Question

M = $\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$ , $\theta$ = $\frac{2\pi}{5}$ then M$^{2026}$ = ?

• A. M$^2$
• B. M
• C. I
• D. M$^{-1}$

Hint:

When asked to find a large power of a rotation matrix, first find its order 'n' (the smallest power for which $M^n = I$). Then, calculate the exponent modulo n. For an angle $\theta = \frac{p}{q} 2\pi$ (in lowest terms), the order of the matrix is 'q'. Here, $\theta = \frac{1}{5} 2\pi$, so the order is 5. We then just need to compute $2026 \pmod 5 = 1$, so $M^{2026} = M^1$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a 2x2 matrix M, which is a standard 2D rotation matrix, and we need to calculate a high power of this matrix, $M^{2026}$.
Step 2: Key Formula or Approach:
A property of 2D rotation matrices is that raising them to a power 'k' is equivalent to rotating by 'k' times the original angle. \[ M^k = \begin{pmatrix} \cos\theta & -\sin\theta
\sin\theta & \cos\theta \end{pmatrix}^k = \begin{pmatrix} \cos(k\theta) & -\sin(k\theta)
\sin(k\theta) & \cos(k\theta) \end{pmatrix} \] An alternative approach is to find the smallest integer 'n' for which $M^n = I$ (the identity matrix) and then use modular arithmetic on the exponent.
Step 3: Detailed Explanation:
Let's use the second approach, which is often simpler. We want to find the order of the matrix M. Let's find the smallest positive integer n such that $n\theta$ is a multiple of $2\pi$. \[ n \times \frac{2\pi}{5} = m \times 2\pi \] \[ \frac{n}{5} = m \] The smallest integer n that satisfies this is n=5 (which gives m=1). So, for n=5, the angle of rotation is $5\theta = 5 \times \frac{2\pi}{5} = 2\pi$. Let's calculate $M^5$: \[ M^5 = \begin{pmatrix} \cos(2\pi) & -\sin(2\pi)
\sin(2\pi) & \cos(2\pi) \end{pmatrix} = \begin{pmatrix} 1 & 0
0 & 1 \end{pmatrix} = I \] Now we can compute $M^{2026}$ by using this property. We can divide the exponent 2026 by 5. \[ 2026 = 5 \times 405 + 1 \] So we can rewrite the expression as: \[ M^{2026} = M^{5 \times 405 + 1} = M^{5 \times 405} \times M^1 = (M^5)^{405} \times M \] Since $M^5 = I$: \[ M^{2026} = (I)^{405} \times M = I \times M = M \] Step 4: Final Answer:
The matrix $M^{2026}$ is equal to the matrix M.