# Let, show that, where I is the identity matrix of order 2 and n ∈ N

Question:

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

Solution:

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

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

We shall prove the result by using the principle of mathematical induction.

For n = 1, we have:

$\mathrm{P}(1):(a I+b A)=a I+b a^{0} A=a I+b A$

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

Let the result be true for $n=k$.

That is,

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

Now, we prove that the result is true for n = k + 1.

Consider

\begin{aligned}(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} I+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} \end{aligned}......(1)

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

From (1), we have:

\begin{aligned}(a I+b A)^{k+1} &=a^{k+1} I+(k+1) a^{k} b A+O \\ &=a^{k+1} I+(k+1) a^{k} b A \end{aligned}

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

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

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