Assume that vector u, vector v, vector w, vector x are linearly independent. Determine whether the following vectors are linearly independent or dependent. If they are linearly dependent then write a dependence relation for them.

2(vector u)-vector v + vector x, 3(vector v) + vector w + 4(vector x), vector x + vector w, 2(vector u) -2(vector v) + vector w + vector x

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If we prepare the matrix corresponding to the above set of vectors we get

[[2,-1,0,1],[0,3,1,4],[0,0,1,1],[2,-2,1,1]]

which can be rearranged to

[[2,-2,1,1],[2,-1,0,1],[0,3,1,4],[0,0,1,1]]`~~` [[1,-1,1/2,1/2],[0,3,1,4],[0,0,1,0],[0,0,1,1]]

Finally we see that determinant of the above matrix is non zero. And so vectors are linearly independent.

Let vect a=2u-v+0w+x

vect b=0u+3v+w+4x

vect c=0u+ov+w+x

vect d=2u-2v+w+x

Let p,q,r,s be real nos, such that

pa+qb+rc+sd=0 (i)

then p=q=r=s=0 .

Then vectors a ,b, c, and d are Linearly independent otherwise linearly dependent.

`[[2,-1,0,1],[0,3,1,4],[0,0,1,1],[2,-2,1,1]][[p],[q],[r],[s]]=[[0],[0],[0],[0]]`

`` write row echlon form of coefficient matrix

`[[1,0,0,1],[0,1,0,1],[0,0,1,1],[0,0,0,0]][[p],[q],[r],[s]]=[[0],[0],[0],[0]]` ,thus rank of the coefficient matrix is 3.

There are infinite numbers of value of p,q,r and s. So vectors a,b,c, and d are linearly dependent.

thus we assume `s!=0` , p+s=0 , q+s=0, q+r=0

solving syestem of equations,we have

p=-s,q=-s and r=s .

Thus from (i) ,we have

pa+qb+rc+sd=0

-sa-sb+sc+sd=0

d=a+b-c which is the dependence relation

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