Question

 If A = $\left[\begin{array}{ccc}3&-4\\1&-1\end{array}\right]$
show that , $A^{n} = \left[\begin{array}{ccc}1+2n&-4n\\n&1-2n\end{array}\right]$
And elaborate Mathematic induction.

Answer 1

$A= \left[\begin{array}{cc}3&-4\\1&-1\end{array}\right] \\\\Characteristic\ Equation=det (\lambda I - A) = 0\\ \\det \left[\begin{array}{cc}{\lambda - 3}&4\\-1&{\lambda +1}\end{array}\right]=0\\ \\\lambda^2-3\lambda+\lambda-3+4=0\\ \\\lambda^2 - 2 \lambda+1 =0$

$Substitute\ [A]\ in\ place\ of \lambda.\\ \\A^2 -2 A + I = 0\\ \\Rank\ of\ A=2\ as\ determinant\ \neq 0\\ \\A^2=2A-I$

$A^2 = 2 \left[\begin{array}{cc}3&-4\\1&-1\end{array}\right] - I = \left[\begin{array}{cc}5&-8\\2&-3\end{array}\right]= \left[\begin{array}{cc}1+2n&-4n\\n&1-2n\end{array}\right]\\as\ \ n=2$

So the given formula is true for n = 2.  Given formula is true for n =1 also as the values match. Can verify directly.

Proof by induction:
$A^2 = 2A - I = 2 (A-I)+I\\\\Let\ A^n=n(A-I)+I,\ \ \ \ ---equation1\\ \\A^{n+1}=n(A-I)A+A=nA^2-nA+A\\\\=2nA-nI-nA+A=(n+1)A-nI=(n+1)(A-I)+I\\ \\So\ equation1\ is\ proved\ by\ induction.$

$A-I= \left[\begin{array}{cc}3&-4\\1&-1\end{array}\right] - \left[\begin{array}{cc}1&0\\0&1\end{array}\right] = \left[\begin{array}{cc}2&-4\\1&-2\end{array}\right]$


$A^n=n\left[\begin{array}{cc}2&-4\\1&-2\end{array}\right]+I$
..