Question

If $B,C$ are square matrices of order $n$ and if $A=B+C=CB,C^2=0$, then which of following is true for any positive integer $N$.

• A. $A^{N+1}=B^N(B+(N+1\left)C\right)$
• B. $A^N=B^N(B+(N+1\left)C\right)$
• C. $A^{N+1}=B(B+(N+1\left)C\right)$
• D. $A^{N+5}=B^N(B+(N+2\left)C\right)$

Answer 1

The correct option is A $A^{N+1}=B^N(B+(N+1\left)C\right)$

Solution - we will prove this solution by induction

and check by A options.

given $B,C$ are square matrix of order n and if

$A=B+C,BC=CB,C^2=0$

$\left(i\right)A^{n+1}=B^N(B+(N+1\left)C\right)$

For $n=1,$ we have

$A^2=B(B+2C)...\left(1\right)$

and given $A=B+C$

So, $A^2=(B+C)(B+C)=B^2+BC+CB+C^2$

$A^2=B^2+2BC$ ($\because BC=CB\&C^2=0)$

$A^2=B(B+2C)$

So, the statement is true for $n=1$

Assume that the statement is true for $n=k$

$\therefore A^{k+1}=B^k(B+(k+1\left)C\right)$

we shall now show that it is true for

$n=k+1$

That is we need to show that

$A^{k+2}=B^{k+1}(B+(K+2\left)C\right)$

Now, $A^{k+2}=A^{k+1}.A$

$=\left(B^k\right(B+(k+1)C\left)\right)(B+C)$ [using induction hypothesis]

$=(B^{k+1}+(k+1\left)B^{k}C\right)(B+C)$

$=B^{k+1}(B+C)+(K+1)B^{kc}(B+C)$

$=B^{k+2}+B^{k+1}C+(k+1)B^{k}CB+(k+1)B^k{C}^2$

$=B^{k+2}+B^{k+1}C+(k+1)B^{k}BC$ [$\because C^2=0\&BC=CB]$

$=B^{k+2}+B^{k+1}C+(k+1)B^{k+1}C$

$=B^{k+2}+(k+2)B^{k+1}C$

$A^{k+2}=B^{k+1}(B+(K+2\left)C\right)$

H.P

So, option (A) is correct

and option (B),(C) and (D) is Discard (Clearly)

because option (B),(C) and (D) is not

true for $n=1$