Answer
See explanations.
Work Step by Step
Step 1. Recall the property of Fibonacci sequence, we have: $F_{n}=F_{n-1}+F_{n-2}$ and $F_1=F_2=1$
Step 2. Prove the statement is true when $n=2$: $LHS=\begin{bmatrix} 1&1\\1&0 \end{bmatrix}\begin{bmatrix} 1&1\\1&0 \end{bmatrix}=\begin{bmatrix} 2&1\\1&1 \end{bmatrix}$, $RHS=\begin{bmatrix} F_3&F_2\\F_2&F_1 \end{bmatrix}=\begin{bmatrix} 2&1\\1&1 \end{bmatrix}=LHS$, thus it is true for $n=2$.
Step 3. Assume the statement is true when $n=k$ ($k\gt2$): we have $\begin{bmatrix} 1&1\\1&0 \end{bmatrix}^k=\begin{bmatrix} F_{k+1}&F_k\\F_k&F_{k-1} \end{bmatrix}$
Step 4. Prove it is true for $n=k+1$: $LHS=\begin{bmatrix} 1&1\\1&0 \end{bmatrix}^{k+1}=\begin{bmatrix} F_{k+1}&F_k\\F_k&F_{k-1} \end{bmatrix}\begin{bmatrix} 1&1\\1&0 \end{bmatrix}=\begin{bmatrix} F_{k+1}+F_k&F_{k+1}\\F_k+F_{k-1}&F_k \end{bmatrix}=\begin{bmatrix} F_{k+2}&F_{k+1}\\F_{k+1}&F_k \end{bmatrix}$
With $n=k+1$, we have $RHS=\begin{bmatrix} F_{k+2}&F_{k+1}\\F_{k+1}&F_k \end{bmatrix}=LHS$
Thus, the statement is also true for $n=k+1$
Step 5. Conclusion: with mathematical induction, we have proved that the statement is true for all natural numbers $n$.
