JEE Main 2024 — 8 April, Shift 1 — Question 21

Let $A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$ We are given that the trace of $A^{13}$ is equal to $3^n$, and we need to find the value of $n$.

To solve this, we diagonalize $A$. First, find the eigenvalues of $A$:

$$\det(A - \lambda I) = \begin{vmatrix} 2 - \lambda & -1 \\ 1 & 1 - \lambda \end{vmatrix} = (2 - \lambda)(1 - \lambda) + 1 = \lambda^2 - 3\lambda + 3$$

Solving $\lambda^2 - 3\lambda + 3 = 0$, we get:

$$\lambda = \frac{3 \pm \sqrt{9 - 12}}{2} = \frac{3 \pm i\sqrt{3}}{2}$$

Let $\lambda_1 = \frac{3 + i\sqrt{3}}{2}$ and $\lambda_2 = \frac{3 - i\sqrt{3}}{2}$. These are complex conjugates.

Now the trace of $A^{13}$ is:

$$\text{tr}(A^{13}) = \lambda_1^{13} + \lambda_2^{13}$$

Let $\lambda_1 = re^{i\theta}$, $\lambda_2 = re^{-i\theta}$, where $r = |\lambda_1| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{3}$

Then:

$$\lambda_1^{13} + \lambda_2^{13} = r^{13}(\cos(13\theta) + i\sin(13\theta) + \cos(13\theta) - i\sin(13\theta)) = 2r^{13}\cos(13\theta)$$

So,$\text{tr}(A^{13}) = 2 \cdot (\sqrt{3})^{13} \cdot \cos(13\theta)$

Given that $\text{tr}(A^{13}) = 3^n$, and since $(\sqrt{3})^{13} = 3^{13/2}$, then:

$$2 \cdot 3^{13/2} \cdot \cos(13\theta) = 3^n$$

Taking logarithms base 3:

$$\log_3(2 \cdot 3^{13/2} \cdot \cos(13\theta)) = n \Rightarrow \log_3(2 \cos(13\theta)) + \frac{13}{2} = n$$

For $n$ to be an integer, $2 \cos(13\theta) = 3^{n - 13/2}$ must also be a power of 3. The only such case is when:

$$2 \cos(13\theta) = 3^{0.5} \Rightarrow \cos(13\theta) = \frac{\sqrt{3}}{2} \Rightarrow n = \frac{13}{2} + \frac{1}{2} = 7$$ $$\boxed{n = 7}$$