Verified Solution • 2025–26 Session moderate

QJEEM23Ma1d22: JEE Main 2023 Mathematics - Class 12 JEEM23Ma1d22 Solution: Let and . If , then equal to :

Let P=[32amp;12−12amp;32],A=[1amp;10amp;1]P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] and Q=PAPTQ=P A P^{T} . If PTQ2007P=[aamp;bcamp;d]P^{T} Q^{2007} P=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] , then 2a+b−3c−4d2 a+b-3 c-4 d equal to :

Multiple Choice Options

  • A 2004
  • B 2006
  • C 2007
  • D 2005

Step-by-Step Verified Solution

 Here, P=[32amp;12−12amp;32],A=[1amp;10amp;1]\text { Here, } P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\\frac{-1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\0 & 1\end{array}\right]

 Here, PPT=[32amp;12−12amp;32][32amp;−1212amp;32]\text { Here, } \mathrm{PP}^{\mathrm{T}}=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\\frac{-1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{-1}{2} \\\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]

=[34+14amp;−34+34−34+34amp;14+34]=[1amp;00amp;1]=I=\left[\begin{array}{cc}\frac{3}{4}+\frac{1}{4} & -\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \\\frac{-\sqrt{3}}{4}+\frac{\sqrt{3}}{4} & \frac{1}{4}+\frac{3}{4}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]=\mathrm{I}

Similarly PTP=1P^T P=1

∵\because Q=PAPTQ=P A P^{T}

Now, Q2007=(PAPT)(PAPT)…2007Q^{2007}=\left(P A P^T\right)\left(P A P^T\right) \ldots 2007 times =PA2007PT=P A^{2007} P^T

amp; As, A=[1amp;10amp;1]amp;⇒A2=[1amp;10amp;1][1amp;10amp;1]=[1+0amp;1+10+0amp;0+1]=[1amp;20amp;1]\begin{aligned}& \text { As, } A=\left[\begin{array}{ll}1 & 1 \\0 & 1\end{array}\right] \\\\& \Rightarrow A^2=\left[\begin{array}{ll}1 & 1 \\0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\0 & 1\end{array}\right]=\left[\begin{array}{ll}1+0 & 1+1 \\0+0 & 0+1\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\0 & 1\end{array}\right]\end{aligned}

A3=[1amp;30amp;1]A^3=\left[\begin{array}{ll}1 & 3 \\0 & 1\end{array}\right]
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A2007=[1amp;20070amp;1]A^{2007}=\left[\begin{array}{cc}1 & 2007 \\0 & 1\end{array}\right]

amp; Hence, PTQ2007P=A2007=[1amp;20070amp;1]amp;⇒[aamp;bcamp;d]=[1amp;20070amp;1]\begin{aligned}& \text { Hence, } \mathrm{P}^{\mathrm{T}} \mathrm{Q}^{2007} \mathrm{P}=\mathrm{A}^{2007}=\left[\begin{array}{cc}1 & 2007 \\0 & 1\end{array}\right] \\\\& \Rightarrow\left[\begin{array}{ll}a & b \\c & d\end{array}\right]=\left[\begin{array}{cc}1 & 2007 \\0 & 1\end{array}\right]\end{aligned}

amp;⇒a=1,b=2007,c=0,d=1amp;∴2a+b−3c−4d=2(1)+2007−3(0)−4(1)amp;=2+2007−4=2005\begin{aligned}& \Rightarrow a=1, b=2007, c=0, d=1 \\\\& \therefore 2 a+b-3 c-4 d=2(1)+2007-3(0)-4(1) \\\\& =2+2007-4=2005\end{aligned}

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