Csir Net 16 September 2022 Linear Algebra Solution | Question ID 378 | Pure Mathematical Academy

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Pure Mathematical Academy

Pure Mathematical Academy

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Пікірлер: 13
@saravana4217
@saravana4217 Жыл бұрын
Excellent explanation sir
@rajeshlahare3487
@rajeshlahare3487 25 күн бұрын
Thank you so much sir 🙏🙏❤❤
@soumyajitkundu7751
@soumyajitkundu7751 2 жыл бұрын
Excellent explanation.
@shamamasthani9842
@shamamasthani9842 Жыл бұрын
Very nice explanation...
@mukeshkumarkaziana2989
@mukeshkumarkaziana2989 Жыл бұрын
I think option c says that one pair of imaginary roots should thati is all roots should not be real 14:38
@gauravtyagitv7538
@gauravtyagitv7538 7 ай бұрын
Yes you are correct
@RUPINDERKAUR-qp8vl
@RUPINDERKAUR-qp8vl Жыл бұрын
U and V ME y,z kis bases pr lyi hai ??
@Ramanarajacademy1986
@Ramanarajacademy1986 2 жыл бұрын
Could you able to give what is the exact linear invertible transformation from T:R^3 to R^3 such that T(U)=V
@puremathcsir2021
@puremathcsir2021 2 жыл бұрын
Acc to my explainaton , take T(x,y,z)=(x-y , z , y/2 )
@JL-uy8hz
@JL-uy8hz Жыл бұрын
How do u construct the map T(1,0,0)=(1,0,0) is not clear. If we don't know the exact linear mapping then can we say that T(1,0,0)=(1,0,0). What is the criteria to write T(1,0,0)=(1,0,0)? Is it only L.I...? Means if T(U)=V then can the extended basis of U (i.e,Here basis of R3) always map extended basis of V(i.e, basis of R3 here) How do you know that T maps (1 0 0) to (1 00) as we don't calculate the mapping
@saravana4217
@saravana4217 Жыл бұрын
Yes You are correct But in exam hall we don't have enough time to find that transformation... That's why he took Linearly independent set maps to Linearly independent sets Hence it is one one as well as invertible....
@hardikmishra3223
@hardikmishra3223 Жыл бұрын
Nice
@deepikajind2245
@deepikajind2245 Жыл бұрын
Second option me T(U) =V hogya kya
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