InterviewSolution
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InterviewSolution
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Let A and B be square matrices of the same order such that `A^(2)=I` and `B^(2)=I`, then which of the following is CORRECT ?A. IF A and B are inverse to each other, then `A=B`.B. If `AB=BA`, then there exists matrix `C=(AB+BA)/2` such that `C^(2)=C`.C. If `AB=BA`, then there exists matrix `D=AB-BA` such that `D^(n)=O` for some `n in N`.D. If `AB=BA` then `(A+B)^(5)=16 (A+B)`. |
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Answer» We have `A^(2)=I` and `B^(2)=I` `:. A=A^(-1)` and `B=B^(-1)` If A and B are inverse to each other, then `A=B^(-1)=B`. `C=(AB+BA)/2` `implies C^(2)=((AB)^(2)+(BA)^(2)+AB.BA+BA.AB)/4` Now, `ABBA=AIA=A A=I` Similarly, `BA AB=I` Also `(AB)^(2)=ABAB` `=A AB B" "("if "AB=BA)` `=A^(2)B^(2)` `=I` Similarly `(BA)^(2)=I` `:. C^(2)=I` Thus, if `AB=BA`, then `C^(2)=I` `D=AB-BA` `implies D^(2)=(AB)^(2)+(BA)^(2)-ABxxBA-BAxxAB` `=I+I-I-I" "("if "AB=BA)` `=O` Thus, if `AB=BA`, then `D^(n)=O`. `(A+B)^(5)=A^(5)+5A^(4)B+10A^(3)B^(2)+10A^(2)B^(3)+5AB^(4)+B^(5)` `=A+5B+10A+10B+5A+B` `=16(A+B)` |
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