If \(A=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right]\), then \((a I+b A)^n\) is (where \(I\) is the identify matrix of order 2)

Given, \(A=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right]\)
Let \(P(n):(a I+b A)^n=a^n I+n a^{n-1} b A\)
For \(n=1, p(1):(a I+b A)^1=a^1 I+1 \cdot a^{1-1} b A\)
\[
a I+b A=a I+b A
\]
It is true for \(n=1\).
Let \(P(n)\) is true for \(n=K\).
So, \(P(K):(a I+b A)^K=a^K I+K a^{K-1} b A\)
Now, for \(n=K+1\)
\[
\begin{aligned}
& P(K+1): \\
& (a I+b A)^{K+1}=a^{K+1} I+(K+1) a^{K+1-1} b A \\
& \text { LHS }=(a I+b A)^{K+1}=(a I+b A)^K(a I+b A) \\
& =\left(a^K I+K a^{K-1} b A\right)(a I+b A) \quad \text { [from Eq. (i)] } \\
& =a^{K+1} I+K a^K I b A+a^K I b A+K a^{K-1} b^2 A^2 \\
& =a^{K+1} I+K a^K b(I A)+a^K b(I A)+K a^{K-1} b^2\left(A^2\right) \\
& =a^{K+1} I+K a^K b A+a^K b A+K a^{K-1} b^2 \times 0 \\
& {[\because I A=A \text { and }} \\
& \left.A^2=A \cdot A=\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=0\right] \\
& =a^{K+1} I+(K+1) a^K b A=\mathrm{RHS} \\
&
\end{aligned}
\]
Hence, \(P(n)\) is true for all \(n \in N\).