Question

Let $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, show that (aI+bA)^n=a^nI+$n{a}^{n-1}$ bA, where, I is the identity matrix of order 2 and $n\in N.$

Answer 1

Given, $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$
We shall prove the result by principle of mathematical induction.
Let $P\left(n\right):(aI+bA{)}^n=a^{n}I+n{a}^{n-1}bA$
Putting $P\left(1\right):(aI+bA{)}^1=a^{1}I+1.a^{-1-1}bA$
aI+bA =aI+bA which is true for n=1.
Let the result be true for n=k.
$\therefore P\left(k\right):(aI+bA{)}^k=a^{k}I+k.a^{k-1}bA.....(i)$
Now, we prove that the result is true for n=k+1.
$\therefore P(k+1):(aI+bA{)}^{k+1}=a^{k+1}I+(k+1).a^{k+1-1}bA$
Now, taking $LHS=(aI+bA{)}^{k+1}$
$=(aI+bA{)}^k(aI+bA\left)\right[\because a^{x+y}=a^x\times a^y]=(a^{k}I+k.a^{k-1}bA\left)\right(aI+bA)$

$[\because$ Using Eq. (i)Putting $(aI+bA{)}^k=a^{k}I+k{a}^{k-1}bA]$
$=a^{k+1}I+k{a}^{k}IbA+a^{k}IbA+k{a}^{k-1}{b}^2{A}^2=a^{k+1}I+k{a}^{k}b\left(IA\right)+a^{k}b\left(IA\right)+k{a}^{k-1}{b}^2\left(A^2\right)(\because AI=Aand{A}^2=AA\Rightarrow \left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0)=a^{k+1}I+k{a}^{k}bA+a^{k}bA+k{a}^{k-1}{b}^2\times 0=a^{k+1}I+(k+1).a^{k}bA$
Therefore, the result is true for n=k +1 whenever it is true for n=k. So, by principle of mathematical induction, it is true for all $n\in N.$