It is given that $A=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right]$

To show:  $\mathrm{P}(n)$ $:(a I+b A)^{n}=a^{n} I+n a^{n-1} b A, n \in N$

We shall prove the result by using the principal of mathematical induction. For $n=1,$ we have:

$P(1):(a I+b A)=a I+b a^{\circ} A=a I+b A$

Therefore, the result is true for $n=1$

Let the result be true for $n=k$

That is,

$P(k):(a I+{b} A)^{k}=a^{k} I+k a^{k \cdot \cdot} {b} A$

Now, we prove that the result is true for $\boldsymbol{n}=\boldsymbol{k}+1$ Consider

$(a I+b A)^{k-1} =(a I+b A)^{k}(a I+b A)$

$=\left(a^{k} I+k a^{k-1} b A\right)(a I+b A)$

$=a^{k-1}+k a^{k} b A I+a^{k} b I A+k a^{k-1} b^{2} A^{2}$

$=a^{k-1} I+(k+1) a^{k} b A+k a^{k-1} b^{2} A^{2} $        ………. $(1)$

Now,      $A^{2}=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$

From $(1)$, we have :

$(a I+b A)^{k+1}=a^{k+1}+(k+1) a^{k} b A+0$

$=a^{k+1}+(k+1) a^{k} b A$

Therefore, the result is true for $n=k+1$.

Thus, by the principal of mathematical induction, we have:

$(a I+b A)^{n}=a^{n} I+n a^{n-1} b A$ where $A=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right], n \in \mathrm{N}$