JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 15)

Let $$P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$$ and $$Q=P A P^{T}$$. If $$P^{T} Q^{2007} P=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$$, then $$2 a+b-3 c-4 d$$ equal to :
2004
2006
2007
2005

Explanation

$$ \text { Here, } P=\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{array}\right], A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] $$

$$ \text { Here, } \mathrm{PP}^{\mathrm{T}}=\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{array}\right]\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{-1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array}\right] $$

$$ =\left[\begin{array}{cc} \frac{3}{4}+\frac{1}{4} & -\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \\ \frac{-\sqrt{3}}{4}+\frac{\sqrt{3}}{4} & \frac{1}{4}+\frac{3}{4} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\mathrm{I} $$

Similarly $P^T P=1$

$$ \because $$ $$Q=P A P^{T}$$

Now, $Q^{2007}=\left(P A P^T\right)\left(P A P^T\right) \ldots 2007$ times $=P A^{2007} P^T$

$$ \begin{aligned} & \text { As, } A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] \\\\ & \Rightarrow A^2=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1+0 & 1+1 \\ 0+0 & 0+1 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right] \end{aligned} $$

$$ A^3=\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] $$
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$$ A^{2007}=\left[\begin{array}{cc} 1 & 2007 \\ 0 & 1 \end{array}\right] $$

$$ \begin{aligned} & \text { Hence, } \mathrm{P}^{\mathrm{T}} \mathrm{Q}^{2007} \mathrm{P}=\mathrm{A}^{2007}=\left[\begin{array}{cc} 1 & 2007 \\ 0 & 1 \end{array}\right] \\\\ & \Rightarrow\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} 1 & 2007 \\ 0 & 1 \end{array}\right] \end{aligned} $$

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

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