Question

Let $A = \begin{bmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{bmatrix}, a , b \in R$. 
If for some $n \in N , A ^{ n }=\begin{bmatrix}1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1\end{bmatrix}$ 
then $n + a + b$ is equal to _________.

Answer 1

Correct Answer: 24

To solve the problem, we need to find the value of $n + a + b$ given that $A^n = \begin{bmatrix}1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1\end{bmatrix}$, where $A = \begin{bmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{bmatrix}$. The matrix $A$ is an upper triangular matrix, and one of the properties of triangular matrices is that their powers maintain the upper triangular form.

Notice the pattern when computing powers of $A$: 
$A^1 = \begin{bmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{bmatrix}$, 
$A^2 = \begin{bmatrix} 1 & 2a & 3a^2+ab \\ 0 & 1 & 2b \\ 0 & 0 & 1\end{bmatrix}$, 
and in general, $A^n = \begin{bmatrix} 1 & na & \frac{n(n-1)}{2}a^2+nab \\ 0 & 1 & nb \\ 0 & 0 & 1\end{bmatrix}$.

Compare $A^n$ with the given matrix:
$\begin{bmatrix}1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1\end{bmatrix}$. This gives us two equations:
1. $na = 48$, 
2. $nb = 96$. Solving for $a$ and $b$, we can divide the second equation by the first to find $b/a = 2$, so $b = 2a$.

Substitute $b = 2a$ into either equation, for simplicity, we use $nb = 96$: 
$2na = 96 \implies na = 48$, which is consistent. Hence, $b = 2a$ and $na = 48$.

From $na = 48$, and knowing $b = 2a$, solve for $n + a + b$:
$a = \frac{48}{n}$ and $b = \frac{96}{n}$ gives $n + a + b = n + \frac{48}{n} + \frac{96}{n} = n + \frac{144}{n}$. Simplifying gives $n + a + b = \left(n + \frac{144}{n}\right)$.

We test if $n = 12$, 
$a = \frac{48}{12} = 4$, 
$b = \frac{96}{12} = 8$.
Then $n + a + b = 12 + 4 + 8 = 24$, fitting within the given range (24,24).

Thus, the value of $n + a + b$ is 24.