Question

If $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$, prove that $A^n=\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$ for all positive integers 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 powers of matrix, we have
$A^1=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=A$
So, the result is true for n = 1.

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

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

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

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.