Question

Let $A=\left[\begin{array}{cc}a& b\\ 0& 1\end{array}\right]$ where $a\ne 0$. Then $A^n=$?

• A. $A^n=\left[\begin{array}{cc}{a}^n& \frac{b(a^n-1)}{(a-1)}\\ 0& 1\end{array}\right]$
• B. $A^n=\left[\begin{array}{cc}\frac{a(b^n-1)}{(b-1)}& \frac{b(a^n-1)}{(a-1)}\\ 0& 1\end{array}\right]$
• C. $A^n=\left[\begin{array}{cc}{a}^n& b^n\\ 0& 1\end{array}\right]$
• D. none of these

Answer 1

The correct option is A $A^n=\left[\begin{array}{cc}{a}^n& \frac{b(a^n-1)}{(a-1)}\\ 0& 1\end{array}\right]$

$A=\left[\begin{array}{cc}a& b\\ 0& 1\end{array}\right]$
$A^2=\left[\begin{array}{cc}a& b\\ 0& 1\end{array}\right]\left[\begin{array}{cc}a& b\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{a}^2& b(a+1)\\ 0& 1\end{array}\right]$
Put $n=2$ and compare with calculated $A^2$, only option A is matching. Hence
this is correct choice.
Another method: Using mathematical Induction.
We have to show by mathematical induction
Step I : For $n=1$
$A^1=\left[\begin{array}{cc}a& b\frac{(a-1)}{(a-1)}\\ 0& 1\end{array}\right]$
$=\left[\begin{array}{cc}a& b\\ 0& 1\end{array}\right]$
Which is true for $n=1$
Step II : Assume it is true for $n=k$
$\therefore A^k=\left[\begin{array}{cc}{a}^k& b\frac{(a^k-1)}{(a-1)}\\ 0& 1\end{array}\right]$
Step III : For $n=k+1$
$A^{k+1}=A^k.A=\left[\begin{array}{cc}{a}^k& b\frac{(a^k-1)}{(a-1)}\\ 0& 1\end{array}\right]\times \left[\begin{array}{cc}a& b\\ 0& 1\end{array}\right]$
$=\left[\begin{array}{cc}{a}^k.a+0& a^k.b+\frac{b(a^k-1)}{(a-1)}\\ 0+0& 0+1.1\end{array}\right]$
$=\left[\begin{array}{cc}{a}^{k+1}& b\left\{\frac{a^{k+1}-a^k+a^k-1}{(a-1)}\right\}\\ 0& 1\end{array}\right]$
$=\left[\begin{array}{cc}{a}^{k+1}& \frac{b(a^{k+1}-1)}{(a-1)}\\ 0& 1\end{array}\right]$
Hence, it is true for all positive integral values of $n\ge 0$.