Answer
**For each integer \(n\ge1\):**
\[
\boxed{
\begin{pmatrix}1 & 1\\[4pt]1 & 0\end{pmatrix}^n
\;=\;
\begin{pmatrix}
F_{n+1} & F_n\\[4pt]
F_n & F_{n-1}
\end{pmatrix},
}
\]
where \(F_0=0,\,F_1=1,\,F_2=1,\,F_3=2,\dots\) are Fibonacci numbers.
Work Step by Step
Below is a standard result linking powers of the \(2 \times 2\) matrix
\(\displaystyle A=\begin{pmatrix}1 & 1\\[6pt]1 & 0\end{pmatrix}\)
to Fibonacci numbers.
---
## 1. Computing \(A^n\) for small \(n\)
Let us define \(A = \begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}\). Compute the first few powers by direct multiplication:
- **\(n=1\)**:
\[
A^1
= \begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}.
\]
- **\(n=2\)**:
\[
A^2
= \begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}
\begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}
= \begin{pmatrix}1+1 & 1+0\\ 1+0 & 1\end{pmatrix}
= \begin{pmatrix}2 & 1\\ 1 & 1\end{pmatrix}.
\]
- **\(n=3\)**:
\[
A^3
= A^2 \cdot A
= \begin{pmatrix}2 & 1\\ 1 & 1\end{pmatrix}
\begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}
= \begin{pmatrix}2\cdot 1 + 1\cdot 1 & 2\cdot 1 + 1\cdot 0\\[4pt]
1\cdot 1 + 1\cdot 1 & 1\cdot 1 + 1\cdot 0\end{pmatrix}
= \begin{pmatrix}3 & 2\\ 2 & 1\end{pmatrix}.
\]
- **\(n=4\)**:
\[
A^4
= A^3 \cdot A
= \begin{pmatrix}3 & 2\\ 2 & 1\end{pmatrix}
\begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}
= \begin{pmatrix}3\cdot 1 + 2\cdot 1 & 3\cdot 1 + 2\cdot 0\\[4pt]
2\cdot 1 + 1\cdot 1 & 2\cdot 1 + 1\cdot 0\end{pmatrix}
= \begin{pmatrix}5 & 3\\ 3 & 2\end{pmatrix}.
\]
- **\(n=5\)**:
\[
A^5
= A^4 \cdot A
= \begin{pmatrix}5 & 3\\ 3 & 2\end{pmatrix}
\begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}
= \begin{pmatrix}5\cdot 1 + 3\cdot 1 & 5\cdot 1 + 3\cdot 0\\[4pt]
3\cdot 1 + 2\cdot 1 & 3\cdot 1 + 2\cdot 0\end{pmatrix}
= \begin{pmatrix}8 & 5\\ 5 & 3\end{pmatrix}.
\]
Observe the pattern of the entries:
\[
A^1 = \begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix},\quad
A^2 = \begin{pmatrix}2 & 1\\ 1 & 1\end{pmatrix},\quad
A^3 = \begin{pmatrix}3 & 2\\ 2 & 1\end{pmatrix},\quad
A^4 = \begin{pmatrix}5 & 3\\ 3 & 2\end{pmatrix},\quad
A^5 = \begin{pmatrix}8 & 5\\ 5 & 3\end{pmatrix},\dots
\]
We recognize **Fibonacci numbers** in these matrices:
- The top‐left entries go \(1,2,3,5,8,\dots\)
- The top‐right and bottom‐left entries also follow Fibonacci patterns.
---
## 2. Conjectured Explicit Formulas for the Entries
Recall the Fibonacci sequence \(\{F_n\}\) defined by
\[
F_0 = 0,\quad F_1 = 1,\quad \text{and} \quad F_{n} = F_{n-1} + F_{n-2}\ \text{for }n\ge2.
\]
Then \(F_2=1,\;F_3=2,\;F_4=3,\;F_5=5,\;F_6=8,\dots\)
Comparing with our matrices, a well‐known identity is:
\[
\boxed{
A^n
= \begin{pmatrix}
F_{n+1} & F_{n}\\[6pt]
F_{n} & F_{n-1}
\end{pmatrix}.
}
\]
To match the initial cases carefully:
- For \(n=1\): \(F_2=1\), \(F_1=1\), \(F_0=0\).
Indeed,
\(\displaystyle A^1 = \begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}.\)
- For \(n=2\): \(F_3=2\), \(F_2=1\), \(F_1=1\).
\(\displaystyle A^2 = \begin{pmatrix}2 & 1\\ 1 & 1\end{pmatrix}.\)
- For \(n=3\): \(F_4=3\), \(F_3=2\), \(F_2=1\).
\(\displaystyle A^3 = \begin{pmatrix}3 & 2\\ 2 & 1\end{pmatrix}.\)
…and so on. The pattern lines up exactly.
Hence the explicit (closed‐form) entries are:
\[
A^n =
\begin{pmatrix}
F_{n+1} & F_n\\[4pt]
F_n & F_{n-1}
\end{pmatrix}.
\]
---
## 3. Proof by Mathematical Induction
We prove by induction on \(n\) that
\(\displaystyle A^n = \begin{pmatrix}F_{n+1} & F_n\\ F_n & F_{n-1}\end{pmatrix}\).
### Base Case
- \(n=1\). Then
\[
A^1 = \begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix},
\]
while the right‐hand side is
\(\displaystyle \begin{pmatrix}F_2 & F_1\\ F_1 & F_0\end{pmatrix}
= \begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}\).
Since \(F_2=1, F_1=1, F_0=0\), they match.
### Inductive Step
Assume for some \(n\ge1\),
\[
A^n
= \begin{pmatrix}F_{n+1} & F_{n}\\[3pt]
F_{n} & F_{n-1}\end{pmatrix}.
\]
We must show
\[
A^{n+1}
= A^n \cdot A
= \begin{pmatrix}F_{n+1} & F_{n}\\ F_{n} & F_{n-1}\end{pmatrix}
\begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}
= \begin{pmatrix}F_{(n+1)+1} & F_{n+1}\\[4pt]F_{n+1} & F_{n+1}-1\end{pmatrix}
\]
(suitably verifying the Fibonacci indices). Let’s do the multiplication carefully:
\[
A^n \cdot A
= \begin{pmatrix}
F_{n+1}\cdot 1 + F_{n}\cdot 1 &\; F_{n+1}\cdot 1 + F_{n}\cdot 0\\[6pt]
F_{n}\cdot 1 + F_{n-1}\cdot 1 &\; F_{n}\cdot 1 + F_{n-1}\cdot 0
\end{pmatrix}
= \begin{pmatrix}
F_{n+1} + F_{n} & F_{n+1}\\[6pt]
F_{n} + F_{n-1} & F_{n}
\end{pmatrix}.
\]
But \(F_{n+1} + F_{n} = F_{(n+1)+1} = F_{n+2}\) (by the Fibonacci recurrence), and \(F_{n} + F_{n-1} = F_{(n)+1} = F_{n+1}\). Hence
\[
A^{n+1}
= \begin{pmatrix}
F_{n+2} & F_{n+1}\\
F_{n+1} & F_{n}
\end{pmatrix}.
\]
That is exactly the form
\(\begin{pmatrix}F_{(n+1)+1} & F_{n+1}\\ F_{n+1} & F_{(n+1)-1}\end{pmatrix}\).
Thus the claim holds for \(n+1\).
By induction, the formula is true for all \(n\ge1\). (One can extend it to \(n=0\) by checking \(A^0 = I\) and noting \(F_1=1, F_{-1}\) can be defined consistently in certain Fibonacci extensions, but usually the statement is for \(n\ge1\).)
