Question

If A = $\left|\begin{array}{cc}1& 0\\ 1& 1\end{array}\right|$B and I =$\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|$ ,then which one of the following holds for all n $\ge$ 1, by
the principle of mathematical indunction

• A. $A^n=nA-(n-1)l$
• B. $A^n=2^{n-1}A-(n-1)l$
• C. $A^n=nA+(n-1)l$
• D. $A^n=2^{n-1}A+(n-1)l$

Answer 1

The correct option is A $A^n=nA-(n-1)l$

$A=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]A^2=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]A^3=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 3& 1\end{array}\right]\therefore A^n=\left[\begin{array}{cc}1& 0\\ n& 1\end{array}\right]$

and $nA=n\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}n& 0\\ n& n\end{array}\right]$

$(n-1)I=(n-1)\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}n-1& 0\\ 0& n-1\end{array}\right]\therefore nA-(n-1)I=\left[\begin{array}{cc}n& 0\\ n& n\end{array}\right]-\left[\begin{array}{cc}n-1& 0\\ 0& n-1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ n& 1\end{array}\right]=A^n$