Question

If $A=\left[\begin{array}{cc}a& b\\ 0& 1\end{array}\right]$, prove that $A^n=\left[\begin{array}{cc}{a}^n& b(a^n-1)/a-1\\ 0& 1\end{array}\right]$ for every positive integer n.

Answer 1

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have
$A^1=\left[\begin{array}{cc}{a}^1& b\left(a^1-1\right)/a-1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}a& b\\ 0& 1\end{array}\right]=A$

So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,
$A^m=\left[\begin{array}{cc}{a}^m& b\left(a^m-1\right)/a-1\\ 0& 1\end{array}\right]$ ...(1)

Now, we shall show that the result is true for $n=m+1$.
Here,
$A^{m+1}=\left[\begin{array}{cc}{a}^{m+1}& b\left(a^{m+1}-1\right)/a-1\\ 0& 1\end{array}\right]$

By definition of integral power of matrix, we have
$$\begin{gathered} A^{m+1}=A^{m}A \\ \Rightarrow A^{m+1}=\left[\begin{array}{cc}{a}^m& b\left(a^m-1\right)/a-1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}a& b\\ 0& 1\end{array}\right]\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow A^{m+1}=\left[\begin{array}{cc}{a}^{m}a+0& \left\{a^{m}b+b\left(a^m-1\right)\right\}/a-1\\ 0+0& 0+1\end{array}\right] \\ \Rightarrow A^{m+1}=\left[\begin{array}{cc}{a}^{m+1}& \left(a^{m+1}b-a^{m}b+a^{m}b-b\right)/a-1\\ 0& 1\end{array}\right] \\ \Rightarrow A^{m+1}=\left[\begin{array}{cc}{a}^{m+1}& b\left(a^{m+1}-1\right)/a-1\\ 0& 1\end{array}\right] \end{gathered}$$

This shows that when the result is true for n = m, it is also true for n = m +1.

Hence, by the principle of mathematical induction, the result is valid for any positive integer n.