Let Vector_1=[Column (1 &-2&1)],Vector_2=[Column(2&-1&1)],Vector_3=[Column(-2&... Let Vector_1=[Column (1 &-2&1)],Vector_2=[Column(2&-1&1)],Vector_3=[Column(-2&-2&0)],Vector_4=[Column(1&-2&-2)]. Show tha

Let Vector_1=[Column (1 &-2&1)],Vector_2=[Column(2&-1&1)],Vector_3=[Column(-2&... Let Vector_1=[Column (1 &-2&1)],Vector_2=[Column(2&-1&1)],Vector_3=[Column(-2&-2&0)],Vector_4=[Column(1&-2&-2)].

Show that S={Vector_1,Vector_2,Vector_3,Vector_4} is linearly dependent.

Show that Vector_3 can be written as a linear combination of Vector_1,Vector_2 and Vector_4.

Show that T={Vector_1,Vector_2,Vector_4} is linearly independent.

Show that the set of all linear combinations of vectors from S is the same as the set of all linear combinations of vectors from T.

Solutions

Expert Solution

Let the given 4 vectors be denoted by v₁,v₂,v₃,v₄ respectively.

1. We know that dimension of R³ is 3. Therefore, the set S = { v₁,v₂,v₃,v₄ } is linearly dependent.

2. Let A be the 3x4 matrix with the given 4 vectors as columns. Then A =

1	2	-2	1
-2	-1	-2	-2
1	1	0	-2

The RREF of A is

1	0	2	0
0	1	-2	0
0	0	0	1

This implies that v₃ = 2v₁-2v₂+0v₄ . Thus, the 3^rd vector v₃ is a linear combination of v₁,v₂,v₄.

3. It may be observed from the RREF of A that the 1^st, 2^nd and 4^th columns of A are linearly independent as none of these can be expressed as linear combinations/scalar multiples of other columns. Therefore, the set T = { v₁,v₂,v₃} is linearly independent.

4. Since v₃ = 2v₁-2v₂+0v₄, hence every linear combination of v₁,v₂,v₃,v₄ is a linear combination of v₁,v₂,v₄. e.g. av₁+bv₂+cv₃+dv₄ = av₁+bv₂+dv₄ +c(2v₁-2v₂ ) = (a+2c)v₁+(b-2c)v₂+ dv₄.( here, a,b,c,d are arbitrary real numbers).

milcah answered 1 year ago

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