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QJEEM23Maad24: JEE Main 2023 Mathematics - Class 12 JEEM23Maad24 Solution: Let . If , then the sum of all the elements of the matrix ∑ n = …

Let A=[1amp;1510amp;1]A=\left[\begin{array}{cc}1 & \frac{1}{51} \\ 0 & 1\end{array}\right] . If B=[1amp;2−1amp;−1]A[−1amp;−21amp;1]\mathrm{B}=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right] A\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right] , then the sum of all the elements of the matrix ∑n=150Bn is equal to

Multiple Choice Options

  • A 50
  • B 75
  • C 100
  • D 125

Step-by-Step Verified Solution

amp; Let C=[1amp;2−1amp;−1],D=[−1amp;−21amp;1]amp;DC=[1amp;2−1amp;−1][−1amp;−21amp;1]=[1amp;00amp;1]=I\begin{aligned}& \text { Let } C=\left[\begin{array}{cc}1 & 2 \\-1 & -1\end{array}\right], \mathrm{D}=\left[\begin{array}{cc}-1 & -2 \\1 & 1\end{array}\right] \\\\& \mathrm{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]=\mathrm{I}\end{aligned}

amp;B=CADamp;Bn=(CAD)(CAD)(CAD)…(CAD)⏟n-times \begin{aligned}& \mathrm{B}=\mathrm{CAD} \\\\& \mathrm{B}^{\mathrm{n}}=\underbrace{(\mathrm{CAD})(\mathrm{CAD})(\mathrm{CAD}) \ldots(\mathrm{CAD})}_{\text {n-times }}\end{aligned}

⇒Bn=CAnD\Rightarrow \mathrm{B}^{\mathrm{n}}=\mathrm{CA}^{\mathrm{n}} \mathrm{D}

A2=[1amp;1510amp;1][1amp;1510amp;1]=[1amp;2510amp;1]\mathrm{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]

A3=[1amp;1510amp;1][1amp;2510amp;1]=[1amp;3510amp;1]\mathrm{A}^3=\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]=\left[\begin{array}{cc}1 & \frac{3}{51} \\0 & 1\end{array}\right]

 Similarly An=[1amp;n510amp;1]\text { Similarly } A^{\mathrm{n}}=\left[\begin{array}{cc}1 & \frac{\mathrm{n}}{51} \\0 & 1\end{array}\right]

Bnamp;=[1amp;2−1amp;−1][1amp;n510amp;1][−1amp;−21amp;1]amp;=[1amp;n51+2−1amp;−n51−1][−1amp;−21amp;1]amp;=[n51+1amp;n51−n51amp;1−n51]\begin{aligned}\mathrm{B}^{\mathrm{n}} & =\left[\begin{array}{cc}1 & 2 \\-1 & -1\end{array}\right]\left[\begin{array}{cc}1 & \frac{\mathrm{n}}{51} \\0 & 1\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\1 & 1\end{array}\right] \\\\& =\left[\begin{array}{cc}1 & \frac{\mathrm{n}}{51}+2 \\-1 & -\frac{\mathrm{n}}{51}-1\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\1 & 1\end{array}\right] \\\\& =\left[\begin{array}{cc}\frac{\mathrm{n}}{51}+1 & \frac{\mathrm{n}}{51} \\-\frac{\mathrm{n}}{51} & 1-\frac{\mathrm{n}}{51}\end{array}\right]\end{aligned}

∑n=150Bn=[50+50⋅512⋅51amp;50⋅512⋅51−50⋅512⋅51amp;50−50⋅512⋅51]=[75amp;25−25amp;25]\sum\limits_{n=1}^{50} B^n=\left[\begin{array}{cc}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{array}\right]=\left[\begin{array}{cc}75 & 25 \\-25 & 25\end{array}\right]

∴\therefore Sum of the elements = 100

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