Let A=[\beginarraycc\frac1\sqrt2 & -2 0 & 1\endarray] and P=[\beginarraycc\cos \theta & -\sin \theta \sin \theta & \cos \theta\endarray], \theta\gt 0 . If B=PAP ^\top, C=P^\top B^10 P and the sum of the diagonal elements of C is \fracmn , where \operatornamegcd(m, n)=1 , then m+n is :

Matrices and Determinants PYQ — JEE Main 2025 · 28 Jan · Shift 2 · Q42 with Solution

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Correct answer: D

\qquad P=\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}$$ Also, $$B=PAP^T, \qquad C=P^T B^{10} P$$ We need the **sum of diagonal elements of** $C$, i.e. $\operatorname{tr}(C)$. --- 2. **Use similarity/orthogonal transformation property** Since $P$ is a rotation matrix, it is orthogonal, so $$P^T P = I$$ Now, $$B=PAP^T$$ Then $$B^{10}=(PAP^T)^{10}=PA^{10}P^T$$ because the middle factors collapse using $P^TP=I$. So, $$C=P^T B^{10} P = P^T(PA^{10}P^T)P = A^{10}$$ Hence, $$\operatorname{tr}(C)=\operatorname{tr}(A^{10})$$ Thus we only need the trace of $A^{10}$. --- 3. **Observe that $A$ is upper triangular** $$A=\begin{bmatrix}\frac{1}{\sqrt2} & -2\\ 0 & 1\end{bmatrix}$$ For any upper triangular matrix, its powers are also upper triangular, and the diagonal entries get raised to the same power. Therefore, $$A^{10} \text{ has diagonal entries } \left(\frac{1}{\sqrt2}\right)^{10} \text{ and } 1^{10}$$ So, $$\operatorname{tr}(A^{10})=\left(\frac{1}{\sqrt2}\right)^{10}+1$$ Now, $$\left(\frac{1}{\sqrt2}\right)^{10}=\frac{1}{(\sqrt2)^{10}}=\frac{1}{2^5}=\frac{1}{32}$$ Hence, $$\operatorname{tr}(C)=1+\frac{1}{32}=\frac{33}{32}$$ Thus, $$m=33,\quad n=32$$ and $$m+n=33+32=65$$ --- 4. **Check options** - A: $127$ - B: $2049$ - C: $258$ - D: $65$ So the correct option is $$\boxed{D}$$ --- 5. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.

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Correct answer: D

\qquad P=\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}$$ Also, $$B=PAP^T, \qquad C=P^T B^{10} P$$ We need the **sum of diagonal elements of** $C$, i.e. $\operatorname{tr}(C)$. --- 2. **Use similarity/orthogonal transformation property** Since $P$ is a rotation matrix, it is orthogonal, so $$P^T P = I$$ Now, $$B=PAP^T$$ Then $$B^{10}=(PAP^T)^{10}=PA^{10}P^T$$ because the middle factors collapse using $P^TP=I$. So, $$C=P^T B^{10} P = P^T(PA^{10}P^T)P = A^{10}$$ Hence, $$\operatorname{tr}(C)=\operatorname{tr}(A^{10})$$ Thus we only need the trace of $A^{10}$. --- 3. **Observe that $A$ is upper triangular** $$A=\begin{bmatrix}\frac{1}{\sqrt2} & -2\\ 0 & 1\end{bmatrix}$$ For any upper triangular matrix, its powers are also upper triangular, and the diagonal entries get raised to the same power. Therefore, $$A^{10} \text{ has diagonal entries } \left(\frac{1}{\sqrt2}\right)^{10} \text{ and } 1^{10}$$ So, $$\operatorname{tr}(A^{10})=\left(\frac{1}{\sqrt2}\right)^{10}+1$$ Now, $$\left(\frac{1}{\sqrt2}\right)^{10}=\frac{1}{(\sqrt2)^{10}}=\frac{1}{2^5}=\frac{1}{32}$$ Hence, $$\operatorname{tr}(C)=1+\frac{1}{32}=\frac{33}{32}$$ Thus, $$m=33,\quad n=32$$ and $$m+n=33+32=65$$ --- 4. **Check options** - A: $127$ - B: $2049$ - C: $258$ - D: $65$ So the correct option is $$\boxed{D}$$ --- 5. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.