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

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Solution

Let $P (n) : {(a I + b A)}^{n} = a^{n} I + n a^{n - 1} b A$
Step $1$ For $n = 1$
L.H.S $= {(a I + b A)}^{1} = a I + b A$
and R.H.S $= a^{1} I + n a^{1 - 1} b A = a I + a^{0} b A = a I + b A$
$\Rightarrow$ L.H.S $=$ R.H.S
$∴, P (1)$ is true.

Step $2$ For $n = k$
L.H.S $= {(a I + b A)}^{k}$
and R.H.S $= a^{k} I + k a^{k - 1} b A$
Here let's assume that $\Rightarrow$ L.H.S $=$ R.H.S
$∴, P (k)$ is true.

Step $3$ For $n = k + 1,$ we have to prove that
$P (k + 1) : {(a I + b A)}^{k + 1} = a^{k + 1} I + (k + 1) a^{k} b A$
L.H.S $= {(a I + b A)}^{k + 1} = {(a I + b A)}^{k} {(a I + b A)}^{1}$
$= {(a I + b A)}^{k} (a I + b A)$
$= (a^{k} I + k a^{k - 1} b A) (a I + b A)$
$= a^{k + 1} I^{2} + a^{k} b (I A) + k a^{k} b (A I) + k a^{k - 1} b^{2} A^{2}$
$= a^{k + 1} I + (k + 1) a^{k} b A + 0$ $∵ A I = A, A^{2} = 0$ and $I^{2} = I$
$= a^{k + 1} I + (k + 1) a^{k} b A =$ R.H.S
$∴, P (k + 1)$ is true.
Hence, by the principle of mathematical induction $P (n)$ is true for all $n \in N$

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