@Rishabh Property SABHI REAL NUNBERS ( POSITIVE AUR NEGATIVE DONO) ke liye satisfy honi chahiye, hame positive aur negative dono numbers lekar check karna hai. Agar hum +2 lenge to property satisfy ho jayegi but -2 lenge tab property satisfy nahi hogi. Maine only -2 hi liya hai ( to save time).
Condition should satisfy for all real numbers including positive and negetive. If we take +2 then condition will satisfy but if we take -2 then condition will not satisfy. To save time i choose -2 only.
Question 12 ka option 3 kyun nahin ho sakta Kyun ki kuchh matrices main subset ka condition satisfy hoga aur kuchh matrices main nahin hoga Please answer
(4,5) , W ka element nahi ho sakta...W ka element hone ke liye second component ka double , first component ke triple ke equal hona chahiye...ex (4, 6) , (7/3, 7/2), (10, 15) etc.
Q.15 Linearly independent ke case main sabhi scalars Zero hone chahiye. Alfa, beta, gama, and delta all are zero. Q.16 Linearly dependent ke case main sabhi scalars zero nahi ho sakte. So alfa, beta and gama all three can not equal to zero.
In 17Q we are not talking about scalars... Set of a single vector is LD if it is a zero vector. So a vector (x,y,z) is LD if its all components are zero i.e. x=y=z=0, then only it becomes a zero vector i.e. (0,0,0). Hope it helps ☺️
a(0,1,1,0)+b(1,2,3,4)+c(1,1,0,5)=(0,0,0,0) (0,a,a,0)+(b,2b,3b,4b)+(c,c,0,5c)=(0,0,0,0) (b+c,a+2b+c,a+3b,4b+5c)=(0,0,0,0) b+c=0 a+2b+c=0 a+3b=0 4b+5c=0 => a=0, b=0, c=0 Since all scalars a, b, c are zero so given vectors are LI.
Dear student, If x= y= z= 0 then S = { (0,0,0) } Now S is a set containing only zero vector and hence S is LD. Note that if a set contains zero vector then the set is Linearly Dependent. For Ex A= { (1,2,-3), (0,0,0), (4,-7,3) } B= { (0,0,0,0) } both sets A and B are Linearly Dependent. Hence in Q.17 option (c) is correct Now let S = { v1, v2, v3 } be a set containing three vectors. If a.v1+ b.v2+ c.v3=0 , a,b,c are scalars and a=b=c=0 then set S is Linearly Independent. For Ex If S= { (1,1,2), (1,-1,1),(1,0,1) } and a(1,1,2)+b(1,-1,1)+c(1,0,1)=(0,0,0) then a=b=c=0. Hence S is Linearly Independent.
Condition should satisfy for EVERY member of W. If sum of any two matrices of W, does not belong to W ,then W is not subspace. Actually it is very easy to check if determinants of two matrices are zero, then it is not necessary that determinant of there sum is also zero. So W is not subspace.
5:23 Q.3 option (c).
Main bhi confused ho gya tha😂
Yes sir..
yes but i understood ur feeling i noted 3 only🙂
Main v bahut confuse ho gaya tha
Ha
Vector space have 2 binary operations naming internal binary operation n external binary operation.. Clear concept
Very nicely explained 👌🏻👌🏻
Nice explanation this video is very helpful please jayada se video dalia
Bhai bhut gajab ka explanation diya hai 🙂🎉
Excellent teaching ....thanks a lot sir ..from tamil nadu
The best ever I saw....
Well explained ❤❤
Nice and clear explanation 👌 👌and very helpful
Very nice explained 👍👍
Great Job sir 👍
This helps me a lot sir thank you so much for this helpful video☺️
brillient sir
gret sir thank you
Informative video
Thanku dear sir please make vedio other topic 💫💫💫🙏🙏
Love from Lucknow UP ❤❤
Thanks Divyanshu ✌️
very useful video
Very very thankyu u cleared my all doubts
Bht ache sir
Very helpful
Q12 right answer is option C
No it's wrong because it doesn't satisfy option a only
No it's wrong because it doesn't satisfy option a only
Sir, thank u very much for videos. Sir, in 12th Q, w is not a subspace in all cases right sir, why not option c is true
Sir ek example ke through hum nehi bol saktahu je ehe subspace hay lekin example ke through not subspace dekha sakatu hu
Exallent sir 🙏
In mcqs 9, I have confusion coz we can take any numbers so we can get either positive or neg numbers
Very good
Thx a lot sir
Question 9 me jab -2 se scalar multiply kr rhe . But agr +2 se kre to condition satisfied hoga...
-2 hi kyo le rhe 11:16 video time pe
@Rishabh
Property SABHI REAL NUNBERS ( POSITIVE AUR NEGATIVE DONO) ke liye satisfy honi chahiye, hame positive aur negative dono numbers lekar check karna hai.
Agar hum +2 lenge to property satisfy ho jayegi but -2 lenge tab property satisfy nahi hogi.
Maine only -2 hi liya hai ( to save time).
Thanks sir .. just loved it. Very nicely explained
Complete linear algebra ki questions post kr do...
Great
Abhi to mcq questions nhi puchhne wale hai exam me so abhi preparation kaise kre
Q.11 m ap n √2 jo k irrational h q multiply krwaia h is ki kia rzn h..?
sir, at 11:23 onwards, why did you taken (-2) while scalar Multiplication ..? Instead of (-2) you can take one (2) right..?
Condition should satisfy for all real numbers including positive and negetive.
If we take +2 then condition will satisfy but if we take -2 then condition will not satisfy.
To save time i choose -2 only.
Sir exam me joh ayega is video me se hi ayega kya ? Meri rtmnu offline mcq exam hai. Kal
Nice Sir👍.. Thank You Sir😊
But one thing, sir plz teach in English 🙏🙏
Explain in English sir... Every one will understand sir... Pls... Thank you🙏
Question 12 ka option 3 kyun nahin ho sakta
Kyun ki kuchh matrices main subset ka condition satisfy hoga aur kuchh matrices main nahin hoga
Please answer
Sabhi metrices ke liye condition satisfy hona chahiye.
@@KTexplains thank you
Thank you so much sir
Sir agar Question 10 me W ke elements (4,5) lete hai to condition satisfy nhi ho raha hai
(4,5) , W ka element nahi ho sakta...W ka element hone ke liye second component ka double , first component ke triple ke equal hona chahiye...ex (4, 6) , (7/3, 7/2), (10, 15) etc.
Sir Q9 me x ki value hum kuch bhi le sakte hai ya kisi particular set se hi lena hai? Plz help
Q main vector space V3(R) given hai.
R means set of real numbers.
So x ki value ko set of real number se hi lena hai.
@@KTexplains thank you
Thank you so much 🥰
Please upload the 2nd part sir as early as possible
ruclips.net/video/3myyc_-9ZAw/видео.html
Sir ye questions ke pdf telegram pe dal dijiye
🎉🎉🎉15 august 2024 me padh rha hun🎉🎉🎉
Qstn 3 option c correct hoga
Well done Sir
Sir isky pdf mil skty?
Sir 16th samajh nhi aaya🤔
Kyoki 15 th m teeno ko zero rkha pr 16 m kyo nhi? Bta dijiyega Mera 27th May ko paper h plz
Q.15
Linearly independent ke case main sabhi scalars Zero hone chahiye.
Alfa, beta, gama, and delta all are zero.
Q.16
Linearly dependent ke case main sabhi scalars zero nahi ho sakte.
So alfa, beta and gama all three can not equal to zero.
Are sir 6 th sem ka koi bhi video nhi banaya hai kya
ruclips.net/p/PL_RrZ82fmpNWvMeHyFnlfK_e8T6Ekk_fa
In 17 th question , the answer is wrong
It is linearly dependent then the scalar are not all zero
Then what is the right option
In 17Q we are not talking about scalars...
Set of a single vector is LD if it is a zero vector.
So a vector (x,y,z) is LD if its all components are zero i.e. x=y=z=0, then only it becomes a zero vector i.e. (0,0,0).
Hope it helps ☺️
Ye questions paper me ajae
the zero vector in vector space vector space v4 is
(0,0,0,0)
@@KTexplains r4 is also 0000
@@Ommohije30 Yes
👌👌👌
Sir 30.20 min
Sir usko LI aur LD batane keliye kya method use karenge
a(0,1,1,0)+b(1,2,3,4)+c(1,1,0,5)=(0,0,0,0)
(0,a,a,0)+(b,2b,3b,4b)+(c,c,0,5c)=(0,0,0,0)
(b+c,a+2b+c,a+3b,4b+5c)=(0,0,0,0)
b+c=0
a+2b+c=0
a+3b=0
4b+5c=0
=> a=0, b=0, c=0
Since all scalars a, b, c are zero so given vectors are LI.
@@KTexplains thank you sir🙏
❤
Question 17 answer is totally wrong. Because all elements are zero then it became a linearly independent
Dear student,
If x= y= z= 0 then S = { (0,0,0) }
Now S is a set containing only zero vector and hence S is LD.
Note that if a set contains zero vector then the set is Linearly Dependent.
For Ex
A= { (1,2,-3), (0,0,0), (4,-7,3) }
B= { (0,0,0,0) }
both sets A and B are Linearly Dependent.
Hence in Q.17 option (c) is correct
Now let S = { v1, v2, v3 } be a set containing three vectors.
If a.v1+ b.v2+ c.v3=0 , a,b,c are scalars and a=b=c=0 then set S is Linearly Independent.
For Ex
If S= { (1,1,2), (1,-1,1),(1,0,1) } and
a(1,1,2)+b(1,-1,1)+c(1,0,1)=(0,0,0)
then a=b=c=0.
Hence S is Linearly Independent.
Tnx sir
Sir pdf chahiyee please......
👍
Sir in question 12....shall we take(4 6.....2 3) and (4 10.... 2 5) matrices.... It can satisfied the sub space condition... Why do we don't take it
Condition should satisfy for EVERY member of W.
If sum of any two matrices of W, does not belong to W ,then W is not subspace.
Actually it is very easy to check if determinants of two matrices are zero, then it is not necessary that determinant of there sum is also zero.
So W is not subspace.
Sir 17 questions,,,, LI - all will be zero
Sir Apne 15 no questions me kaha ki L.I Hoga jab all are 0. And 17 me kahe Rahe ho ki L.D Hoga jab all element zero ho
In Q15 we are talking about scalars but in Q17 we are talking about components of vector.
Very good
Thank u so much sir.