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Question Number 187186 by 073 last updated on 14/Feb/23

Commented by 073 last updated on 14/Feb/23

please solution??

pleasesolution??

Commented by Rasheed.Sindhi last updated on 15/Feb/23

k=4 (please confirm the answer)

k=4(pleaseconfirmtheanswer)

Commented by mr W last updated on 15/Feb/23

k=4 is correct.

k=4iscorrect.

Commented by Rasheed.Sindhi last updated on 16/Feb/23

sir mr W, when you have some leisure  please see my answer to Q#186966  for its health!

sirmrW,whenyouhavesomeleisureYou can't use 'macro parameter character #' in math modeforitshealth!

Commented by 073 last updated on 16/Feb/23

thanks

thanks

Commented by 073 last updated on 16/Feb/23

thanks

thanks

Answered by Rasheed.Sindhi last updated on 15/Feb/23

 determinant ((★,1,2,3,4,5),(1,5,1,2,3,4),(2,1,2,3,4,5),(3,2,3,4,5,1),(4,3,4,5,1,2),(5,4,5,1,2,3))  A={1,2,3,4,5}  x★e=x ∀x∈A⇒e=2  1^(−1) =3   [∵ 1★3=2]  2^(−1) =2   [∵ 2★2=2]  3^(−1) =1    [∵ 3★1=2]  4^(−1) =5    [∵ 4★5=2]  5^(−1) =4   [∵ 5★4=2]    (2★k)^(−1) ★5=3  k^(−1) ★5=3   [∵2★k=k]  k^(−1) ★5★5^(−1) =3★5^(−1)   k^(−1) ★(5★5^(−1) )=3★4  [∵5^(−1) =4]  k^(−1) ★2=3★4     [∵5★5^(−1) =2]  k^(−1) =5     [∵ k^(−1) ★2=k^(−1) ]  k=4

12345151234212345323451434512545123A={1,2,3,4,5}xe=xxAe=211=3[13=2]21=2[22=2]31=1[31=2]41=5[45=2]51=4[54=2](2k)15=3k15=3[2k=k]k1551=351k1(551)=34[51=4]k12=34[551=2]k1=5[k12=k1]k=4

Commented by 073 last updated on 16/Feb/23

thanks alot sir

thanksalotsir

Answered by mr W last updated on 15/Feb/23

x∗e=x ⇒e=2  x∗x^(−1) =e=2 ⇒1^(−1) =3, 2^(−1) =2, 3^(−1) =1, 4^(−1) =5, 5^(−1) =4  5∗5=3 ⇒(2∗k)^(−1) =5 ⇒2∗k=4 ⇒k=4 ✓

xe=xe=2xx1=e=211=3,21=2,31=1,41=5,51=455=3(2k)1=52k=4k=4

Commented by 073 last updated on 16/Feb/23

nice solution

nicesolution

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