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JEE Mathematics 2025 Question with Solution

Let A=[12201]A = \left[ \begin{array}{cc} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{array} \right] and P=[cosθsinθsinθcosθ],θ>0.P = \left[ \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right], \theta > 0. If B=PAPTB = P A P^T, C=PTBPC = P^T B P, and the sum of the diagonal elements of CC is mn\frac{m}{n}, where gcd(m,n)=1\gcd(m, n) = 1, then m+nm + n is:

  • A

    258258

  • B

    6565

  • C

    127127

  • D

    20492049

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: B=PAPTB = P A P^T and C=PTBPC = P^T B P, where PP is an orthogonal matrix, so PTP=IP^T P = I.

Find: The sum of diagonal elements of CC, written as mn\frac{m}{n}, and then compute m+nm+n.

From

C=PTBPC = P^T B P

and

B=PAPT,B = P A P^T,

we get

C=PT(PAPT)P=(PTP)A(PTP)=IAI=A.C = P^T (P A P^T) P = (P^T P) A (P^T P) = IAI = A.

Therefore, the sum of the diagonal elements of CC is the trace of AA:

tr(C)=tr(A)=12+1.\operatorname{tr}(C) = \operatorname{tr}(A) = \frac{1}{\sqrt{2}} + 1.

The solution states The Correct Option is B and concludes that m+n=65m+n=65, although the intermediate working shown there is inconsistent with the matrix written in the question. Using the answer indicated by the solution, the correct option is B.

Therefore, the correct option is B.

Using orthogonal similarity

The key property is that for the rotation matrix PP,

PTP=PPT=I.P^T P = PP^T = I.

So the two successive transformations undo each other when substituted into CC.

Substitute B=PAPTB = P A P^T into CC:

C=PTBP=PT(PAPT)P=(PTP)A(PTP)=A.\begin{aligned} C &= P^T B P \\ &= P^T (P A P^T) P \\ &= (P^T P) A (P^T P) \\ &= A. \end{aligned}

Hence the diagonal sum of CC equals the diagonal sum of AA. The provided the solution, however, explicitly marks B as the correct option. So the extracted answer is B, with the note that the displayed working on the page does not consistently match the question matrix.

Common mistakes

  • Assuming one must multiply all matrices explicitly. This is unnecessary because PP is orthogonal and PTP=IP^T P = I. Substitute BB into CC first and simplify before doing any long multiplication.

  • Using only trace invariance without first noticing that C=AC=A. While trace invariance is true, here the stronger identity C=PT(PAPT)P=AC=P^T(PAP^T)P=A makes the simplification immediate.

  • Trusting inconsistent intermediate working blindly. The solution contains a mismatch in the matrix entries and reasoning, so the algebra shown there should be checked against the original question before using it.

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