3 and 4 .Determinants and Matrices
normal

Let $A$ be a non-zero periodic matrix with period $4$ and $A^{12} + B =I$, where $I$ is identity matrix and $B$ is any square matrix of same order as of $A$. Matrix product $AB$ is equal to

A

$I$

B

$A$

C

$A + I$

D

null matrix

Solution

$\mathrm{A}=\mathrm{A}^{4+1} \Rightarrow \mathrm{A}=\mathrm{A}^{5}=\mathrm{A}^{9}=\Lambda^{13}$

$\therefore \mathrm{A}^{12}+\mathrm{B}=\mathrm{I}$

$\mathrm{A}\left(\mathrm{A}^{12}+\mathrm{B}\right)=\mathrm{A} \Rightarrow \mathrm{A}^{13}+\mathrm{AB}=\mathrm{A} \ldots \Rightarrow \mathrm{AB}=0$

 

Standard 12
Mathematics

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