Let P = [ 3 2 1 2 - 1 2 3 2 ] , A = [ 1 1 0 1 ] and Q = P A P T . If P T Q 2007 P = [ a b c d ] , then 2 a + b - 3 c - 4 d is equal to [2023]

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Solution

Q.
Let $P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}], A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] and Q = P A P^{T} .$ If $P^{T} Q^{2007} P = [\begin{matrix} a & b \\ c & d \end{matrix}],$ then $2 a + b - 3 c - 4 d$ is equal to [2023]

1 2006

2 2004

3 2005

4 2007

Ans.
(3)

We have, $P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}], A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$Q = P A P^{T}$

Also, $P^{T} P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{- 1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}] [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{- 1}{2} & \frac{\sqrt{3}}{2} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$Q^{2007} = (P A P^{T}) (P A P^{T}) \dots (2007 times) = P A^{2007} P^{T} [∵ P^{T} P = I]$

$∴ P^{T} Q^{2007} P = P^{T} P A^{2007} P^{T} P = A^{2007}$

Now, $A^{2} = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$A^{3} = A \cdot A^{2} = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix}]$

Continuing in the same way, we get:

$A^{2007} = [\begin{matrix} 1 & 2007 \\ 0 & 1 \end{matrix}]$

$∴ P^{T} Q^{2007} P = [\begin{matrix} 1 & 2007 \\ 0 & 1 \end{matrix}] = [\begin{matrix} a & b \\ c & d \end{matrix}] (Given)$

$\Rightarrow a = 1, b = 2007, c = 0, d = 1$

$∴ 2 a + b - 3 c - 4 d = 2 (1) + 2007 - 3 (0) - 4 (1) = 2005$

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