Question-186963

Question Number 186963 by 073 last updated on 12/Feb/23

Commented by 073 last updated on 12/Feb/23

$$\mathrm{solotion}\:\mathrm{please}?? \\ $$$$\mathrm{need}\:\mathrm{it} \\ $$

Answered by Rasheed.Sindhi last updated on 13/Feb/23

determinant ((∗,K,E,M,A,L),(K,E,M,A,L,K),(E,M,A,L,K,E),(M,A,L,K,E,M),(A,L,K,E,M,A),(L,K,E,M,A,L)) K^3 ∗(L^(−5) ∗E)=? K^3 =(K∗K)∗K=E∗K=M L^(−5) =(L^(−1) )^5 =(L^5 )^(−1) (X^(−1) means inverse of X) It can be observed that X∗L=X , where X∈{K,E,M,A,L} ∴ L is an identity element w r t ∗ ∵ K∗A=L ∴ K^(−1) =A Similarly L^(−1) =L L^(−5) =(L^(−1) )^5 =L^5 =L K^3 ∗(L^(−5) ∗E)=M∗(L∗E) =M∗E=L

$$\begin{array}{|c|c|c|c|c|c|}{\ast}&\hline{\mathrm{K}}&\hline{\mathrm{E}}&\hline{\mathrm{M}}&\hline{\mathrm{A}}&\hline{\mathrm{L}}\\{\mathrm{K}}&\hline{\mathrm{E}}&\hline{\mathrm{M}}&\hline{\mathrm{A}}&\hline{\mathrm{L}}&\hline{\mathrm{K}}\\{\mathrm{E}}&\hline{\mathrm{M}}&\hline{\mathrm{A}}&\hline{\mathrm{L}}&\hline{\mathrm{K}}&\hline{\mathrm{E}}\\{\mathrm{M}}&\hline{\mathrm{A}}&\hline{\mathrm{L}}&\hline{\mathrm{K}}&\hline{\mathrm{E}}&\hline{\mathrm{M}}\\{\mathrm{A}}&\hline{\mathrm{L}}&\hline{\mathrm{K}}&\hline{\mathrm{E}}&\hline{\mathrm{M}}&\hline{\mathrm{A}}\\{\mathrm{L}}&\hline{\mathrm{K}}&\hline{\mathrm{E}}&\hline{\mathrm{M}}&\hline{\mathrm{A}}&\hline{\mathrm{L}}\\\hline\end{array} \\ $$$$\mathrm{K}^{\mathrm{3}} \ast\left(\mathrm{L}^{−\mathrm{5}} \ast\mathrm{E}\right)=? \\ $$$$\mathrm{K}^{\mathrm{3}} =\left(\mathrm{K}\ast\mathrm{K}\right)\ast\mathrm{K}=\mathrm{E}\ast\mathrm{K}=\mathrm{M} \\ $$$$\mathrm{L}^{−\mathrm{5}} =\left(\mathrm{L}^{−\mathrm{1}} \right)^{\mathrm{5}} =\left(\mathrm{L}^{\mathrm{5}} \right)^{−\mathrm{1}} \\ $$$$\left(\mathrm{X}^{−\mathrm{1}} \:{means}\:{inverse}\:{of}\:\:\mathrm{X}\right) \\ $$$${It}\:{can}\:{be}\:{observed}\:{that} \\ $$$$\mathrm{X}\ast\mathrm{L}=\mathrm{X}\:,\:{where}\:\mathrm{X}\in\left\{\mathrm{K},\mathrm{E},\mathrm{M},\mathrm{A},\mathrm{L}\right\} \\ $$$$\therefore\:\mathrm{L}\:{is}\:{an}\:\boldsymbol{{identity}}\:\boldsymbol{{element}}\:{w}\:{r}\:{t}\:\ast \\ $$$$\because\:\mathrm{K}\ast\mathrm{A}=\mathrm{L} \\ $$$$\therefore\:\mathrm{K}^{−\mathrm{1}} =\mathrm{A} \\ $$$${Similarly}\:\mathrm{L}^{−\mathrm{1}} =\mathrm{L} \\ $$$$\mathrm{L}^{−\mathrm{5}} =\left(\mathrm{L}^{−\mathrm{1}} \right)^{\mathrm{5}} =\mathrm{L}^{\mathrm{5}} =\mathrm{L} \\ $$$$\: \\ $$$$\mathrm{K}^{\mathrm{3}} \ast\left(\mathrm{L}^{−\mathrm{5}} \ast\mathrm{E}\right)=\mathrm{M}\ast\left(\mathrm{L}\ast\mathrm{E}\right) \\ $$$$=\mathrm{M}\ast\mathrm{E}=\mathrm{L} \\ $$

Commented by 073 last updated on 14/Feb/23

$$\mathrm{nice}\:\mathrm{solution} \\ $$$$\mathrm{thanks}\:\mathrm{alot} \\ $$

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