JEE Main 2025 Jan 28 Shift 2, Mathematics Q10: Algebra Of Matrices

JEE Main2025Jan 28, Shift 2Mathematics

Let $A = \frac{1}{2} - 2 01$ and $P = [cos θ - sin θ sin θ cos θ]$ , $θ > 0$ . 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 $g cd (m, n) = 1$ , then $m + n$ is :

A
$65$
B
$127$
C
$258$
D
$2049$

Solution

$P$ is orthogonal ( $P^{T} P = I$ ). From $B = P A P^{T}$ we get $B^{k} = P A^{k} P^{T}$ for all $k \geq 1$ , hence $C = P^{T} B^{10} P = A^{10}$ .

Now $A$ is upper triangular with diagonal entries $\frac{1}{2}$ and $1$ . For an upper-triangular matrix, $A^{n}$ remains upper triangular with diagonal entries equal to the $n$ -th powers of $A$ 's diagonal entries.

So the diagonal of $A^{10}$ is $(\frac{1}{2})^{10} = \frac{1}{32}$ and $1^{10} = 1$ .

Trace $C = \frac{1}{32} + 1 = \frac{33}{32}$ . With $g cd (33, 32) = 1$ , $m + n = 33 + 32 = 65$ .

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JEE Main 2025 Jan 28 Shift 2, Mathematics Q10: Algebra Of Matrices

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