GATE ME 2009

\text{If }\quad A^{T}=A^{-1}
then A is orthogonal matrix.
Therefore A . A^{-1}=A^{-1} A=I
\text{and }A^{T} A=A A^{T}=I
since \mathrm{M} is orthogonal matrix
M^{T} M=I
=\left[\begin{array}{cc}\frac{3}{5} & x \\ \frac{4}{5} & \frac{3}{5}\end{array}\right]\left[\begin{array}{cc}\frac{3}{5} & \frac{4}{5} \\ x & \frac{3}{5}\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]
\left[\begin{aligned} \left(\frac{3}{5}\right)^{2}+x^{2} \quad\left(\frac{3}{5} \cdot \frac{4}{5}\right)+\frac{3}{5} x \\ \left(\frac{4}{5} \cdot \frac{3}{5}\right)+\frac{3}{5} \cdot x\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2} \end{aligned} \right]
=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \Rightarrow \text{Compare both sides} a_{12}
a_{12}=\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)+\left(\frac{3}{5}\right) x=0
\Rightarrow \quad \frac{3}{5} x=-\frac{3}{5} \cdot \frac{4}{5}
\Rightarrow x=-\frac{4}{5}

\text{Vector field,}
\vec{f}=3 x z \hat{i}+2 x y \hat{j}-y z^{2} \hat{k}
\vec{f}=\left[f_{1} \hat{i}+f_{2} \hat{j}+f_{3} \hat{k}\right]
\begin{aligned}&\text{Divergence of vector field} \vec{f}=\nabla \cdot \vec{f} \\ &\text{Div}(f)=\left(i \frac{\partial}{\partial x}+\hat{j} \frac{\partial}{\partial y}+\hat{k} \frac{\partial}{\partial z}\right) \cdot\left(t_{1} \hat{i}+f_{2} \hat{j}+f_{3} \hat{k}\right) \\ &\text{Div}(f)=\frac{\partial f_{1}}{\partial x}+\frac{\partial f_{2}}{\partial y}+\frac{\partial f_{3}}{\partial z} \\ &\text{Div}(f)=\frac{\partial}{\partial x}[3 x z]+\frac{\partial}{\partial y}[2 x y]+\frac{\partial}{\partial z}\left[-2 y z^{2}\right] \\ &\text{Div}(f)=3 z+2 x-2 z y \\ &\text{Div}(f)|_{(1,1,1)}=3(1)+2(1)-2(1)(1)=3 \end{aligned}

\begin{aligned} L^{-1}\left(\frac{1}{s^{2}+s}\right) &=? \\ \frac{1}{s^{2}+s} &=\frac{1}{s(s+1)}=\frac{1}{s}-\frac{1}{s+1} \\ L^{-1}\left(\frac{1}{s^{2}+s}\right) &=L^{-1}\left(\frac{1}{s}\right)--L^{-1}\left(\frac{1}{s+1}\right) \\ &=1-e^{-t} \end{aligned}
[Using standard formulae]
Standard formula:
\begin{aligned} L^{-1}\left(\frac{1}{s}\right) &=1 \\ L^{-1}\left(\frac{1}{s+a}\right) &=e^{-a t} \\ L^{-1}\left(\frac{1}{s-a}\right) &=e^{a t} \end{aligned}

Three coins are tossed simultaneously, the total
number of ways =2^{3}=8
Number of ways to get all tails = 1
Probability to get all tails
=\frac{\text { Number of ways for tails }}{\text { Total number of ways }}=\frac{1}{8}
Therefore probability to get at least one head is
=1-\frac{1}{8}=\frac{7}{8}
or, Total number of ways with at least one head is
\left.\begin{matrix} & &\left.\begin{array}{lll} H & T & T \\ T & H & T \\ T & T & H \end{array} \right] \\ &&\left.\begin{array}{lll} H & T & T \\ T & H & T \\ T & T & H \end{array} \right] \\ &&\left.\begin{array}{lll} H & H & H \end{array} \right] \end{matrix}\right\}
Therefore probability to get at least one head =\frac{7}{8}

If a closed system is undergoing an irreversible process, the entropy of the system can increase, decrease or remain constant