Exam

JEE Main

Year

2025

Date

28 January

Shift

Evening Shift

Subject

Mathematics

Chapter

Matrices

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Matrices

Let A=[ array cc 1 2 & -2 0 & 1 array ] and P=[ array cc θ & - θ θ...

Let A=\left[$\begin{array}{cc}\frac{1}{\sqrt{2}} & -2 \\ 0 & 1\end{array}$\right] andP=\left[$$\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}$$\right], \theta\gt0$. If$\mathrm{B}=\mathrm{PAP}^{\mathrm{T}}, $\mathrm{C}=\mathrm{P}$^{\mathrm{T}} \mathrm{B}^{10} \mathrm{P} and the sum of the diagonal elements of $C$ is $\frac{\mathrm{m}}{\mathrm{n}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is :

Step-by-step solution

$$\begin{aligned} & \mathrm{P}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ & \because \mathrm{P}^{\mathrm{T}} \mathrm{P}=\mathrm{I} \\ & \mathrm{~B}=\mathrm{PAPT} \end{aligned}$$ Pre multiply by $\mathrm{P}^{\mathrm{T}}$ ( Given) $\mathrm{P}^{\mathrm{T}} \mathrm{~B}=\mathrm{P}^{\mathrm{T}} \mathrm{P} \mathrm{AP}^{\mathrm{T}}=\mathrm{AP}^{\mathrm{T}}$ Now post multiply by P $\mathrm{P}^{\mathrm{T}}$\mathrm{BP}=\mathrm{AP}$^{\mathrm{T}} \mathrm{P}=\mathrm{A}$
$\mathrm{A}^2=\mathrm{P}^{\mathrm{T}} \mathrm{~B}^2 \mathrm{P}$ Similarly $A^{10}=P^T B^{10} P=C$$\begin{aligned} & A=\left[\begin{array}{cc} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{array}\right] \text{ (Given) } \\ & \Rightarrow A^2=\left[\begin{array}{cc} \frac{1}{2} & -\sqrt{2}-2 \\ 0 & 1 \end{array}\right] \end{aligned}$$Similarly check$A^3$and so on since$C=A^{10} \Rightarrow$Sum of diagonal elements of C is \left(\frac{1}{\sqrt{2}}\right)^{10}+1$ $$\begin{aligned} & =\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \operatorname{g~cd}(\mathrm{m}, \mathrm{n})=1(\text{ Given }) \\ & \Rightarrow \mathrm{m}+\mathrm{n}=65\end{aligned}$$