geduiwelh

2021-08-11

Determine whether the vectors ${\upsilon }_{1}=\left(-1,2,-3\right),{\upsilon }_{2}=\left(3,2,1\right),{\upsilon }_{3}=\left(5,6,-1\right)$ are linearly dependent or independent in ${\mathbb{R}}^{3}.$

grbavit

Here we have to determine the vectors ${\upsilon }_{1}=\left(-1,2,-3\right),{\upsilon }_{2}=\left(3,2,1\right),{\upsilon }_{3}=\left(5,6,-1\right)$are linearly dependent or independent in${\mathbb{R}}^{3}$
. Let us consider the relation
${c}_{1}{\upsilon }_{1}+{c}_{2}{\upsilon }_{2}+{c}_{3}{\upsilon }_{3}=0$
$⇒{c}_{1}\left(-1,2,13\right)+{c}_{2}\left(3,2,1\right)+{c}_{3}\left(5,6,-1\right)=\left(0,0,0\right)$
$⇒-{c}_{1}+3{c}_{2}+5{c}_{3}=0$
$2{c}_{1}+2{c}_{2}+6{c}_{3}=0$
$-3{c}_{1}+{c}_{2}-{c}_{3}=0$
This above system has no zero solution since the determinant of the matrix
$A=\left[\left[-1,3,5\right]\left[2,2,6\right]\left[-3,1,-1\right]\right]$ is zero.
Since, the matrix is singular then there exists a non-zero vector $c=\left({c}_{1},{c}_{2},{c}_{3}\right)$  such that Ac=0.
Thus, the vectors ${\upsilon }_{1}=\left(-1,2,-3\right)$,
${\upsilon }_{2}=\left(3,2,1\right),{\upsilon }_{3}=\left(5,6,-1\right)$ are linearly dependent.

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