1. Core Concept: The Cayley-Hamilton Theorem and Matrix Powers

The problem involves powers of a $2 \times 2$ matrix $A$. For such matrices, the Cayley-Hamilton Theorem is an extremely powerful tool. It states that every square matrix satisfies its own characteristic equation. This theorem allows us to express higher powers of a matrix in terms of lower powers, often leading to a simple general form for $A^n$.

Given the matrix $A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}$.

First, we find the characteristic equation of $A$. The characteristic equation is given by $\det(A - \lambda I) = 0$, where $I$ is the identity matrix and $\lambda$ represents the eigenvalues.

$\det \left( \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) = 0$ $\det \left( \begin{bmatrix} 2-\lambda & -1 \\ 1 & -\lambda \end{bmatrix} \right) = 0$ $(2-\lambda)(-\lambda) - (-1)(1) = 0$ $-2\lambda + \lambda^2 + 1 = 0$ $\lambda^2 - 2\lambda + 1 = 0$ This is $(\lambda-1)^2 = 0$, meaning $\lambda=1$ is an eigenvalue with algebraic multiplicity 2.

By the Cayley-Hamilton Theorem, the matrix $A$ must satisfy this equation: $A^2 - 2A + I = O$ where $O$ is the zero matrix. This equation is fundamental as it allows us to simplify any power of $A$. From this, we can write: $A^2 = 2A - I$

2. Deriving a General Expression for $A^n$

Our goal is to find a general formula for $A^n$ that holds for all integers $n$. Let's look for a pattern using the relation $A^2 = 2A - I$.

• For $n=0$: $A^0 = I$
• For $n=1$: $A^1 = A$
• For $n=2$: $A^2 = 2A - I$ (from Cayley-Hamilton)
• For $n=3$: $A^3 = A \cdot A^2 = A(2A - I) = 2A^2 - A = 2(2A - I) - A = 4A - 2I - A = 3A - 2I$
• For $n=4$: $A^4 = A \cdot A^3 = A(3A - 2I) = 3A^2 - 2A = 3(2A - I) - 2A = 6A - 3I - 2A = 4A - 3I$

Observing the pattern, it appears that $A^n = nA - (n-1)I$ for $n \ge 1$. Let's check for $n=0$: $A^0 = 0A - (0-1)I = I$. This holds.

Now, let's verify this formula for negative integer powers. From $A^2 - 2A + I = O$, we can multiply by $A^{-1}$ (since $\det(A) = (2)(0) - (-1)(1) = 1 \ne 0$, $A^{-1}$ exists): $A - 2I + A^{-1} = O$ So, $A^{-1} = 2I - A$.

Let's check our conjectured formula $A^n = nA - (n-1)I$ for $n=-1$: $A^{-1} = (-1)A - (-1-1)I = -A - (-2)I = -A + 2I$. This matches perfectly!

The formula $A^n = nA - (n-1)I$ is valid for all integers $n \in \mathbf{Z}$. This is a crucial simplification for the problem.

Tip: For $2 \times 2$ matrices, finding a general form for $A^n$ using the Cayley-Hamilton Theorem is often the most efficient method, especially when $A$ is diagonalizable or has a simple characteristic polynomial like in this case.

3. Substituting into the Matrix Equation

The given matrix equation is $A^{m^2}+A^m=3 I-A^{-6}$. We will use our general formula $A^n = nA - (n-1)I$ for each term:

• For $A^{m^2}$: Replace $n$ with $m^2$. $A^{m^2} = m^2 A - (m^2-1)I$
• For $A^m$: Replace $n$ with $m$. $A^m = mA - (m-1)I$
• For $A^{-6}$: Replace $n$ with $-6$. $A^{-6} = (-6)A - (-6-1)I = -6A - (-7)I = -6A + 7I$

Now, substitute these expressions back into the original equation: $[m^2 A - (m^2-1)I] + [mA - (m-1)I] = 3I - [-6A + 7I]$

Combine terms on the left side: $(m^2+m)A - (m^2-1 + m-1)I = 3I + 6A - 7I$ $(m^2+m)A - (m^2+m-2)I = 6A - 4I$

4. Equating Coefficients and Solving for $m$

We now have an equation of the form $XA + YI = PA + QI$. Since $A$ and $I$ are linearly independent matrices (unless $A$ is a scalar multiple of $I$, which is not the case here), we can equate the coefficients of $A$ and $I$ on both sides of the equation.

Equating coefficients of $A$: $m^2+m = 6$ Equating coefficients of $I$: $-(m^2+m-2) = -4$

Let's solve the first equation: $m^2+m-6 = 0$ This is a quadratic equation. We can factor it: $(m+3)(m-2) = 0$ This gives two possible integer values for $m$: $m=-3$ or $m=2$.

Now, let's check if these values also satisfy the second equation. From the second equation: $m^2+m-2 = 4$ $m^2+m-6 = 0$ This is the exact same quadratic equation as the first one! This gives us confidence in our solution. So, the values of $m$ that satisfy both conditions are $m=-3$ and $m=2$.

Common Mistake: Forgetting to check if the solutions derived from one coefficient match the other. If the equations for coefficients of $A$ and $I$ yielded different sets of solutions, then there would be no integer $m$ satisfying the given matrix equation.

5. Determining $n(S)$

The set $S$ is defined as $S=\left\{m \in \mathbf{Z}: A^{m^2}+A^m=3 I-A^{-6}\right\}$. We found that the integer values of $m$ that satisfy the equation are $m=-3$ and $m=2$. Therefore, $S = \{-3, 2\}$.

The number of elements in $S$, denoted by $n(S)$, is 2.

Final Answer: The final answer is $\boxed{2}$.

6. Summary and Key Takeaway

This problem beautifully demonstrates the utility of the Cayley-Hamilton Theorem for simplifying matrix expressions involving powers. The key steps were:

1. Derive the characteristic equation for matrix $A$.
2. Use Cayley-Hamilton Theorem to find a recurrence relation for $A^n$ (e.g., $A^2 = 2A-I$).
3. Establish a general formula for $A^n$ (in this case, $A^n = nA - (n-1)I$) valid for all integers $n$. This is often the most challenging but rewarding step.
4. Substitute this general formula into the given matrix equation.
5. Simplify and equate coefficients of $A$ and $I$ (due to their linear independence) to form scalar equations.
6. Solve the resulting scalar equations for $m$ and identify the integer solutions.

Mastering the Cayley-Hamilton Theorem and recognizing patterns in matrix powers is essential for solving such problems efficiently in competitive exams like JEE.