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Let $B, C$ be $n \times n$ matrices such that $A=B+C, B C=C B$ and $C^2$ is a null matrix. Then, $B^{2020}[B+(2021) C]=$
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Verified Answer
The correct answer is:
$A^{2021}$
Given that,
$A=B+C$
$B C=C B$ and $C^2=0$
From Eq. (i)
$\begin{aligned} & A^{n+1}=(B+C)^{n+1} \\ & \Rightarrow \quad A^{n+1}={ }^{n+1} C_0 \cdot B^{n+1}+{ }^{n+1} C_1 C \cdot B^n+\ldots \\ & \text { But, } \quad C^2=0, C^3=C^4 \ldots=C^r=0 \\ & \therefore \quad A^{n+1}={ }^{n+1} C_0 B^{n+1}+{ }^{n+1} C_1 B^n \cdot C \\ & =B^{n+1}+(n+1) B^n C \\ & A^{n+1}=B^n[B+(n+1) C] \\ & \therefore B^{2020}[B+(2020+1) C]=A^{2020+1} \\ & \Rightarrow \quad B^{2020}[B+(2021) C]=A^{2021} \\ & \end{aligned}$
$A=B+C$
$B C=C B$ and $C^2=0$
From Eq. (i)
$\begin{aligned} & A^{n+1}=(B+C)^{n+1} \\ & \Rightarrow \quad A^{n+1}={ }^{n+1} C_0 \cdot B^{n+1}+{ }^{n+1} C_1 C \cdot B^n+\ldots \\ & \text { But, } \quad C^2=0, C^3=C^4 \ldots=C^r=0 \\ & \therefore \quad A^{n+1}={ }^{n+1} C_0 B^{n+1}+{ }^{n+1} C_1 B^n \cdot C \\ & =B^{n+1}+(n+1) B^n C \\ & A^{n+1}=B^n[B+(n+1) C] \\ & \therefore B^{2020}[B+(2020+1) C]=A^{2020+1} \\ & \Rightarrow \quad B^{2020}[B+(2021) C]=A^{2021} \\ & \end{aligned}$
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