Question

In: Math

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 3rd vector v3 is a linear combination of v1,v2,v4.

3. It may be observed from the RREF of A that the 1st, 2nd and 4th 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 scalars).


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