3 and 4 .Determinants and Matrices
hard

माना $\mathrm{A}=\left[\begin{array}{cc}1 & \frac{1}{51} \\ 0 & 1\end{array}\right]$ है। यदि $\mathrm{B}=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right] \mathrm{A}\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$ है, तो $\sum_{\mathrm{n}=1}^{50} \mathrm{~B}^{\mathrm{n}}$ के सभी अवयवों का योग है :

A

$100$

B

$50$

C

$75$

D

$125$

(JEE MAIN-2023)

Solution

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

$DC =\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]= I$

$\begin{aligned} & B = CAD \\ & B ^{ n }=\underbrace{( CAD )( CAD )( CAD ) \ldots( CAD )}_{ n -\text { times }}\end{aligned}$

$\Rightarrow B ^{ n }= CA ^{ n } D$

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

$A^3=\left[\begin{array}{cc}1 & \frac{3}{51} \\ 0 & 1\end{array}\right]$

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

$B ^{ n }=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right]\left[\begin{array}{cc}1 & \frac{ n }{51} \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$

$=\left[\begin{array}{cc}1 & \frac{ n }{51}+2 \\ -1 & -\frac{ n }{51}-1\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$

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

$\sum \limits_{ n =1}^{50} B ^{ n }=\left[\begin{array}{cc}25+50 & 25 \\ -25 & -25+50\end{array}\right]=\left[\begin{array}{cc}75 & 25 \\ -25 & 25\end{array}\right]$

Sum of the elements $=100$

Standard 12
Mathematics

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