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Question:

If $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$, prove that $A^{n}=\left[\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right]$ for all positive integers $n$.

Solution:

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

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

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

$A^{m+1}=A^{m} A$

$=\left[\begin{array}{cc}1 & m \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$     $[$ From eq. $(1)]$

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

 

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

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

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

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