Question

If I is the identity matrix of order 2 and A = $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$, then for $n\ge 1$, mathematical induction gives

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

Answer 1

Answer: (A)

$A^n=nA-(n-1)I$

$A=\left|\begin{array}{cc}1& 1\\ 0& 1\end{array}\right|$

$A^2=\left|\begin{array}{cc}1& 1\\ 0& 1\end{array}\right|\left|\begin{array}{cc}1& 1\\ o& 1\end{array}\right|=\left|\begin{array}{cc}1& 2\\ 0& 1\end{array}\right|$

$A^3=\left|\begin{array}{cc}1& 2\\ 0& 1\end{array}\right|\left|\begin{array}{cc}1& 1\\ 0& 1\end{array}\right|=\left|\begin{array}{cc}1& 3\\ 0& 1\end{array}\right|$

$\therefore A^n=\left|\begin{array}{cc}1& n\\ 0& 1\end{array}\right|$

$nA=\left|\begin{array}{cc}n& n\\ 0& n\end{array}\right|$

$(n-1)I=\left|\begin{array}{cc}n-1& 0\\ 0& n-1\end{array}\right|$

$\therefore A^n=nA-(n-1)I$