Prove that $(a I + b A)^{n} = a^{n} I + n a^{n - 1} b A$ is true for all the value $n ϵ N .$

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Solution

Let P (n) be tha statement.
$P (n) : (a I + b A)^{n} = a^{n} I + n a^{n - 1} b A P (1) : a l + b A = a i + b A H e n c e P (1) i s t r u e . P (k) : (a I + b A)^{k} = a^{k} I + n a^{k - 1} b A . . . . . (1) N o w w e h a v e t o p r o v e t h a t p (k + 1) i s t r u e P (k + 1) : (a I + b A)^{k + 1} = a^{k + 1} I + (k + 1) a^{k} b A L . H . S o f P (k + 1) (a I + b A)^{k + 1} = (a I + b A)^{k} (a I + b A) = (a^{k} I + k a^{k - 1}) (u s i n g (1)) = (a^{k + 1}) I^{2} + k a^{k} b A l + k a^{k - 1} b^{2} A^{2} = (a^{k + 1}) + (k + 1) a^{k} b A l + O [∵ A^{2} = A . A = [\begin{matrix} 0 & 0 0 & 0 \end{matrix}]] = d^{k + 1} I + (k + 1) d^{k} b A [∵ A I = A] = R . H . S ∴ P (k + 1) i s t r u e .$
Hence P(n) is true for all the value $n ϵ N .$

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Prove that (aI+bA)n=anI+nan−1bA is true for all the value nϵN.

Prove that $(a I + b A)^{n} = a^{n} I + n a^{n - 1} b A$ is true for all the value $n ϵ N .$