28 Replies
,rotate
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.57 btw
Look at it this way: n(n+1) is even so if A²=I they will lie on every T_n. (As I^(anything)=I) and plugging n=1 we get T_1 consisting of the A²=I sets only.
So the required intersection is elements of T_1 only
Ok this feels like 58
Oh sorry
Sorry
Wait a minute I forgot how to write matrices in latex, I'll brb.
From S we can see that $S^{2}=\begin{pmatrix} 1 & a(b-1) \ 0 & b^{2} \end{pmatrix}$
SirLancelotDuLac
For this to be Identity matrix set b=1 and a can be anything, correct?
So the number of elements is 100. (1 choice for b and 100 choices for A)
@Phalawor
Not getting the Tn wala part
For any even index of A we get an element for Tn why did you just settle on 1
T_2 gives us A^6 why did we not consider that
Is it because that the case already happened in T_1 ?
We need only the intersections right? Now, $T{1}$ wale elements have $S^{2}=I$ and nothing else. Also n(n+1) is even so these elements will also have $S^{n(n+1)}=I$ for all n and hence lie on every $T{n}$
SirLancelotDuLac
But consider some S which has A^6=I and not A^2=I. Since these are not in T_1, these will not be counted in the intersection.
Achchaa
From here, the maximum things you can "take common'' out of all sets is all of the elements of T_1
The "air 1" flair from jeeneetards ðŸ˜:psyduck:
bro this is some next level stuff
so basically T_1 is the biggest of them all
sorry smallest
the most fundamental isiliye intersection me keval use hi leke we go forward??
baaki ye to saajh gaya mai (sirf yahi samjh paaya apne aap 😠)
Yep. We don't need to concern ourselves with the other sets once one finds the fact that T_1 is the fundamental set. (Like its the largest stuff that you can take common from all sets)
NICEE
GOTEEM THENN
but lke
is this mathematically correct
oh wait nvm makes sense
ok cool
achcha chalo ye to hua bhai 58 kaise karenge @SirLancelotDuLac
like how do we go about the case working??
Okay, so we have 0,1,2,3,4,5 and the prime we require lies in {3,5,7}. So make 1 case as p=3
AHAAA OKOK
same way i was going yayyy
Which can be written as 1+1+1+0=3 (giving us 4 matrices)
2+1+0+0 giving us 12 or smth
cool cool
thanks
+solved @SirLancelotDuLac
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