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Tutor profile: Jasveer J.

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Jasveer J.
Maths tutor for 5 years
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Subject: Linear Algebra

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Question:

Let V be the vector space of all $$2\times2$$ matrices over the field F. Prove that V has dimension 4 by exhibiting a basis for V which has 4 elements.

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Jasveer J.
Answer:

Let $$V(F)= \begin{bmatrix}a & b \\c & d \end{bmatrix}$$ such that $$a,b,c,d$$ belongs to F Let $$\alpha_1=\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}, \alpha_2=\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix}, \alpha_3=\begin{bmatrix}0 & 0 \\1 & 0 \end{bmatrix}, \alpha_4=\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}$$ be four elements of V. Let $$S=[\alpha_1,\alpha_2,\alpha_3,\alpha_4]\subseteq V$$ 1) First we have to show that S is Linearly Independent If $$a_1,a_2,a_3,a_4$$ belongs to F then $$a_1\alpha_1+a_2\alpha_2+a_3\alpha_3+a_4\alpha_4=0$$ $$\Rightarrow a_1\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}+a_2\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix}+a_3\begin{bmatrix}0 & 0 \\1 & 0 \end{bmatrix}+a_4\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix}a_1 & a_2 \\a_3 & a_4 \end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}$$ 2) Now we have to show that$$L(S)= V$$ We know that $$L(S)\subseteq V.................(1)$$ Let $$\alpha= \begin{bmatrix}a & b \\c & d\end{bmatrix} $$ belongs to V then $$\begin{bmatrix}a & b \\c & d\end{bmatrix}=a\begin{bmatrix}1 & 0 \\0 & 0\end{bmatrix}+b\begin{bmatrix}0 & 1 \\0 & 0\end{bmatrix}+c\begin{bmatrix}0 & 0 \\1 & 0\end{bmatrix}+d\begin{bmatrix}0 & 0 \\0 & 1\end{bmatrix}$$ $$=a\alpha_1+b\alpha_2+c\alpha_3+d\alpha_4$$ belongs to $$L(S)$$ hence $$\alpha$$ belongs to L(S) hence V $$\subseteq L(S).............(2)$$ Hence from (1) and (2) $$V=L(S)$$ Hence S is a basis of V. hence dimension of V =4

Subject: Calculus

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Question:

Discuss the maxima and minima of the function $$u=sinxsinysinz$$ where $$x,y,z$$ are the angles of a triangle.

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Jasveer J.
Answer:

Let f be a function with two or more variables with continuous second order partial derivatives f xx , f yy and f xy at a critical point (a,b). Let $$D = fxx(a,b) fyy(a,b) - fxy^2(a,b)=rt-s^2$$ a) If $$D > 0$$ and $$f xx (a,b) > 0$$, then f has a relative minimum at (a,b). b) If $$D > 0$$ and $$f xx (a,b) < 0$$, then f has a relative maximum at (a,b). c) If $$D < 0$$, then f has a saddle point at (a,b). d) If $$D = 0$$, then no conclusion can be drawn. Since $$x,y,z$$ are the angles of triangle, hence $$x+y+z=\pi..............(1)$$ Given $$u=sinxsinysinz...........(2)$$ For maxima or minima we will partially differentiate equation (2) on both the sides. $$du=cosxsinysinzdx+sinxcosysinzdy+sinxsinycoszdz...........(3)$$ Also from equation (1) we have $$dx+dy+dz=0........................(4)$$ Multiplying (3) by 1 and (4) by $$\lambda$$ and adding, we get $$(cosxsinysinz+\lambda)dx+(sinxcosysinz+\lambda)dy+(sinxsinycosz+\lambda)dz=0$$ Equating the coefficient of $$dx,dy,dz$$ to zero, we get $$cosxsinysinz+\lambda=0$$, $$sinxcosysinz+\lambda=0$$ $$sinxsinycosz+\lambda=0.......................(5)$$ Hence $$cosxsinysinz=sinxcosysinz=sinxsinycosz=-\lambda$$ (or) $$cotx=coty=cotz$$, dividing each term by $$sinxsinysinz$$ (or) $$x=y=z=\frac{\pi}{3}$$, from (1) Let $$x$$ and $$y$$ be independent and z be dependent, as the relation(1) exists to find r,s and t. from (2) $$\frac{\delta u}{\delta x}=cosxsinysinz+sinxsinycosz\frac{\delta z}{\delta x}$$ But from (1) we get $$1+\frac{\delta z}{\delta x}=0$$ i.e. $$\frac{\delta z}{\delta x}=-1$$ Hence $$\frac{\delta u}{\delta x}=cosxsinysinz-sinxsinycosz$$ and $$\frac{\delta^2 u}{\delta x^2}=(-sinxsinysinz+cosxsinycosz\frac{\delta z}{\delta x})-(cosxsinycosz-sinxsinysinz\frac{\delta z}{\delta x})$$ $$=(-sinxsinysinz-cosxsinycosz)-(cosxsinycosz+sinxsinysinz)$$ that is, $$r=-2(sinxsinysinz+cosxsinycosz)$$ and $$\frac{\delta^2 u}{\delta y \delta x}=cosxcosysinz+cosxsinycosz\frac{\delta z}{\delta y}-sinxcosycosz+sinxsinysinz\frac{\delta z}{\delta x}$$ where $$\frac{\delta z}{\delta y}=-1$$ (as $$ x+y+z=\pi\Rightarrow1+\frac{\delta z}{\delta y}=0\Rightarrow\frac{\delta z}{\delta y}=-1)$$ $$\Rightarrow s=cosxcosysinz-cosxsinycosz-sinxcosycosz-sinxsinysinz$$ putting $$x=y=z=\frac{\pi}{3}$$, we get $$r=-2(\frac{1}{2}.\sqrt3.\frac{1}{2}.\sqrt3.\frac{1}{2}.\sqrt3+\frac{1}{2}.\frac{1}{2}.\sqrt3.\frac{1}{2})=-\sqrt3$$ $$s= \frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\sqrt3-\frac{1}{2}.\frac{1}{2}.\sqrt3.\frac{1}{2}-\frac{1}{2}.\sqrt3.\frac{1}{2}.\frac{1}{2}-\frac{1}{2}.\sqrt3.\frac{1}{2}.\sqrt3.\frac{1}{2}.\sqrt3=-\frac{1}{2}.\sqrt3$$ By symmetry $$t=\frac{\delta^2 u}{\delta y^2}=-\sqrt3$$ at $$x=y=z=\frac{\pi}{3}$$ $$rt-s^2=(-\sqrt3)(-\sqrt3)-(-\frac{1}{2}\sqrt3)^2$$ $$=3-\frac{9}{4}=\frac{9}{4}$$ Also $$r=-\sqrt3$$ which is less than 0 Hence there is maximum at $$x=y=z=\frac{\pi}{3}$$

Subject: Differential Equations

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Question:

Solve $$x^2 \frac{d^2y}{dx^2}-3x\frac{dy}{dx}+y=\frac{logxsin(logx)+1}{x}$$

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Jasveer J.
Answer:

Here the given differential equation is non-homogeneous. Hence for finding out the general solution to the given second-order differential equation we have to first find out both Complementary Function and Particular Integral. Our general solution will be the sum of both the complementary function and particular integral. $$y(x)=y_c+y_p$$ Now the given equation is not having constant coefficients so, we have to convert it into a differential equation that has constant coefficients. Please have a look at the method below how we converted it into a constant-coefficient equation. Given $$(x^2D^2-3xD+1)y=\frac{logxsin(logx)+1}{x}$$................1 Let $$x=e^z$$ so that $$z=logx$$ and let $$D_1=\frac{d}{dz}$$ then $$xD=D_1$$ and $$x^2D^2=D_1(D_1-1)$$ Hence 1 will become $$[D_1(D_1-1)-3D_1+1]y=\frac{zsinz+1}{e^z}$$ $$\Longrightarrow(D_1^2-4D_1+1)y=\frac{1}{e^z}$$+$$\frac{zsinz}{e^z}$$.......2 Auxiliary equation of 2 is $$D_1^2-4D_1+1=0$$ $$\Longrightarrow D_1=2+\sqrt3,2-\sqrt3$$ Hence Complementary Function $$C.F(y_c).=e^{2z}[C_1cosh\sqrt3z+C_2sinh\sqrt3z]$$ $$=x^2[C_1cosh\sqrt3logx+C_2sinh\sqrt3logx]$$ where cosh and sinh are hyperbolic functions Particular integral$$(y_p)$$ corresponding to $$e^{-z}:$$ $$=\frac{e^{-z}}{D_1^2-4D_1+1}=\frac{e^{-z}}{1+4+1}=\frac{x^{-1}}{6}=\frac{1}{6x}$$ here we have put the coefficient of exponential function in place of $$D_1$$ in order to operate $$e^{-z}$$ Particular Integral corresponding to $$e^{-z}zsinz:$$ Now we take $$e^{-z}$$ outside the operator and replace $$D_1$$ by $$D_1-1$$ where $$-1$$ is the coefficient of $$e^{-z}$$. $$=\frac{e^{-z}zsinz}{D_1^2-4D_1+1}=e^{-z}\frac{1}{(D_1-1)^2-4(D_1-1)+1}zsinz=e^{-z}\frac{1}{(D_1^2-6D_1+6)}zsinz$$ $$=e^{-z}[z\frac{1}{D_1^2-6D_1+6}sinz-\frac{2D_1-6}{(D_1^2-6D_1+6)^2}sinz]$$ as $$\frac{1}{F(D)}xV=x\frac{1}{F(D)}V-\frac{F'(D)}{[F(D)]^2}V$$ $$=e^{-z}[z\frac{1}{-1-6D_1+6}sinz-(2D_1-6)\frac{1}{(-1-6D_1+6)^2}sinz]$$ $$=e^{-z}[z\frac{1}{5-6D_1}sinz-(2D_1-6)\frac{1}{(5-6D_1)^2}sinz]$$ $$=e^{-z}[z\frac{5+6D_1}{25-36D_1^2}sinz-(2D_1-6)\frac{1}{(25-60D_1+36D_1)^2}sinz]$$ Here we are putting $$-1$$ in place of $$D_1^2$$ as the angle of sinz has coefficient 1 $$=e^{-z}[z\frac{5+6D_1}{25+36}sinz-(2D_1-6)\frac{1}{(25-60D_1-36)}sinz]$$ $$=e^{-z}[z\frac{5sinz+6cosz}{61}+\frac{(2D_1-6)}{(11+60D_1)}sinz]$$ $$=e^{-z}[z\frac{5sinz+6cosz}{61}+\frac{(2D_1-6)(60D_1-11)}{(3600D_1^2-121)}sinz]$$ $$=e^{-z}[z\frac{5sinz+6cosz}{61}+\frac{(120D_1^2-382D_1+66)}{(-3600-121)}sinz]$$ $$=e^{-z}[z\frac{5sinz+6cosz}{61}+\frac{(120(-sinz)-382cosz+66sinz)}{-3721}]$$ $$=e^{-z}[z\frac{5sinz+6cosz}{61}+\frac{(54sinz+382cosz)}{3721}]$$ Now put back the value of $$x=e^z$$ $$=\frac{1}{x}[logx\frac{5sin(logx)+6cos(logx)}{61}+\frac{(54sin(logx)+382cos(logx))}{3721}]$$ Now the particular Integral would be $$y_p=\frac{1}{x}[logx\frac{5sin(logx)+6cos(logx)}{61}+\frac{(54sin(logx)+382cos(logx))}{3721}]+\frac{1}{6x}$$ Hence the general solution would be $$y(x)=y_c+y_p$$ $$=x^2[C_1cosh\sqrt3logx+C_2sinh\sqrt3logx]+\frac{1}{x}[logx\frac{5sin(logx)+6cos(logx)}{61}+\frac{(54sin(logx)+382cos(logx))}{3721}]+\frac{1}{6x}$$

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