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

\[P\left( 1 \right) : A^2 = B\left( B + 2C \right)\]
\[\]
\[Here, \]
\[LHS = A^2 \]
\[ = \left( B + C \right)\left( B + C \right)\]
\[ = B\left( B + C \right) + C\left( B + C \right)\]
\[ = B^2 + BC + CB + C^2 \]
\[ = B^2 + 2BC \left[ \because BC = \text{CB and} C^2 = O \right]\]
\[ = B\left( B + 2C \right) = RHS\]

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,

\[P\left( n \right)\]is true for all

\[n \in N\]