Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]$. If the sum of the diagonal elements of $A^{13}$ is $3^n$, then $n$ is equal to_________

Imagine you are sitting in the examination hall. You see the matrix and the question asks for the sum of the diagonal elements of $A^{13}$.

Your first instinct might be to reach for your pen and start multiplying. $A^2$, then $A^3$, then $A^4$... stop! If you do that, you have fallen into the trap.

The JEE Advanced examiners are not testing your ability to perform tedious arithmetic; they are testing your ability to see the soul of the matrix. Let us embark on a journey to solve this not with brute force, but with the elegance of linear algebra.

The Characteristic Equation

The secret to understanding any matrix lies in its eigenvalues. These are the scalars that define how the matrix stretches space.

For any $2\times 2$ matrix, we do not need to perform complex row reductions. We have a beautiful shortcut: the characteristic equation

Let us calculate the trace and determinant of our matrix $A$. The trace, $\text{tr}(A)$, is the sum of the diagonal elements: $2+1=3$. The determinant, $\det (A)$, is $(2)(1)-(-1)(1)=3$.

Substituting these into our equation, we get This quadratic equation is the key that unlocks the entire problem.

Solving for $\lambda$, we use the quadratic formula:

We have arrived at two complex conjugate eigenvalues. Do not be intimidated by the $i$. In the world of matrices, complex eigenvalues often appear when the transformation involves a rotation.

To handle these, we must step out of the Cartesian plane and into the world of Euler. We convert these eigenvalues into polar form: $\lambda =r{e}^{\pm i\theta }$.

The modulus $r$ is

The argument $\theta$ is

Thus, our eigenvalues are $\sqrt{3}{e}^{\pm i\frac{\pi }{6}}$.

Now, we invoke the Spectral Mapping Theorem. We know that the trace of $A^{13}$ is simply the sum of the 13th powers of the eigenvalues:

Substituting our polar forms, we get

Factoring out $(\sqrt{3}{)}^{13}$, we are left with

This is where the magic happens. We know that $e^{i\theta }+e^{-i\theta }=2\cos (\theta )$. Our expression becomes

Let us simplify the terms. $(\sqrt{3}{)}^{13}$ is $3^{6.5}$, which is $3^6\cdot \sqrt{3}$.

The cosine term, $\cos (\frac{13\pi }{6})$, is equivalent to $\cos (2\pi +\frac{\pi }{6})$, which is simply $\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}$.

Putting it all together:

The $2$s cancel, and $\sqrt{3}\cdot \sqrt{3}$ becomes $3$. We are left with $3^6\cdot 3=3^7$.

The problem asks for $n$ where the trace is $3^n$. By comparing our result, we see clearly that $n=7$.

We did not multiply the matrix once. We used the power of eigenvalues, the beauty of Euler's formula, and the symmetry of complex conjugates to solve a problem that would have taken others twenty minutes of painful arithmetic. This is the JEE Advanced mindset: think first, calculate later.