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

\(\begin{array}{l}A=\begin{bmatrix} 1& 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \\\end{bmatrix} + \begin{bmatrix}0 & a & a \\0 & 0 & b \\0 & 0 & 0 \\\end{bmatrix} = I + B\end{array}\)
\(\begin{array}{l}B^2=\begin{bmatrix} 0& a & a \\0 & 0 & b \\0 & 0 & 0 \\\end{bmatrix}+\begin{bmatrix}0 & a & a \\0 & 0 & b \\0 & 0 & 0 \\\end{bmatrix}=\begin{bmatrix} 0& 0 & ab \\0 & 0 & 0 \\0 & 0 & 0 \\\end{bmatrix}\end{array}\)
\(B^3 = 0\)
∴ Aⁿ = (1 + B)ⁿ = ⁿC₀I + ⁿC₁ B + ⁿC₂ B² + ⁿC₃ B³ + ….
\(\begin{array}{l}=\begin{bmatrix} 1& 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \\\end{bmatrix}+\begin{bmatrix}0 & na & na \\0 & 0 & nb \\0 & 0 & 0 \\\end{bmatrix}+\begin{bmatrix} 0& 0 & \frac{n(n-1)ab}{2} \\0 & 0 & 0 \\0 & 0 & 0 \\\end{bmatrix}\end{array}\)
\(\begin{array}{l}=\begin{bmatrix} 1& na & na+\frac{n(n-1)}{2}ab \\0 & 1 & nb \\0 & 0 & 1 \\\end{bmatrix}=\begin{bmatrix}1 & 48 & 2160 \\0 & 1 & 48 \\0 & 0 & 1 \\\end{bmatrix}\end{array}\)
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