Mathematics - Matrices Question with Solution | TestHub

Given,

$A=\left[\begin{array}{cc}1& \frac{1}{51}\\ 0& 1\end{array}\right]$,

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

So, $B=MAN$

Now let $M=\left[\begin{array}{cc}1& 2\\ -1& -1\end{array}\right] \& N=\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]$

Now solving,

$\stackrel{\left(M\right)}{\left[\begin{array}{cc}1& 2\\ -1& -1\end{array}\right]} \stackrel{\left(N\right)}{\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\therefore MN=I=NM$

Now using, $B=MAN$ we get

$B^n={\left(MAN\right)}^n=\left(MAN\right) \left(MAN\right) .......\left(MAN\right)$

$\Rightarrow B^n=MA\underset{I}{\underbrace{NM}}A\underset{I}{\underbrace{NM}} .......\underset{I}{\underbrace{NM}}AN$

$\Rightarrow B^n=M\underset{n \text{times}}{\underbrace{AAAAA.....A}}N =M{A}^{n}N$

Now, using $A=\left[\begin{array}{cc}1& \frac{1}{51}\\ 0& 1\end{array}\right]$

 $\Rightarrow A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}0& \frac{1}{51}\\ 0& 0\end{array}\right]=I+E$

Now finding, $E^2=\left[\begin{array}{cc}0& \frac{1}{51}\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& \frac{1}{51}\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

 $\Rightarrow E^2=0$, so all higher power will also be zero,

$\therefore A^n={\left(I+E\right)}^n=I+nE+\underset{0}{\underset{⏝}{{}^n{C}_2{E}^2}}+\underset{0}{\underbrace{{}^n{C}_3{E}^3}}+\underset{0}{\underbrace{{}^n{C}_4{E}^4}}...........$

$\Rightarrow A^n=I+nE$

$\Rightarrow A^n=\left[\begin{array}{cc}1& \frac{n}{51}\\ 0& 1\end{array}\right]$

So, $B^n=M{A}^{n}N=\left[\begin{array}{cc}1& 2\\ -1& -1\end{array}\right] \left[\begin{array}{cc}1& \frac{n}{51}\\ 0& 1\end{array}\right]\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]$$\Rightarrow B^n= \left[\begin{array}{cc}1& \frac{n}{51}+2\\ -1& \frac{-n}{51}-1\end{array}\right]\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]$ 

$\Rightarrow B^n=\left[\begin{array}{cc}1+\frac{n}{51}& \frac{n}{51}\\ \frac{-n}{51}& 1-\frac{n}{51}\end{array}\right]$

$\sum _{n=1}^{50}{B}^n=\left[\begin{array}{cc}50+\frac{50·51}{2·51}& \frac{{\displaystyle 50·51}}{{\displaystyle 2·51}}\\ \frac{{\displaystyle -50·51}}{{\displaystyle 2·51}}& 50-\frac{{\displaystyle 50·51}}{{\displaystyle 2·51}}\end{array}\right]=\left[\begin{array}{cc}75& 25\\ -25& 25\end{array}\right]$

$\therefore$ Sum $=100$