1. Key Concepts and Formulas

• Decomposition of Matrices: An upper triangular matrix with ones on the main diagonal can often be decomposed into the sum of an identity matrix $I$ and a strictly upper triangular matrix $N$, i.e., $A = I + N$.
• Nilpotent Matrices: A square matrix $N$ is called nilpotent if $N^k = \mathbf{0}$ (the zero matrix) for some positive integer $k$. For a $3 \times 3$ strictly upper triangular matrix, its cube ($N^3$) is always the zero matrix.
• Binomial Theorem for Matrices: If two matrices $X$ and $Y$ commute (i.e., $XY = YX$), then for any positive integer $n$: $(X+Y)^n = \binom{n}{0}X^n + \binom{n}{1}X^{n-1}Y + \binom{n}{2}X^{n-2}Y^2 + \dots + \binom{n}{n}Y^n$ Since the identity matrix $I$ commutes with any matrix $N$ ($IN = NI = N$), this theorem can be applied to $(I+N)^n$. If $N$ is nilpotent, the series expansion terminates, simplifying the calculation of $A^n$.

2. Step-by-Step Solution

Step 1: Decompose matrix A into I + N We begin by expressing the given matrix $A$ as the sum of an identity matrix $I$ and another matrix $N$. This is a standard technique for matrices of this form, as it allows us to utilize the Binomial Theorem effectively. $A=\left[\begin{array}{lll} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] + \left[\begin{array}{lll} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{array}\right] = I + N$ Here, the matrix $N$ is: $N = \left[\begin{array}{lll} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{array}\right]$

Step 2: Determine the nilpotency of N Next, we calculate the powers of $N$ to find its index of nilpotency. This is crucial because it determines how many terms will be non-zero in the binomial expansion. $N^2 = N \cdot N = \left[\begin{array}{lll} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{array}\right] \left[\begin{array}{lll} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 0 & ab \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$ Now, calculate $N^3$: $N^3 = N^2 \cdot N = \left[\begin{array}{ccc} 0 & 0 & ab \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \left[\begin{array}{lll} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{array}\right] = \left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = \mathbf{0}$ Since $N^3 = \mathbf{0}$, $N$ is a nilpotent matrix of index 3. This means any higher power of $N$ ($N^k$ for $k \ge 3$) will also be the zero matrix, which simplifies the binomial expansion significantly.

Step 3: Apply the Binomial Theorem to find A^n Now we can apply the Binomial Theorem to $A^n = (I+N)^n$. Since $I$ and $N$ commute, and $N^3 = \mathbf{0}$, the expansion will only have three non-zero terms: $(I+N)^n = \binom{n}{0}I^n + \binom{n}{1}I^{n-1}N + \binom{n}{2}I^{n-2}N^2 + \binom{n}{3}I^{n-3}N^3 + \dots$ Since $I^k = I$ for any positive integer $k$, and $N^k = \mathbf{0}$ for $k \ge 3$: $A^n = I + nN + \frac{n(n-1)}{2}N^2 + \mathbf{0} + \dots$ Substitute the expressions for $I$, $N$, and $N^2$: $A^n = \left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] + n \left[\begin{array}{lll} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{array}\right] + \frac{n(n-1)}{2} \left[\begin{array}{ccc} 0 & 0 & ab \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$ Perform the scalar multiplications: $A^n = \left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] + \left[\begin{array}{ccc} 0 & na & na \\ 0 & 0 & nb \\ 0 & 0 & 0 \end{array}\right] + \left[\begin{array}{ccc} 0 & 0 & \frac{n(n-1)}{2}ab \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$ Adding these matrices element-wise, we get the general form of $A^n$: $A^n = \left[\begin{array}{ccc} 1 & na & na + \frac{n(n-1)}{2}ab \\ 0 & 1 & nb \\ 0 & 0 & 1 \end{array}\right]$

Step 4: Compare elements and solve for n, a, b We are given the value of $A^n$: $A^{n}=\left[\begin{array}{ccc} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{array}\right]$ By comparing the corresponding elements of our derived $A^n$ with the given $A^n$, we form a system of equations:

1. Element $(1,2)$: $na = 48$
2. Element $(2,3)$: $nb = 96$
3. Element $(1,3)$: $na + \frac{n(n-1)}{2}ab = 2160$

Now, we solve this system: From equation (1), $a = \frac{48}{n}$. From equation (2), $b = \frac{96}{n}$.

Substitute $na=48$ into equation (3): $48 + \frac{n(n-1)}{2}ab = 2160$ $\frac{n(n-1)}{2}ab = 2160 - 48$ $\frac{n(n-1)}{2}ab = 2112$

Now substitute the expressions for $a$ and $b$ (in terms of $n$) into this equation: $\frac{n(n-1)}{2} \left(\frac{48}{n}\right) \left(\frac{96}{n}\right) = 2112$ $\frac{n(n-1)}{2} \frac{48 \times 96}{n^2} = 2112$ Simplify the expression: $\frac{(n-1)}{2n} (48 \times 96) = 2112$ $\frac{(n-1)}{n} (24 \times 96) = 2112$ $\frac{(n-1)}{n} (2304) = 2112$

To find $n$, isolate $\frac{n-1}{n}$: $\frac{n-1}{n} = \frac{2112}{2304}$ Simplify the fraction. Both numerator and denominator are divisible by 192: $2112 = 192 \times 11$ $2304 = 192 \times 12$ So, $\frac{n-1}{n} = \frac{11}{12}$

Now, solve for $n$: $12(n-1) = 11n$ $12n - 12 = 11n$ $12n - 11n = 12$ $n = 12$

Now that we have $n=12$, we can find $a$ and $b$: $a = \frac{48}{n} = \frac{48}{12} = 4$ $b = \frac{96}{n} = \frac{96}{12} = 8$ These values are consistent with $n \in \mathbb{N}$ and $a, b \in \mathbb{R}$.

Step 5: Calculate n+a+b Finally, we calculate the required sum: $n+a+b = 12 + 4 + 8 = 24$

3. Common Mistakes & Tips

• Commutativity for Binomial Theorem: Always ensure that the matrices $X$ and $Y$ commute ($XY=YX$) before applying the Binomial Theorem for $(X+Y)^n$. For $A=I+N$, this condition is always met.
• Nilpotency Calculation: Carefully calculate powers of $N$. For a $3 \times 3$ strictly upper triangular matrix, $N^3$ will always be the zero matrix. Knowing this can save time and prevent errors.
• Algebraic Precision: The system of equations often involves fractions and larger numbers. Double-check all algebraic manipulations to avoid arithmetic errors, especially when simplifying fractions.

4. Summary

This problem demonstrates a powerful technique for finding powers of specific types of matrices. By decomposing the given matrix $A$ into the sum of an identity matrix $I$ and a nilpotent matrix $N$, we can leverage the Binomial Theorem for matrices. The nilpotency of $N$ causes the binomial expansion to terminate quickly, yielding a concise formula for $A^n$. Equating the elements of this formula with the given $A^n$ leads to a system of equations that can be solved to find $n, a,$ and $b$. Finally, summing these values provides the answer.

The final answer is $\boxed{24}$.