Let A = [1/√2, -2; 0, 1] and P = [cos θ, -sin θ; sin θ, cos θ], θ 0. If B = PAP^T, C = P^T B^{10}

Q: Wait, why is $P$ orthogonal? Is this always true for rotation matrices?

Yes! A matrix $P$ is orthogonal if $P^T P = I$. The given matrix $P = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ is a standard rotation matrix. It preserves lengths and angles, making it inherently orthogonal. This is a fundamental property you should memorize for JEE Advanced.

Q: Can't we just multiply $A$ ten times to find $A^{10}$?

You could, but you would be trapped in a calculation nightmare! In JEE Advanced, we look for patterns. The similarity transformation $B = PAP^T$ is designed to simplify the matrix power. Always look for the structure before you start calculating. The beauty of this problem lies in the cancellation of $P$ and $P^T$.

Q: Why is the trace of $A^{10}$ the sum of the eigenvalues raised to the power of 10?

Great question! If $\lambda$ is an eigenvalue of $A$, then $\lambda^k$ is an eigenvalue of $A^k$. Since the trace of any matrix is the sum of its eigenvalues, the trace of $A^{10}$ is simply the sum of the 10th powers of the eigenvalues of $A$. It is a powerful shortcut that saves you from tedious matrix multiplication.

Question

Let $A=\begin{bmatrix}\frac{1}{\sqrt{2}}&-2\ 0&1\end{bmatrix}$ and $P=\begin{bmatrix}\cos~	heta&-\sin~	heta\ \sin~	heta&\cos~	heta\end{bmatrix}$, $	heta>0$. If $B=PAP^{T}$, $C=P^{T}B^{10}P$ and the sum of the diagonal elements of C is $\frac{m}{n}$ where $\gcd(m,n)=1,$ then $m+n$ is:

Options:

(A) 127

(B) 258

(C) 65

(D) 2049

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are a conductor of an orchestra, and the matrices $A$ and $P$ are your instruments. At first glance, the expression $C = P^T B^{10} P$ with $B = PAP^T$ looks like a chaotic, complex sequence of transformations.

In the world of linear algebra, there is a hidden rhythm, a beautiful symmetry waiting to be uncovered. Let us embark on this journey together.

### The Orthogonal Key

First, look at matrix $P = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$. This is not just any matrix; it is a rotation matrix.

Its most profound property is that it is orthogonal, meaning $P^T P = I$. This is our golden ticket.

Whenever you see a matrix sandwiched between its transpose and itself, like $P^T \dots P$, think of it as a simplification waiting to happen.

### The Power Chain

We are given $B = PAP^T$. We need to find $B^{10}$. If we try to compute $B^{10}$ directly, we will be lost in a sea of indices.

Instead, let us look at the pattern. $B^2 = (PAP^T)(PAP^T)$. Because of matrix associativity, this becomes $PA(P^T P)AP^T$.

Since $P^T P = I$, this simplifies to $PA^2 P^T$. If we continue this, $B^3 = PA^3 P^T$, and by induction:

$$B^{10} = PA^{10} P^T$$

We have successfully tamed the beast!

### The Simplification

Now, let us look at $C = P^T B^{10} P$. Substituting our expression for $B^{10}$, we get $C = P^T (PA^{10} P^T) P$.

Regrouping the terms, we see $C = (P^T P) A^{10} (P^T P)$. Since $P^T P = I$, the entire expression collapses into:

$$C = A^{10}$$

The scary-looking matrix $C$ is just $A^{10}$ in disguise.

### The Eigenvalue Shortcut

Now, we need the sum of the diagonal elements of $C$, which is the trace of $A^{10}$. Do not reach for your pen to multiply $A$ ten times!

Matrix $A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \ 0 & 1 \end{bmatrix}$ is an upper triangular matrix. The eigenvalues of an upper triangular matrix are simply its diagonal elements:

$$\lambda_1 = \frac{1}{\sqrt{2}}, \quad \lambda_2 = 1$$

The eigenvalues of $A^{10}$ are:

$$\lambda_1^{10} = \left( \frac{1}{\sqrt{2}} ight)^{10} = \frac{1}{32}, \quad \lambda_2^{10} = 1^{10} = 1$$

The trace of $A^{10}$ is the sum of its eigenvalues:

$$	ext{Tr}(A^{10}) = \frac{1}{32} + 1 = \frac{33}{32}$$

Thus, $m = 33$ and $n = 32$. The final answer, $m + n$, is:

**$33 + 32 = 65$**

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [\frac{1}{2} 0 - 2 1]$ 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:

Select Answer:

Visualized Solution

Introduction to Matrices $A$ and $P$

Identifying the Orthogonal Nature of $P$

Analyzing the Matrix $B$

Calculating $B^{2}$

Generalizing to $B^{10}$

Substituting $B^{10}$ into Matrix $C$

Simplifying Matrix $C$

Eigenvalues of Matrix $A$

Eigenvalues of $A^{10}$

Calculating the Trace of $C$

Finding $m + n$

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Orthogonal Key

The Power Chain

The Simplification

The Eigenvalue Shortcut

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Introduction to Matrices .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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