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# If $A=\begin{bmatrix}0 & 1\\0 &0\end{bmatrix}$ prove that for all $n\in N.(aI+bA)^n=a^nI+na^{n-1}bA.$where I is the identity matrix of order 2.

Toolbox:
• Here we will use the principle of mathematical induction.
Let $P(n)=(aI+bA)^n=a^nI+na^{n-1}bA$
Where A=$\begin{bmatrix}0 &1\\0 & 0\end{bmatrix}$ and I=$\begin{bmatrix}1 & 0\\0 &1\end{bmatrix}$
When n=1
LHS=$(aI+bA)^1$
$\qquad=aI+bA.$
RHS=$a^1I+na^0bA$
$\qquad=aI+1.a^0bA$
$\qquad=aI+bA$
$\qquad=LHS$
P(1) is true.
Let P(m) be true.
$\Rightarrow (aI+bA)^m=a^mI+ma^{m-1}bA$------(1)
To prove P(m+1) is true.
(i.e)$(aI+bA)^{m+1}=a^{m+1}I+(m+1)a^mbA.$
Multiply both sides of (1) by aI+bA we get
$(aI+bA)^{m+1}=(aI+A).(aI+bA)^m$
$\qquad\qquad\quad=(aI+bA).(a^mI+ma^{m-1}bA)$
$\qquad\qquad\quad=aI(a^mI+ma^{m-1}bA)+bA(a^m1+ma^{m-1}bA)$
$\qquad\qquad\quad=a^{m+1}+ma^mbA+a^mbA+ma^{m-1}b^2A^2 \quad( I.I=I,IA=A=AI)$
$\qquad\qquad\quad=a^{m+1}I+(m+1)a^mbA.\quad( A^2=0)$