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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).
Solution
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
\[P\left( 1 \right) : A^2 = B\left( B + 2C \right)\]
\[\]
\[Here, \]
\[LHS = A^2 \]
\[ = \left( B + C \right)\left( B + C \right)\]
\[ = B\left( B + C \right) + C\left( B + C \right)\]
\[ = B^2 + BC + CB + C^2 \]
\[ = B^2 + 2BC \left[ \because BC = \text{CB and} C^2 = O \right]\]
\[ = B\left( B + 2C \right) = RHS\]
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,
\[\]\[A^{k + 2} = A^{k + 1} A\]
\[ = \left[ B^k \left( B + \left( k + 1 \right)C \right) \right]\left( B + C \right) \left[\text{From eq} . \left( 1 \right) \right]\]
\[ = \left[ B^{k + 1} + \left( k + 1 \right) B^k C \right]\left( B + C \right)\]
\[ = B^{k + 1} \left( B + C \right) + \left( k + 1 \right) B^k C\left( B + C \right)\]
\[ = B^{k + 2} + B^{k + 1} C + \left( k + 1 \right) B^k CB + \left( k + 1 \right) B^k C^2 \]
\[ = B^{k + 2} + B^{k + 1} C + \left( k + 1 \right) B^k BC \left[ \because BC = \text{CB and} C^2 = 0 \right]\]
\[ = B^{k + 2} + B^{k + 1} C + \left( k + 1 \right) B^{k + 1} C\]
\[ = B^{k + 2} + \left( k + 2 \right) B^{k + 1} C\]
\[ = B^{k + 1} \left[ B + \left( k + 2 \right)C \right]\]
So the statement is true for n = k+1.
Hence, by the principle of mathematical induction,
\[n \in N\]
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