25 Sep 2008 lecture by Sir Faazil
Linear Algebra
Let v1, v2…… vk be vectors in a vector space V,A vector v in V is called a linear combination of v1, v2…… vk if v = c1v1+c2v2+…….+ckvk
For some real numbers c1, c2, c3…..ck.
Linear combination of two vectors:-
V=2i^ + 4j^ + 3k^ (ya unit vectors han actually in ki cap sahee trha sa dalna ni ata.is ley side pa ha.)
V=2(2i^ + 4j^ + 3k^)
Example:-
Show that
V= (2, 1, 5) is a linear combination of
V1= (1, 2, 1), v2= (1, 0, 2)
And v3= (1, 1, 0)(is ma hamna ya show krana ha v1, v2, v3 ka combination V k equal ho ga.is cheez ko proof krna k ley koi bhi combination use kren ga.)
Solution:-
V is a linear combination of v1,v2,v3 if we can find real numder c1,c2, and c3
So that
V= c1v1+c2v2+…….+ckvk ----------------(1)
idher c1,c2,c3 ko paranthesis k ander multiply kr dea ha.teenon steps isi k han.
(2, 1, 5) = c1(1,2,1)+c2(1,0,2)+c3(1,1,0)
(2, 1, 5) = (c1,2c1,c1)+(c2,0,2c2)+(c3,c3,0)
(2, 1, 5) = (c1+c2+c3, 2c1+c3, c1+2c2)
yellow shading ma right side pa jtni values han respectively in ko add krna ha.phir next step ma equation k left side ki values ko right side wali values k sath equate kr dena ha.jesa k ab dea gaya ha….
2= c1+c2+c3 --------- (2)
1=2c1+c3 --------- (3)
5= c1+2c2 --------- (4)
Now we have to find the values of c1, c2 and c3
Equation (2) - Equation(3) (subtract the equation 2 and 3)
2 = c1 + c2 + c3
1 = 2c1 + c3
1 = -c1 +c2
ð1 = c2 - c1 -------------(A)
Now Eq (4) + eq (A)
5 = c1 + 2c2
1 = -c1 + c2
6 = +3c2
6= 3c2
c2=6/3
c2=2 -------------(B)
From equation (A)
1=c2-c1
1=2-c1
c1=1 ---------------(C)
From equation (3)
1=2c1+c3
1=2(1)+c3
c3=-1 ----------------------(D)
==>
2= c1+c2+c3 --------- (2)
1=2c1+c3 --------- (3)
5= c1+2c2 --------- (4)
c1=1 ,c2=2 ,c3=-1
so equation (1) becomes
v=v1+2v2-v3
(2, 1, 5) = c1(1,2,1)+c2(1,0,2)+c3(1,1,0)
(2, 1, 5) = (1, 2, 1) +2(1, 0, 2)-(1, 1, 0) ( by putting the values of c1,c2,c3 )
(2, 1, 5) = (1, 2, 1) +(2, 0, 4)-(1, 1, 0)
(2, 1, 5) = (3,2,5) -(1, 1, 0)
(2, 1, 5) =(2, 1, 5) hence proved.
Thanks Behna...