Solve this

Solution:

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

$A^{1}=\left[\begin{array}{cc}a^{1} & b\left(a^{1}-1\right) / a-1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}a & b \\ 0 & 1\end{array}\right]=A$

Step 2: Let the result be true for $n=m$. Then,

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

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

$\Rightarrow A^{m+1}=\left[\begin{array}{cc}a^{m} & b\left(a^{m}-1\right) / a-1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}a & b \\ 0 & 1\end{array}\right]$    $[$ From eq. (1)]

$\Rightarrow A^{m+1}=\left[\begin{array}{cc}a^{m} a+0 & \left\{a^{m} b+b\left(a^{m}-1\right)\right\} / a-1 \\ 0+0 & 0+1\end{array}\right]$

$\Rightarrow A^{m+1}=\left[\begin{array}{cc}a^{m+1} & \left(a^{m+1} b-a^{m} b+a^{m} b-b\right) / a-1 \\ 0 & 1\end{array}\right]$

 

$\Rightarrow A^{m+1}=\left[\begin{array}{cc}a^{m+1} & b\left(a^{m+1}-1\right) / a-1 \\ 0 & 1\end{array}\right]$

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.