If \(A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\) and I is the identity matrix of order 2, then which one of the following holds for every natural number n?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(A^n = nA - (n - 1)I\)
The first few powers of A show the pattern:
\( \displaystyle A^2 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \)
\( \displaystyle A^3 = A^2A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} \)
Each multiplication by A adds 1 to the element in the second row and first column, so
\( \displaystyle A^n = \begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix} \)
Now the expression in Option (1):
\( \displaystyle nA - (n - 1)I = \begin{bmatrix} n & 0 \\ n & n \end{bmatrix} - \begin{bmatrix} n - 1 & 0 \\ 0 & n - 1 \end{bmatrix} \)
\( \displaystyle = \begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix} \)
This equals \(A^n\) for every natural number n.
Why the other options are incorrect
Option (2) \(A^n = 2^{n-1}A + (n - 1)I\) — Incorrect. For n = 2 it gives \(2A + I = \begin{bmatrix} 3 & 0 \\ 2 & 3 \end{bmatrix}\), but \(A^2 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\).
Option (3) \(A^n = nA + (n - 1)I\) — Incorrect. For n = 2 it also gives \(2A + I = \begin{bmatrix} 3 & 0 \\ 2 & 3 \end{bmatrix}\), not \(A^2\).
Option (4) \(A^n = 2^{n-1}A - (n - 1)I\) — Incorrect. It is right for n = 1 and n = 2, but for n = 3 it gives \(4A - 2I = \begin{bmatrix} 2 & 0 \\ 4 & 2 \end{bmatrix}\), while \(A^3 = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}\).