Let A=[top(1 1) bottom (0 1)] be a matrix consisting of real numbers (we are not interpreting this as a bit matrix). Find a formula for An, where n ϵ Z+, and use Mathematical Induction to prove that your formula is correct.

Let $A=\left[\begin{array}{cc} 1 & 1 \\ 0 & 1\end{array}\right].$

Let us prove by Mathematical Induction the following formula:

$A^n=\left[\begin{array}{cc} 1 & n \\ 0 & 1\end{array}\right].$

Base case:

$n=1:$

$A^1=\left[\begin{array}{cc} 1 & 1 \\ 0 & 1\end{array}\right]$

$n=2:$

$A^2=\left[\begin{array}{cc} 1 & 1 \\ 0 & 1\end{array}\right]\cdot \left[\begin{array}{cc} 1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc} 1 & 2 \\ 0 & 1\end{array}\right]$

Inductive step:

Assume that for $n=k$ it is true that $A^k=\left[\begin{array}{cc} 1 & k \\ 0 & 1\end{array}\right]$ and prove for $n=k+1:$

$A^{k+1}=A^k\cdot A=\left[\begin{array}{cc} 1 & k \\ 0 & 1\end{array}\right]\cdot\left[\begin{array}{cc} 1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc} 1 & k+1 \\ 0 & 1\end{array}\right]$

Conclusion:

Therefore, by Mathematical Induction $A^n=\left[\begin{array}{cc} 1 & n \\ 0 & 1\end{array}\right]$ for each $n \in\mathbb Z_+$