Let S = {m ∈ Z : A^{m^2} + A^m = 3I - A^{-6}} where A = [2, -1; 1, 0]. Then n(S) is equal to

Q: Wait, why do we reject the negative root here? Or do we?

Actually, we do not reject any roots! The question asks for the number of integers $m$ that satisfy the equation. Our quadratic equation $m^2 + m - 6 = 0$ gave us $m = -3$ and $m = 2$. Both are integers, so both are valid solutions. Always check if the question specifies 'positive integers' or just 'integers'!

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure linear algebra problem, not physics! While it looks like a system of equations, the 'energy' here is the matrix structure. The 'conservation' is the Cayley-Hamilton theorem, which keeps the matrix powers within the span of $I$ and $N$.

Q: Is $N$ always nilpotent? How do I know?

A matrix $N$ is nilpotent if $N^k = O$ for some integer $k$. In this problem, we found $N = A - I$ and showed $N^2 = O$. This happened because the characteristic equation of $A$ was $(\lambda - 1)^2 = 0$. If the eigenvalues were different, $N$ might not be nilpotent. Always check the characteristic equation first!

Question

Let $S=\{m\in Z:A^{m^2}+A^{m}=3I-A^{-6}\}$ where $A=\begin{bmatrix}2&-1\ 1&0\end{bmatrix}$. Then $n(S)$ is equal to

Options:

(A) 1

(B) 2

(C) 3

(D) 0

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, intimidating matrix equation: 
$$A^{m^2} + A^m = 3I - A^{-6}$$

You might be tempted to start multiplying the matrix $A = \begin{bmatrix} 2 & -1 \ 1 & 0 \end{bmatrix}$ by itself over and over. But stop! In the world of JEE Advanced, brute force is rarely the intended path. There is a hidden elegance here, a secret structure waiting to be uncovered.

### The Master Key: The Characteristic Equation

Whenever you see high powers of a matrix, your brain should immediately pivot to the characteristic equation. It is the DNA of the matrix.

We find it by solving $\det(A - \lambda I) = 0$. Let us set up the determinant:
$$\det \begin{bmatrix} 2 - \lambda & -1 \ 1 & -\lambda \end{bmatrix} = 0$$

Expanding this, we get $(2 - \lambda)(-\lambda) - (-1)(1) = 0$, which simplifies to $\lambda^2 - 2\lambda + 1 = 0$. This is a perfect square: $(\lambda - 1)^2 = 0$. Our eigenvalues are $1$ and $1$. This repeated eigenvalue is our first clue that something special is happening.

### The Magic of Cayley-Hamilton

The Cayley-Hamilton theorem tells us that every square matrix satisfies its own characteristic equation. So, $(A - I)^2 = O$, where $O$ is the null matrix.

Let us define a new matrix $N = A - I$. Calculating this, we get:
$$N = \begin{bmatrix} 2 - 1 & -1 \ 1 & 0 - 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \ 1 & -1 \end{bmatrix}$$

From our theorem, we know $N^2 = O$. This is the definition of a nilpotent matrix!

### The Binomial Collapse

Now, the magic happens. We can write $A = I + N$. If we want to find $A^k$, we can use the binomial expansion:
$$A^k = (I + N)^k = I + kN + \frac{k(k-1)}{2}N^2 + \dots$$

But wait! Since $N^2 = O$, all terms involving $N^2, N^3, \dots$ vanish into thin air. The entire infinite series collapses into just $A^k = I + kN$.

This works for any integer $k$, even negative ones! So, we have:
$$A^{m^2} = I + m^2 N$$
$$A^m = I + mN$$
$$A^{-6} = I - 6N$$

### The Final Victory

Now, let us substitute these back into our original, scary equation:
$$(I + m^2 N) + (I + mN) = 3I - (I - 6N)$$

Simplifying the left side, we get $2I + (m^2 + m)N$. Simplifying the right side, we get $2I + 6N$.

The $2I$ terms cancel out, leaving us with $(m^2 + m)N = 6N$. Since $N$ is not the null matrix, we can equate the coefficients:
$$m^2 + m = 6 \Rightarrow m^2 + m - 6 = 0$$

Factoring this quadratic, we get $(m + 3)(m - 2) = 0$. Thus, **$m = -3$ or $m = 2$**.

We have found our two integer solutions! The number of elements in our set $S$ is **$2$**. You see? With the right tools, even the most intimidating problems become a beautiful, logical dance.

Animated Solution for Mathematics - Matrices and Determinants: Let $S = {m \in Z : A^{m^{2}} + A^{m} = 3 I - A^{- 6}}$ where $A = [21 - 1 0]$ . Then $n (S)$ is equal to

Select Answer:

Visualized Solution

The Matrix Equation Setup

The Characteristic Equation

Solving for Eigenvalues

Cayley-Hamilton Theorem

Defining the Nilpotent Matrix $N$

Binomial Expansion of $A^{k}$

General Formula for $A^{k}$

Substituting into the Main Equation

Simplifying the Matrix Equation

Equating the Coefficients

Solving the Quadratic Equation

Determining the Number of Solutions

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Master Key

The Magic of Cayley-Hamilton

The Binomial Collapse

The Final Victory

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={m∈Z:Am2+Am=3I−A−6} where A=[21​−10​]. Then n(S) is equal to