JEE Main 2023 — Mathematics Algebra

Given,

$A=[\begin{matrix}1 & \frac{1}{51} \\ 0 & 1\end{matrix}]$A=[10​511​1​],

$$B=[\begin{matrix}1 & 2 \\ -1 & -1\end{matrix}]A[\begin{matrix}-1 & -2 \\ 1 & 1\end{matrix}]$$B=[1−1​2−1​]A[−11​−21​]

So, $B=MAN$B=MAN

Now let M=[\begin{matrix}1 & 2 \\ -1 & -1\end{matrix}]&N=[\begin{matrix}-1 & -2 \\ 1 & 1\end{matrix}]

Now solving,

$$\overset{(M)}{[\begin{matrix}1 & 2 \\ -1 & -1\end{matrix}]}\overset{(N)}{[\begin{matrix}-1 & -2 \\ 1 & 1\end{matrix}]}=[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}]$$[1−1​2−1​](M)​[−11​−21​](N)​=[10​01​]

$$\therefore MN=I=NM$$∴MN=I=NM

Now using, $B=MAN$B=MAN we get

$${B}^{n}={(MAN)}^{n}=(MAN)(MAN).......(MAN)$$Bn=(MAN)n=(MAN)(MAN).......(MAN)

$$\Rightarrow {B}^{n}=MA\underset{I}{\underset{⏟}{NM}}A\underset{I}{\underset{⏟}{NM}}.......\underset{I}{\underset{⏟}{NM}}AN$$⇒Bn=MAI⏟NM​​AI⏟NM​​.......I⏟NM​​AN

$$\Rightarrow {B}^{n}=M\underset{n\text{times}}{\underset{⏟}{AAAAA.....A}}N=M{A}^{n}N$$⇒Bn=Mntimes⏟AAAAA.....A​​N=MAnN

Now, using $A=[\begin{matrix}1 & \frac{1}{51} \\ 0 & 1\end{matrix}]$A=[10​511​1​]

$$\Rightarrow A=[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}]+[\begin{matrix}0 & \frac{1}{51} \\ 0 & 0\end{matrix}]=I+E$$⇒A=[10​01​]+[00​511​0​]=I+E

Now finding, ${E}^{2}=[\begin{matrix}0 & \frac{1}{51} \\ 0 & 0\end{matrix}][\begin{matrix}0 & \frac{1}{51} \\ 0 & 0\end{matrix}]=[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}]$E2=[00​511​0​][00​511​0​]=[00​00​]

$\Rightarrow {E}^{2}=0$⇒E2=0, so all higher power will also be zero,

$$\therefore {A}^{n}={(I+E)}^{n}=I+nE+\underset{0}{\underset{⏝}{{}^{n}{C}_{2}{E}^{2}}}+\underset{0}{\underset{⏟}{{}^{n}{C}_{3}{E}^{3}}}+\underset{0}{\underset{⏟}{{}^{n}{C}_{4}{E}^{4}}}...........$$∴An=(I+E)n=I+nE+0⏝nC2​E2​​+0⏟nC3​E3​​+0⏟nC4​E4​​...........

$$\Rightarrow {A}^{n}=I+nE$$⇒An=I+nE

$$\Rightarrow {A}^{n}=[\begin{matrix}1 & \frac{n}{51} \\ 0 & 1\end{matrix}]$$⇒An=[10​51n​1​]

So, ${B}^{n}=M{A}^{n}N=[\begin{matrix}1 & 2 \\ -1 & -1\end{matrix}][\begin{matrix}1 & \frac{n}{51} \\ 0 & 1\end{matrix}][\begin{matrix}-1 & -2 \\ 1 & 1\end{matrix}]$Bn=MAnN=[1−1​2−1​][10​51n​1​][−11​−21​]$\Rightarrow {B}^{n}=[\begin{matrix}1 & \frac{n}{51}+2 \\ -1 & \frac{-n}{51}-1\end{matrix}][\begin{matrix}-1 & -2 \\ 1 & 1\end{matrix}]$⇒Bn=[1−1​51n​+251−n​−1​][−11​−21​]

$$\Rightarrow {B}^{n}=[\begin{matrix}1+\frac{n}{51} & \frac{n}{51} \\ \frac{-n}{51} & 1-\frac{n}{51}\end{matrix}]$$⇒Bn=[1+51n​51−n​​51n​1−51n​​]

$$\sum _{n=1}^{50}{B}^{n}=[\begin{matrix}50+\frac{50\cdot 51}{2\cdot 51} & \frac{50\cdot 51}{2\cdot 51} \\ \frac{-50\cdot 51}{2\cdot 51} & 50-\frac{50\cdot 51}{2\cdot 51}\end{matrix}]=[\begin{matrix}75 & 25 \\ -25 & 25\end{matrix}]$$n=1∑50​Bn=[50+2⋅5150⋅51​2⋅51−50⋅51​​2⋅5150⋅51​50−2⋅5150⋅51​​]=[75−25​2525​]

$\therefore$∴ Sum $=100$=100