Let . If and the sum of the diagonal elements of C is , where gcd (m, m) = 1, then m + n is : [2025]

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Solution

Q.
Let $A = [\begin{matrix} \frac{1}{\sqrt{2}} & - 2 \\ 0 & 1 \end{matrix}] and P = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}],$ $θ > 0 .$ . If $B = P A P^{T}, C = P^{T} B^{10} P$ and the sum of the diagonal elements of C is $\frac{m}{n}$ , where gcd (m, n) = 1, then m + n is : [2025]

1 127

2 2049

3 258

4 65

Ans.
(4)

We have, $P = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$

$P^{T} = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$

$∴ P P^{T} = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$

$= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I$ ... (i)

Now, $B = P A P^{T}$

Pre multiply by $P^{T}$ , we get

$P^{T} B = P^{T} P A P^{T} = A P^{T}$ [using (i)]

Now, post multiply by $P^{T}$ , we get

$P^{T} B P = A P^{T} P = A$

Now, $A^{2} = (P^{T} B P) (P^{T} B P) = P^{T} B^{2} P$

Similarly, $A^{10} = P^{T} B^{10} P = C$ [ $∵ C = P^{T} B^{10} P$ ]

Now, $A^{2} = [\begin{matrix} \frac{1}{\sqrt{2}} & - 2 \\ 0 & 1 \end{matrix}] [\begin{matrix} \frac{1}{\sqrt{2}} & - 2 \\ 0 & 1 \end{matrix}] = [\begin{matrix} {(\frac{1}{\sqrt{2}})}^{2} & - \sqrt{2} - 2 \\ 0 & 1 \end{matrix}]$

Similarly for $A^{3}, A^{4}, . . ., A^{10} = C$

So, sum of diagonal elements of $C = {(\frac{1}{\sqrt{2}})}^{10} + 1$

$= \frac{1}{32} + 1 = \frac{33}{32} = \frac{m}{n}$

$g c d (m, n) \Rightarrow g c d (33, 32) = 1$

So, m + n = 33 + 32 = 65.

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