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

$\begin{array}{l}A=\begin{bmatrix} 1& 0 & 0 \\\0 & 1 & 0 \\\0 & 0 & 1 \\\\\end{bmatrix} + \begin{bmatrix}0 & a & a \\\0 & 0 & b \\\0 & 0 & 0 \\\\\end{bmatrix} = I + B\end{array}$
$\begin{array}{l}B^2=\begin{bmatrix} 0& a & a \\\0 & 0 & b \\\0 & 0 & 0 \\\\\end{bmatrix}+\begin{bmatrix}0 & a & a \\\0 & 0 & b \\\0 & 0 & 0 \\\\\end{bmatrix}=\begin{bmatrix} 0& 0 & ab \\\0 & 0 & 0 \\\0 & 0 & 0 \\\\\end{bmatrix}\end{array}$
$B^3 = 0$
∴ A n = (1 + B)n = n C 0 I + n C 1 B + n C 2 B 2 + n C 3 B 3 + ….
$\begin{array}{l}=\begin{bmatrix} 1& 0 & 0 \\\0 & 1 & 0 \\\0 & 0 & 1 \\\\\end{bmatrix}+\begin{bmatrix}0 & na & na \\\0 & 0 & nb \\\0 & 0 & 0 \\\\\end{bmatrix}+\begin{bmatrix} 0& 0 & \frac{n(n-1)ab}{2} \\\0 & 0 & 0 \\\0 & 0 & 0 \\\\\end{bmatrix}\end{array}$
$\begin{array}{l}=\begin{bmatrix} 1& na & na+\frac{n(n-1)}{2}ab \\\0 & 1 & nb \\\0 & 0 & 1 \\\\\end{bmatrix}=\begin{bmatrix}1 & 48 & 2160 \\\0 & 1 & 48 \\\0 & 0 & 1 \\\\\end{bmatrix}\end{array}$
On comparing we get na = 48, nb = 96 and
$\begin{array}{l}na+\frac{n(n-1)}{2}ab=2160\end{array}$
⇒ a = 4, n = 12 and b = 8
n + a + b = 24