}{2} \times 0}} + {{\rm{c}}_2}{\rm{\;}}{{\rm{e}}^{ - \frac{3}{2} \times 0}}\)

\( - 4 = - \frac{3}{2}{\rm{\;}}{{\rm{c}}_1} + {{\rm{c}}_2}\)

\( - 4 = - \frac{3}{2} \times 1 + {{\rm{c}}_2}\)

\({{\rm{c}}_2} = - 4 + \frac{3}{2} = - \frac{5}{2}\)

Now put C₁ & C₂ values in equation (i)

\({\rm{y}}\left( {\rm{x}} \right) = \left( {1 - \frac{5}{2}} \right){{\rm{e}}^{ - \frac{3}{2}{\rm{x}}}}\)

\({\rm{y}}\left( 1 \right) = \left( {1 - \frac{5}{2}} \right){{\rm{e}}^{ - \frac{3}{2} \times 1}}\)

\({\rm{y}}\left( 1 \right) = - \frac{3}{2}{{\rm{e}}^{ - \frac{3}{2}}}\)

Concept:

If limiting expression \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}}\) leads to \(\frac{0}{0}\) form, then apply L-Hospital’s rule to find the limiting value.. Applying L-Hospital’s rule we get,

\(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\) 

Calculation:

\(\mathop {\lim }\limits_{x \to 1} \frac{{{x^x} - x}}{{x - 1 - \log x}};\frac{0}{0}\) form, so applying L-Hospital’s rule,

\(\mathop {\lim }\limits_{x \to 1} \frac{{{x^x} - x}}{{x - 1 - \log x}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left( {{x^x} - x} \right)}}{{\frac{d}{{dx}}\left( {x - 1 - \log x} \right)}} \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}{x^x} - \frac{d}{{dx}}x}}{{\frac{d}{{dx}}\left( {x - 1} \right) - \frac{d}{{dx}}\left( {\log x} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^x}\left( {1 + \log x} \right) - 1}}{{1 - \frac{1}{x}}};\left( {\frac{0}{0}\;form} \right)\)

Again applying L-Hospital’s rule,

\(\mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left( {{x^x}} \right) \cdot \left( {1 + \log x} \right) + {x^x}\left( {\frac{1}{x}} \right) - 0}}{{\frac{1}{{{x^2}}}}}\)

\( = \mathop {\lim }\limits_{x \to 1} \frac{{{x^x}{{\left( {1 + \log x} \right)}^2} + {x^x}\left( {\frac{1}{x}} \right)}}{{\frac{1}{{{x^2}}}}} = \frac{{1{{\left( {1 + 0} \right)}^2} + \frac{1}{1}}}{1}\) = 2

Important point:

Concept:

Absolute minimum value of the function is the least value of function in the given domain which occurs at boundary of the domain or critical points (within the domain) of the function.

Calculation:

f(x) = 12x^{4 / 3} – 6x^{1 / 3}; xϵ [-1, 1]

\(f'\left( x \right) = 12 \cdot \frac{4}{3}{x^{\frac{1}{3}}} - 6 \cdot \frac{1}{3}{x^{ - \frac{2}{3}}} = 16{x^{\frac{1}{3}}} - \frac{2}{{{x^{\frac{2}{3}}}}}\)

\( \Rightarrow f'\left( x \right) = \frac{{2\left( {8x - 1} \right)}}{{{x^{\frac{2}{3}}}}}\)

\(f'\left( x \right) = 0 \Rightarrow x = \frac{1}{8}\)

\(f\left( {\frac{1}{8}} \right) = 12{\left( {\frac{1}{8}} \right)^{\frac{4}{3}}} - 6{\left( {\frac{1}{8}} \right)^{\frac{1}{3}}} = - \frac{9}{4}\)

f(-1) = 12(-1)^{4 / 3} – 6(-1)^{1 / 3} = 18

f(1) = 12(1)^{4 / 3} – 6(1)^{1 / 3} = 12 – 6 = 6

Absolute minimum value \( = \min \left\{ {18,\;6, - \frac{9}{4}} \right\} = - \frac{9}{4} = - 2.25\)

Important point:

The maxima and minima occurs alternatively and , if f(x) be a function then critical points can be obtained by the solution of f’(x) = 0

Concept:

Cayley Hamilton’s Theorem states that any matrix satisfies its characteristic equation.

i.e. if ‘A’ is any matrix then A will satisfy its characteristic equation |A – λI| = 0 where I is identity matrix.

Calculation:

\(A - \lambda I = \left[ {\begin{array}{*{20}{c}} 0&2&{ - 3}\\ 0&3&2\\ 0&0&{ - 2} \end{array}} \right] - \lambda \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {1 - \lambda }&2&{ - 3}\\ 0&{3 - \lambda }&2\\ 0&0&{ - 2 - \lambda } \end{array}} \right]\)

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

⇒ (1 - λ)(3 - λ)(-2 - λ) = 0

⇒ λ = 1, 3, -2

Eigen values of A³ = λ³ = 1³, 3³, (-2)³ = 1, 27, -8

Eigen values of A² = λ² = 1², 3², (-2)² = 1, 9, 4

Eigen values of A = λ = 1, 3, -2

Eigen values of B,

First eigen value = 3(1)³ + 5(1)² – 6(1) + 2(1) = 4

Second eigen value = 3(27) + 5(9) – 6(3) + 2(1) = 110

Third eigen value = 3(-8) + 5(4) – 6(-2) + 2(1) = 10

Hence, the required eigen values are 4, 110 & 10

Determinant of matrix B = λ₁ λ₂ λ₃ = 4 * 110 * 10 = 4400

Imporant Point:

i) Eigen value of Aⁿ = λⁿ

Concept:

If f(x), g(x) & h(x) be three different functions then Wronskian is given by \(\left| {\begin{array}{*{20}{c}}{f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\{f'\left( x \right)}&{g'\left( x \right)}&{h'\left( x \right)}\\{f''\left( x \right)}&{g''\left( x \right)}&{h''\left( x \right)}\end{array}} \right|\)

Calculation:

f(x) = x²

g(x) = 3x + 2

h(x) = 2x + 3

Wronskian \(= \left| {\begin{array}{*{20}{c}}{{x^2}}&{3x + 2}&{2x + 3}\\{\frac{d}{{dx}}\left( {{x^2}} \right)}&{\frac{d}{{dx}}\left( {3x + 2} \right)}&{\frac{d}{{dx}}\left( {2x + 3} \right)}\\{\frac{{{d^2}}}{{d{x^2}}}\left( {{x^2}} \right)}&{\frac{{{d^2}}}{{d{x^2}}}\left( {3x + 2} \right)}&{\frac{{{d^2}}}{{d{x^2}}}\left( {2x + 3} \right)}\end{array}} \right|\) 

\(\Rightarrow Wronskian = \left| {\begin{array}{*{20}{c}}{{x^2}}&{3x + 2}&{2x + 3}\\{2x}&3&2\\2&0&0\end{array}} \right|\) 

Expanding along 3^rd row, Wronskian = 2[2(3x + 2) - 3(2x + 3)] = 2 (6x + 4 – 6x - 9) = -10

Concept:

The probability of ‘r’ success in ‘n’ trial is given as \(p\left( {x = r} \right) = {}_{}^n{C_r}{p^r}{q^{n - r}}\) 

Where, p = probability of success

d = 1 – p = probability of failure

Calculation:

Probability of defective item, \(p = \frac{{20}}{{100}} = 0.2\)

Probability of non-defective item, d = 1 – p = 0.8

Probability that exactly 3 of the chosen items are defective \( = {}_{}^{20}{C_3}{\left( p \right)^3}{\left( d \right)^{20 - 3}} = {}_{}^{20}{C_3}{\left( {0.2} \right)^3}{\left( {0.8} \right)^{17}}\)

= 0.205

Important Point:

\({}_{}^n{C_r} = \frac{{n!}}{{r!\left( {n - r} \right)!}}\)

Concept:

If the system has more than one solution then |A| = 0; where A is the coefficient matrix.

For the system,

x + y + z = 0

0.x + y + 2z = 0

αx + 0.y + z = 0

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

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

⇒ α (2 - 1) + 1 (1 - 0) = 0

⇒ α + 1 = 0

⇒ α = 1

Hence, α = -1

Property:

\(\mathop{\int }_{\text{a}}^{\text{b}}\text{f}\left( \text{x} \right)\text{ }\!\!~\!\!\text{ dx }\!\!~\!\!\text{ }=\mathop{\int }_{\text{a}}^{\text{b}}\text{f}\left( \text{a}+\text{b}-\text{x} \right)\text{ }\!\!~\!\!\text{ dx}\)

Calculation:​​​

\(\text{I}=\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\text{x }\!\!~\!\!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }\!\!~\!\!\text{ dx}=\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\left( \text{ }\!\!\pi\!\!\text{ }-\text{x} \right)\text{ }\!\!~\!\!\text{ }{{\left( \cos \left( \text{ }\!\!\pi\!\!\text{ }-\text{x} \right) \right)}^{2}}\text{ }\!\!~\!\!\text{ dx}\)

\(\text{I}=\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\pi\!\!\text{ }\!\!~\!\!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }\!\!~\!\!\text{ dx}-\text{ }\!\!~\!\!\text{ }\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\text{x }\!\!~\!\!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }\!\!~\!\!\text{ dx}\)

\(\text{I}=\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\pi\!\!\text{ }\!\!~\!\!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }\!\!~\!\!\text{ dx}-\text{ }\!\!~\!\!\text{ I}\)

\(2\times \text{I}=\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\pi\!\!\text{ }\!\!~\!\!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }\!\!~\!\!\text{ dx}=\text{ }\!\!\pi\!\!\text{ }\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\frac{1+\cos 2\text{x}}{2}\text{ }\!\!~\!\!\text{ dx}\)

\(2\times \text{I}=\text{ }\!\!\pi\!\!\text{ }\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\frac{1}{2}\text{ }\!\!~\!\!\text{ dx}+\text{ }\!\!\pi\!\!\text{ }\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\frac{\cos 2\text{x}}{2}\text{ }\!\!~\!\!\text{ dx}\)

\(2\times \text{I}=\text{ }\!\!~\!\!\text{ }\frac{\text{ }\!\!\pi\!\!\text{ }}{2}\left( \text{ }\!\!\pi\!\!\text{ }-0 \right)+\frac{1}{2}\left( \sin 2\text{ }\!\!\pi\!\!\text{ }-\sin 0 \right)=\frac{{{\text{ }\!\!\pi\!\!\text{ }}^{2}}}{2}\)

\(\therefore \mathbf{I}=\mathop{\int }_{0}^{\mathbf{\pi }}\mathbf{x}~{{\left( \cos \mathbf{x} \right)}^{2}}~\mathbf{dx}=\frac{{{\mathbf{\pi }}^{2}}}{4}\)

Explanation:

Let f(x) = x⁴ – x – 10

f'(x) = 4x³ – 1

Newton-Raphson’s formula is

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

Putting n = 0, first approximation is given by

\({x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}}\)     ---(1

Since, x₀ = 2

f(x₀) = 2⁴ – 2 – 10 = 4

f’(x₀) = 31

Put in (1)

\({x_1} = 2 - \frac{4}{{31}}\)

∴ x₁ = 1.8709

Also,

\({x_2} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}}\)     ---(2)

f(x₁) = (1.8709)⁴ – 1.8709 – 10 = 0.3809

f’(x₁) = 4(1.8709)³ – 1 = 25.1945

\({x_2} = 1.8709 - \frac{{0.3809}}{{25.1945}} = 1.8557\)

Explanation:

(i) P(X ≤ a) = P(X > a)

\(\therefore \;\mathop \smallint \limits_0^a 3{x^2}dx = \mathop \smallint \limits_a^1 3{x^2}dx\)

a³ = 1 – a³

\({a^3} = \frac{1}{2}\)

∴ a = 0.7937

(ii) P(X > b) = 0.05

\(\mathop \smallint \limits_b^1 3{x^2}dx = 0.05\)

b³ = 0.95

∴ b = 0.9830

Concept:

Diagonalization of the matrix:

If a square matrix Q of order n has n linearly independent Eigenvectors, 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 Eigenvalues 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 Eigenvectors.

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 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:

Option 1: In a lower triangular matrix or in an upper triangular matrix, the diagonal elements itself are the Eigenvalues.

Therefore, the product of the diagonal elements is equal to the product of Eigenvalues.

Option 2:

For n × n matrices A and B i.e. for square matrices

AB = I

⇒ B = A^-1

Now, BA = A^-1A = I

If AB = I, then BA = I. It is always true.

Therefore, the given statement is false.

Note: But the above statement is valid for non-square matrices.

Example: \(A = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\\0&0\end{array}} \right]\)

Option 3:

If there exists an invertible matrix Q such that A = QBQ^-1, the det A = det B.

From the diagonalization of matrix A, the above statement is true.

Option 4:

A is an invertible matrix. So, matrix A is a nonsingular matrix and hence the rank of a matrix is n.

Therefore, the rows of A are linearly independent.

Concept:

A vector F is said to be solenoidal when ∇. F = 0

A vector F is said to be irrotational when ∇ × F = 0

Calculation:

\(\vec V = \left( {x + y + az} \right)i + \left( {bx + 2y - z} \right)j + + \left( { - x + cy + 2z} \right)k\)

\(\nabla \times \vec V = \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {\left( {x + y + az} \right)}&{\left( {bx + 2y - z} \right)}&{\left( { - x + cy + 2z} \right)} \end{array}} \right| = 0\)

⇒ i (c + 1) – j (-1 – a) + k (b – 1) = 0

⇒ a = -1, b = 1, and c = -1

\(\vec V = \left( {x + y - z} \right)i + \left( {x + 2y - z} \right)j + \left( { - x - y + 2z} \right)k\)

Divergence of the given vector is

\(div\;\vec V = \nabla .\vec V = \frac{\partial }{{\partial x}}\left( {x + y - z} \right) + \frac{\partial }{{\partial y}}\left( {x + 2y - z} \right) + \frac{\partial }{{\partial z}}\left( { - x - y + 2z} \right)\)