Related Formula
For orthogonal matrix P$P$, P^T P = P P^T = I$P^T P = P P^T = I$.
If B = PAP^T$B = PAP^T$, then:
B^k = (PAP^T)(PAP^T) (PAP^T) = PA^kP^T$$B^k = (PAP^T)(PAP^T)\dots(PAP^T) = PA^kP^T$$
Core Logic
Given C = P^T B¹⁰ P$C = P^T B^{10} P$. Substitute B¹⁰ = P A¹⁰ P^T$B^{10} = P A^{10} P^T$ into the expression:
C = P^T (P A¹⁰ P^T) P$$C = P^T (P A^{10} P^T) P$$
C = (P^T P) A¹⁰ (P^T P)$$C = (P^T P) A^{10} (P^T P)$$
Since P$P$ is an orthogonal rotation matrix, P^T P = I$P^T P = I$, meaning:
C = I · A¹⁰ · I = A¹⁰$$C = I \cdot A^{10} \cdot I = A^{10}$$
Therefore, the sum of diagonal elements of C$C$ is simply the trace of A¹⁰$A^{10}$.
Step 1: Analyze Powers of Upper Triangular Matrix A
Matrix A$A$ is upper triangular:
A = bmatrix 1√(2) & -2 0 & 1 bmatrix$$A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix}$$
For any upper triangular matrix, any integer power k$k$ preserves the main diagonal entries as simply the powers of the individual diagonal elements:
A¹⁰ = bmatrix ( 1√(2))¹⁰ & * 0 & 1¹⁰ bmatrix = bmatrix (1)/(32) & * 0 & 1 bmatrix$$A^{10} = \begin{bmatrix} \left(\frac{1}{\sqrt{2}}\right)^{10} & * \\ 0 & 1^{10} \end{bmatrix} = \begin{bmatrix} \frac{1}{32} & * \\ 0 & 1 \end{bmatrix}$$
Step 2: Calculate the Trace and sum m+n
Sum of diagonal elements = Trace(C) = Trace(A¹⁰) = (1)/(32) + 1 = (33)/(32)$$\text{Sum of diagonal elements } = \text{Trace}(C) = \text{Trace}(A^{10}) = \frac{1}{32} + 1 = \frac{33}{32}$$
Given (m)/(n) = (33)/(32)$\frac{m}{n} = \frac{33}{32}$ with (33, 32) = 1$\gcd(33, 32) = 1$:
m = 33, n = 32$$m = 33, \quad n = 32$$
m + n = 33 + 32 = 65$$m + n = 33 + 32 = 65$$
Pattern Recognition
Traces of matrices are invariant under cyclic permutations, so Tr(P^T B¹⁰ P) = Tr(P P^T B¹⁰) = Tr(B¹⁰)$\text{Tr}(P^T B^{10} P) = \text{Tr}(P P^T B^{10}) = \text{Tr}(B^{10})$. Furthermore, Tr(P A¹⁰ P^T) = Tr(A¹⁰)$\text{Tr}(P A^{10} P^T) = \text{Tr}(A^{10})$. This identity bypasses the need to evaluate any outer matrix multiplication.
Chapter Mix
Class 12 Mathematics: Matrices and Determinants