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Choose the correct option.
If A = (1 b; 0 1), then Aⁿ (n ∈ mathbbN) is equal to
If $A = \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}$, then $A^n$ ($n \in \mathbb{N}$) is equal to
- (a)$\begin{pmatrix} 1 & nb \\ 0 & 1 \end{pmatrix}$
- (b)$\begin{pmatrix} 1 & b^n \\ 0 & 1 \end{pmatrix}$
- (c)$\begin{pmatrix} 1 & n^b \\ 0 & 1 \end{pmatrix}$
- (d)$\begin{pmatrix} 1 & nb \\ 0 & 0 \end{pmatrix}$
Answer
Answer
Correct option: (a)
Answer: (a) $\begin{pmatrix} 1 & nb \\ 0 & 1 \end{pmatrix}$
$A^2 = \begin{pmatrix} 1 & 2b \\ 0 & 1 \end{pmatrix}$, $A^3 = \begin{pmatrix} 1 & 3b \\ 0 & 1 \end{pmatrix}$, and by induction $A^n = \begin{pmatrix} 1 & nb \\ 0 & 1 \end{pmatrix}$.
Options (a) and (c) are printed identically in the paper.