Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{ccc}2& -1& -1\\ 1& 0& -1\\ 1& -1& 0\end{array}\right]$ and $B=A-I$. If $\omega =\frac{\sqrt{3}i-1}{2}$, then the number of elements in the set $\{n\in \{1,2,\dots ,100\}:A^n+(\omega B{)}^n=A+B\}$ is equal to

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dissect a problem that, at first glance, looks like a chaotic mess of matrix powers and complex numbers.

But beneath the surface, there is a profound elegance waiting to be uncovered. Let us begin with matrix:

When you first encounter a matrix raised to the power of $n$, your instinct might be to panic. But stop. Breathe. In JEE Advanced, whenever you see a matrix power, there is almost always a hidden pattern.

Let us calculate $A^2$. When you perform the multiplication $A\cdot A$, you will find that:

This is not a coincidence; it is the hallmark of an idempotent matrix. Once a matrix satisfies $A^2=A$, it is essentially 'fixed' in the space of matrix powers. By induction, $A^n=A$ for all $n\in \mathbb{N}$. We have just tamed the first beast of this problem.

Now, let us turn our attention to $B=A-I$. We know $A^2=A$, so let us see how $B$ behaves when squared.

Substituting our knowledge that $A^2=A$, we get:

But wait, $I-A$ is just $-B$! This is a beautiful discovery. If $B^2=-B$, then:

The powers of $B$ cycle with a period of two: $B^1=B$, $B^2=-B$, $B^3=B$, $B^4=-B$. We can generalize this as:

We have now reduced the entire matrix side of the equation to simple, manageable terms.

With our simplifications, the original equation $A^n+(\omega B{)}^n=A+B$ becomes much friendlier. Substituting $A^n=A$ and $B^n=(-1{)}^{n-1}B$, we get:

The $A$ on both sides cancels out instantly, leaving us with:

Comparing the coefficients, we arrive at the scalar condition:

This is where the complex number $\omega =\frac{\sqrt{3}i-1}{2}$ comes into play. Recall that $\omega$ is a cube root of unity, meaning ${\omega }^3=1$. Its powers cycle through $1,\omega ,{\omega }^2$.

We must now split our analysis into two cases based on the parity of $n$. If $n$ is odd, $(-1{)}^{n-1}=1$, so our equation becomes ${\omega }^n=1$.

Since ${\omega }^3=1$, $n$ must be a multiple of $3$. Given $n$ is odd, $n$ must be an odd multiple of $3$, such as $3,9,15,\dots ,99$.

If $n$ is even, $(-1{)}^{n-1}=-1$, so our equation becomes $-{\omega }^n=1$, or ${\omega }^n=-1$. As we discussed, the powers of $\omega$ only yield $1,\omega ,{\omega }^2$. None of these are $-1$, so this case yields no solutions.

We are left with the arithmetic progression $3,9,15,\dots ,99$. Using the formula for the $k$-th term of an AP, $a_k=a+(k-1)d$, we set:

Solving this, we find $96=6(k-1)$, which leads to $16=k-1$, and finally $k=17$.

There are exactly 17 values of $n$ that satisfy the condition. You see? What seemed like a terrifying matrix equation was just a beautiful, rhythmic dance of patterns. Keep looking for these patterns, and you will master any problem.