'HIPPOPOTAMUS' শব্দটির বর্ণগুলো থেকে 1টি স্বরবর্ণ ও 2টি ব্যঞ্জনবর্ণ নিয়ে কতগুলো শব্দ গঠন করা সম্ভব যেন স্বরবর্ণটি সবসময় মাঝখানে থাকে।

$A=\left[\begin{array}{cc}2& 1\\ 3& 2\end{array}\right]$

$\Rightarrow \left|A\right|=\left|\begin{array}{cc}2& 1\\ 3& 2\end{array}\right|=4-5=-1$

$\therefore A^{-1}=\frac{1}{1}\left[\begin{array}{cc}2& -1\\ -3& 2\end{array}\right]=\left[\begin{array}{cc}2& -1\\ -3& 2\end{array}\right]$

$C=\left[\begin{array}{cc}-2& 1\\ 5& -2\end{array}\right]$

$\Rightarrow \left|C\right|=\left|\begin{array}{cc}-2& 1\\ 5& -2\end{array}\right|=4-5=-1$

$\therefore C^{-1}=\frac{1}{-1}\left[\begin{array}{cc}-2& -1\\ -5& -2\end{array}\right]=\left[\begin{array}{cc}2& 1\\ 5& 2\end{array}\right]$

এখন, $ABC=\left[\begin{array}{cc}-2& 4\\ 3& -1\end{array}\right]$

$\Rightarrow B=A^{-1}\left[\begin{array}{cc}-2& 4\\ 3& -1\end{array}\right]C^{-1}$

$\Rightarrow B=\left[\begin{array}{cc}2& -1\\ -3& 2\end{array}\right]\left[\begin{array}{cc}-2& 4\\ 3& -1\end{array}\right]\left[\begin{array}{cc}2& 1\\ 5& 2\end{array}\right]$

$\Rightarrow B=\left[\begin{array}{cc}-4-3& 8+1\\ 6+6& -12-2\end{array}\right]\left[\begin{array}{cc}2& 1\\ 5& 2\end{array}\right]$

$\Rightarrow B=\left[\begin{array}{cc}-7& 9\\ 12& -14\end{array}\right]\left[\begin{array}{cc}2& 1\\ 5& 2\end{array}\right]$

$\Rightarrow B=\left[\begin{array}{cc}-4+45& -7+18\\ 24-70& 12-38\end{array}\right]$

$\therefore B=\left[\begin{array}{cc}31& 11\\ -46& -16\end{array}\right]$

$\therefore \left|\begin{array}{ccc}1& 1& 1\\ 2& 0& -4\\ 1& \lambda & 3\end{array}\right|=0$

$\Rightarrow 4\lambda -10+2\lambda =0 \therefore \lambda =\frac{5}{3}$

লম্ব একক ভেক্টর = $=\pm \frac{\overrightarrow{a}\times \overrightarrow{b}}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}$

$\therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& 0& -4\end{array}\right|=\hat{i}\left(-4-0\right)-\hat{j}\left(-4-2\right)+\hat{k}\left(0-2\right)$

                                            $=-4\hat{i}+6\hat{j}-2\hat{k}$

∴ লম্ব একক ভেক্টর $=\pm \frac{\overrightarrow{a}\times \overrightarrow{b}}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}=\pm \frac{-4\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{{\left(-4\right)}^2+6^2+{\left(-2\right)}^2}}$

$=\pm \frac{-4\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{56}}=\pm \frac{-4\hat{i}+6\hat{j}-2\hat{k}}{2\sqrt{14}}=\pm \frac{2\hat{i}-3\hat{j}+\hat{k}}{\sqrt{14}}$

y = ae^x + be^-x ⇒ $\frac{dy}{dx}=$ ae^x - be^-x ⇒ $\frac{d^{2}y}{d{x}^2}=$ ae^x + be^-x

গুরুমান ও লঘুমানের জন্য, $\frac{dy}{dx}=0$

$\Rightarrow a{e}^x-\frac{b}{e^x}=0\Rightarrow \frac{a{\left(e^x\right)}^2-b}{e^x}=0\Rightarrow {\left(e^x\right)}^2=\frac{b}{a} \therefore e^x=\pm \sqrt{\frac{b}{a}}$

y² = 4ax এবং y =2x হতে পাই,

4x² = 4ax ⇒ x² - ax = 0 ⇒ x(x - a) = 0

∴ x = 0 এবং x = a যা সীমা নির্দেশ করে। 

$Z=-1+\sqrt{3i}$

এখানে, $r=\sqrt{{\left(-1\right)}^2+{\left(\sqrt{3}\right)}^2}=2; \theta =\pi -{\tan }^{-1}\sqrt{3}=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$

$\therefore Z=2\left(\cos \frac{2\pi }{3}+i \sin \frac{2\pi }{3}\right)$

যা, $Z=-1+\sqrt{3i}$ এর পোলার আকার 

এখন $\left|Z\right|=r=2, Argz=\theta =\frac{2\pi }{3}$ Ans.