Question #290592

 Study whether the vectors v1(1, 1, 2), v2(2, 3, 0), v3(0, 1, 2) in R 3 are linearly independent.


1
Expert's answer
2022-01-25T17:08:57-0500

A sequence of vectors v1,v2,...,vk\vec v_1, \vec v_2,..., \vec v_k  from a vector space VV is said to be linearly dependent, if there exist scalars a1,a2,...,aka_1, a_2, ..., a_k not all zero, such that


a1v1+a2v1+...+akvk=0a_1\vec v_1+a_2\vec v_1+...+a_k\vec v_k=\vec 0

where 0\vec 0 denotes the zero vector.

Consider the set of vectors v1=(1,1,2),v2=(2,3,0),v3=(0,1,2)\vec v_1=(1,1,2), \vec v_2=(2,3,0), \vec v_3=(0,1,2) then the condition for linear dependence seeks a set of non-zero scalars, such that


[120131202][a1a2a3]=[000]\begin{bmatrix} 1 & 2 & 0 \\ 1 & 3 & 1 \\ 2 & 0 & 2 \\ \end{bmatrix}\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

Augmented matrix


[120 0131 0202 0]\begin{bmatrix} 1 & 2 & 0 & \ 0 \\ 1 & 3 & 1 & \ 0 \\ 2 & 0 & 2 & \ 0 \\ \end{bmatrix}

R2=R2R1R_2=R_2-R_1


[120 0011 0202 0]\begin{bmatrix} 1 & 2 & 0 & \ 0 \\ 0 & 1 & 1 & \ 0 \\ 2 & 0 & 2 & \ 0 \\ \end{bmatrix}

R3=R32R1R_3=R_3-2R_1


[120 0011 0042 0]\begin{bmatrix} 1 & 2 & 0 & \ 0 \\ 0 & 1 & 1 & \ 0 \\ 0 & -4 & 2 & \ 0 \\ \end{bmatrix}

R1=R12R2R_1=R_1-2R_2


[102 0011 0042 0]\begin{bmatrix} 1 & 0 & -2 & \ 0 \\ 0 & 1 & 1 & \ 0 \\ 0 & -4 & 2 & \ 0 \\ \end{bmatrix}

R3=R3+4R2R_3=R_3+4R_2


[102 0011 0006 0]\begin{bmatrix} 1 & 0 & -2 & \ 0 \\ 0 & 1 & 1 & \ 0 \\ 0 & 0 & 6 & \ 0 \\ \end{bmatrix}

R3=R3/6R_3=R_3/6


[102 0011 0001 0]\begin{bmatrix} 1 & 0 & -2 & \ 0 \\ 0 & 1 & 1 & \ 0 \\ 0 & 0 & 1 & \ 0 \\ \end{bmatrix}

R1=R1+2R3R_1=R_1+2R_3


[100 0011 0001 0]\begin{bmatrix} 1 & 0 & 0 & \ 0 \\ 0 & 1 & 1 & \ 0 \\ 0 & 0 & 1 & \ 0 \\ \end{bmatrix}

R2=R2R3R_2=R_2-R_3


[100 0010 0001 0]\begin{bmatrix} 1 & 0 & 0 & \ 0 \\ 0 & 1 & 0 & \ 0 \\ 0 & 0 & 1 & \ 0 \\ \end{bmatrix}

Then a1=a2=a3=0.a_1=a_2=a_3=0.

Therefore the vectors v1=(1,1,2),v2=(2,3,0),v3=(0,1,2)\vec v_1=(1,1,2), \vec v_2=(2,3,0), \vec v_3=(0,1,2) in R3R ^3 are linearly independent.


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