Animated Solution for Mathematics - Matrices and Determinants: The number of elements in the set $\{A=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right):a,b,d\in \{-1,0,1\}\text{ and }(I-A{)}^3=I-A^3\}$, where $I$ is $2\times 2$ identity matrix, is :

Welcome, fellow traveler on the road to JEE Advanced mastery. Today, we are going to dissect a problem that, at first glance, looks like a standard matrix exercise but hides a beautiful, elegant structure beneath its surface.

We are dealing with a $2\times 2$ matrix $A=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$, where the elements $a,b,d$ are restricted to the set $\{-1,0,1\}$. The condition given is $(I-A{)}^3=I-A^3$.

Many students, in their haste, might look at $(I-A{)}^3=I-A^3$ and think, "Oh, this is just like the scalar identity $(x-y{)}^3=x^3-y^3$." But stop! In the world of matrices, we must be careful.

The binomial expansion is only valid if the matrices commute. Fortunately, the identity matrix $I$ is the best friend of any matrix $A$; it commutes with everything! So, we can safely expand the left-hand side using the binomial theorem:

Since $I^n=I$ and $IA=AI=A$, this simplifies to:

Now, let's set this equal to the right-hand side of our original equation, $I-A^3$. The equation becomes:

Look at the symmetry here. We have an $I$ on both sides and a $-A^3$ on both sides. They cancel out with such satisfying precision, leaving us with:

Dividing by $3$, we arrive at the core of the problem: $A^2=A$. This is the definition of an idempotent matrix. It is a matrix that, when multiplied by itself, remains unchanged.

Now that we know $A^2=A$, we need to find the specific values of $a,b,$ and $d$ that satisfy this. Let's compute $A^2$ explicitly:

Equating $A^2$ to $A$, we get three equations:

1. $a^2=a$ 2. $d^2=d$ 3. $ab+bd=b$

From the first two, $a^2-a=0$ implies $a(a-1)=0$, so $a\in \{0,1\}$. Similarly, $d\in \{0,1\}$. We reject $-1$ because $(-1)^2 = 1 eq -1$. This gives us $2\times 2=4$ possible pairs for $(a,d)$.

Now, consider the third equation: $ab+bd=b$. Rearranging, we get $b(a+d-1)=0$. This is where the problem branches into two cases.

If $b=0$, the equation $b(a+d-1)=0$ is satisfied for any $a$ and $d$. Since $a$ can be $0$ or $1$ and $d$ can be $0$ or $1$, we have $2\times 2=4$ possible matrices.

If $b eq 0$, then $b$ must be $1$ or $-1$ (2 choices). For the equation to hold, we must have $a+d-1=0$, or $a+d=1$.

Given $a,d\in \{0,1\}$, the pairs $(a,d)$ that satisfy this are $(0,1)$ and $(1,0)$. That is 2 pairs. So, we have $2\text{ (choices for }b)\times 2\text{ (pairs for }a,d)=4$ possible matrices.

Adding the results from our two cases, $4+4=8$. We have found that there are exactly 8 such matrices.

This problem is a beautiful reminder that even in complex matrix algebra, the most powerful tool is often a simple, systematic case analysis. Keep practicing, keep visualizing, and most importantly, keep falling in love with the logic behind the math!