Question

If B, C are n rowed square matrices and if A = B + C, BC = CB, C² = O, then show that for every n ∈ N, Aⁿ⁺¹ = Bⁿ (B + (n + 1) C).

Answer 1

Let $P\left(n\right)$ be the statement given by $P\left(n\right):A^{n+1}=B^n\left(B+\left(n+1\right)C\right)$.

For n = 1, we have
$$\begin{gathered} P\left(1\right):A^2=B\left(B+2C\right) \\ \mathrm{Here}, \\ \mathrm{LHS}=A^2 \\ =\left(B+C\right)\left(B+C\right) \\ =B\left(B+C\right)+C\left(B+C\right) \\ =B^2+BC+CB+C^2 \\ =B^2+2BC\left[\because BC=CB\mathrm{and}{C}^2=O\right] \\ =B\left(B+2C\right)=\mathrm{RHS} \end{gathered}$$

Hence, the statement is true for n = 1.

If the statement is true for n = k, then
$P\left(k\right):A^{k+1}=B^k\left(B+\left(k+1\right)C\right)$ ...(1)

For $P\left(k+1\right)$ to be true, we must have
$P\left(k+1\right):A^{k+2}=B^{k+1}\left(B+\left(k+2\right)C\right)$

Now,
$$\begin{gathered} A^{k+2}=A^{k+1}A \\ =\left[B^k\left(B+\left(k+1\right)C\right)\right]\left(B+C\right)\left[\mathrm{From}\mathrm{eqn}.\left(1\right)\right] \\ =\left[B^{k+1}+\left(k+1\right)B^{k}C\right]\left(B+C\right) \\ =B^{k+1}\left(B+C\right)+\left(k+1\right)B^{k}C\left(B+C\right) \\ =B^{k+2}+B^{k+1}C+\left(k+1\right)B^{k}CB+\left(k+1\right)B^k{C}^2 \\ =B^{k+2}+B^{k+1}C+\left(k+1\right)B^{k}BC\left[\because BC=CBand{C}^2=0\right] \\ =B^{k+2}+B^{k+1}C+\left(k+1\right)B^{k+1}C \\ =B^{k+2}+\left(k+2\right)B^{k+1}C \\ =B^{k+1}\left[B+\left(k+2\right)C\right] \end{gathered}$$

So the statement is true for n = k+1.
Hence, by the principle of mathematical induction, $P\left(n\right)$ is true for all $n\in N$.