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thnku ..sir. ji..for this series......it is very helpful for every student..❤❤❤
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Sir vector space ke bhi short trik vedio upload kijiye na ....
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sir pls upload short trick videos on real analysis
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Jab determinant 0 ho jayega tho or eigen value kaise posible hogi
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A✔️
Ans A
Sir please 3 d geometry padha dije Conicoid se
Only a is true
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A option
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option a
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A
I DONE MY BETECH ON ELECTRICAL AND ELECTRONICS . BIOTECHNOLOGY ME EXAM KO ATTENDKAR SAKTI HU KYA SIR?
Opt A
A ans
In Q no.4. 2, -1 not necessarily be eigenvalues of M if the other condition has not been given, so need to explain it properly
CALEY HAMILTON THEORM - EVERY SQUARE MATRIX SATISFY HIS CHARACTERISTICS EQUATIONS ... So value of lambda's are eigen value
@@ankitkumardip90 the characteristics equation of identity matrix is (x-1)^3=0, but it satisfies (x-1)^2(x-2)=0 as well, and then by your logic, 2 also must be an eigen value of identity matrix, which is not possible
@@mdhar7825 you are wrong ... You are going reversely... Sufficient condition is not true
@@mdhar7825 (x-1)^3 ko jo jo satisfy karenge wahi unka eigen value honge ...
@@ankitkumardip90 in question where is it been mentioned that this is the characteristics equation of the matrix, you can't assume on your own
Aaa
Sir iska answer A h
A,c
Option A is correct
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thnku ..sir. ji..for this series......it is very helpful for every student..❤❤❤
Key points are outstanding sir 🙏 🙏 🙏
Option A is correct answer 👍👍
Thank you so much sir very helpful video 🙏🙏🔥🤗
Option A correct✅✅
Thank you sir 🙏🙏🙏🙏
Option (A ) is correct.
Thank you so much Sir . It is helping us a lot. 🤗
No words are enough to complement U sir
Thanks 😊😊😊✌️✌️👌🐦
Thank you sir , option A is correct answer
Thanks sir . You are my inspiration.
LOVE YOU SIR 😍 💗
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Thank u guruji
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Opt a
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Thank you sir most helpful lectures
Thank you sir. Love from Pakistan❤
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Thank you sir🙏. Ans:A
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Sir vector space ke bhi short trik vedio upload kijiye na ....
Sir please start 2d geometry with starting
A is correct 💯
sir pls upload short trick videos on real analysis
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Option ( a ) ....
Jab determinant 0 ho jayega tho or eigen value kaise posible hogi
Option A
A✔️
Ans A
Sir please 3 d geometry padha dije
Conicoid se
Only a is true
A is correct in 5 sec
A option
Only 1 is right
option a
A is truu
A
I DONE MY BETECH ON ELECTRICAL AND ELECTRONICS . BIOTECHNOLOGY ME EXAM KO ATTEND
KAR SAKTI HU KYA SIR?
Opt A
A ans
A
In Q no.4. 2, -1 not necessarily be eigenvalues of M if the other condition has not been given, so need to explain it properly
CALEY HAMILTON THEORM - EVERY SQUARE MATRIX SATISFY HIS CHARACTERISTICS EQUATIONS ...
So value of lambda's are eigen value
@@ankitkumardip90 the characteristics equation of identity matrix is (x-1)^3=0, but it satisfies (x-1)^2(x-2)=0 as well, and then by your logic, 2 also must be an eigen value of identity matrix, which is not possible
@@mdhar7825 you are wrong ... You are going reversely... Sufficient condition is not true
@@mdhar7825 (x-1)^3 ko jo jo satisfy karenge wahi unka eigen value honge ...
@@ankitkumardip90 in question where is it been mentioned that this is the characteristics equation of the matrix, you can't assume on your own
Aaa
Sir iska answer A h
A,c
A is correct
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Option A
A
A
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A
A
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A
A
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A
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A
A
A
A
A
A
A
A