Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{cc}1+i& 1\\ -i& 0\end{array}\right]$ where $i=\sqrt{-1}$. Then, the number of elements in the set $\{n\in \{1,2,\dots ,100\}:A^n=A\}$ is

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a matrix problem; we are uncovering a hidden rhythm.

We are given a matrix $A=\left[\begin{array}{cc}1+i& 1\\ -i& 0\end{array}\right]$ and asked to find the number of integers $n$ between $1$ and $100$ such that $A^n=A$.

This is not a brute-force calculation; it is a search for periodicity. Imagine the matrix as a dancer on a stage, repeating its steps in a cycle. Our goal is to find how many times this dancer returns to its starting position within a hundred steps.

Invertibility

Before we start multiplying matrices, we must ask: is this matrix invertible? The determinant is our gatekeeper.

Calculating the determinant of $A$, we get:

Since $i eq 0$, the matrix $A$ is invertible. This is a massive relief! It means we can multiply both sides of our equation $A^n=A$ by $A^{-1}$ without any fear.

This transforms our problem into $A^n\cdot A^{-1}=A\cdot A^{-1}$, which simplifies to $A^{n-1}=I$. Now, the problem is reduced to finding when the power of $A$ becomes the identity matrix $I$.

We need to find the period. Let's start by calculating $A^2$.

Multiplying $A$ by itself:

Performing the row-by-column multiplication: - The element $(1,1)$ is $(1+i{)}^2+1(-i)=(1+2i-1)-i=i$. - The element $(1,2)$ is $(1+i)(1)+1(0)=1+i$. - The element $(2,1)$ is $-i(1+i)+0(-i)=-i-i^2=1-i$. - The element $(2,2)$ is $-i(1)+0(0)=-i$.

So, $A^2=\left[\begin{array}{cc}i& 1+i\\ 1-i& -i\end{array}\right]$. This is not the identity matrix, so the cycle continues.

Instead of calculating $A^3$, let's be strategic and calculate $A^4$ by squaring $A^2$:

Let's compute the elements carefully: - The top-left element is $(i)(i)+(1+i)(1-i)=i^2+(1-i^2)=-1+2=1$. - The top-right element is $(i)(1+i)+(1+i)(-i)=i(1+i)-i(1+i)=0$. - The bottom-left element is $(1-i)(i)+(-i)(1-i)=i(1-i)-i(1-i)=0$. - The bottom-right element is $(1-i)(1+i)+(-i)(-i)=(1-i^2)+i^2=2-1=1$.

Behold! $A^4=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$. We have found the period!

Since $A^4=I$, the condition $A^{n-1}=I$ implies that $n-1$ must be a multiple of $4$. Let $n-1=4k$, where $k$ is an integer.

Thus, $n=4k+1$. We are given $1\le n\le 100$, so:

Subtracting $1$ gives $0\le 4k\le 99$, and dividing by $4$ gives $0\le k\le 24.75$. Since $k$ must be an integer, $k$ can take values from $0$ to $24$.

That is $24-0+1=25$ values. There are exactly 25 such integers $n$. You have successfully navigated the cycle and found the answer.