Question #206212

determine whether the vectors (1,3,-1,4),(3,8,-5,7),(2,9,4,23) in R^4 are linearly independent or linearly dependent 


1
Expert's answer
2021-06-15T16:33:47-0400

Let us determine whether the vectors (1,3,1,4), (3,8,5,7), (2,9,4,23)(1,3,-1,4),\ (3,8,-5,7),\ (2,9,4,23) in R4\R^4 are linearly independent or linearly dependent. Let us consider the equality

a(1,3,1,4)+b(3,8,5,7)+c(2,9,4,23)=(0,0,0,0).a(1,3,-1,4)+b (3,8,-5,7)+c (2,9,4,23)=(0,0,0,0).

It follows that (a+3b+2c,3a+8b+9c,a5b+4c,4a+7b+23c)=(0,0,0,0),(a+3b+2c, 3a+8b+9c,-a-5b+4c,4a+7b+23c)=(0,0,0,0), and hence we have the following system:


{a+3b+2c=03a+8b+9c=0a5b+4c=04a+7b+23c=0\begin{cases} a+3b+2c=0\\ 3a+8b+9c=0 \\ -a-5b+4c=0 \\ 4a+7b+23c=0\end{cases}


After adding to the second equation the first multiplied by -3, to the third equation the first, and to the fourth equation the first multiplied by -4, we conclude that the last system is equivalent to following the system:


  {a+3b+2c=0b+3c=02b+6c=05b+15c=0\begin{cases} a+3b+2c=0\\ -b+3c=0 \\ -2b+6c=0 \\ -5b+15c=0\end{cases}


After dividing the third equation by 2 and the fourth equation by 5, we conclude that the second equation coinside with the third equation and the fourth equation. Therefore, we have the following equivalent systems:


{a+3b+2c=0b=3ccR\begin{cases} a+3b+2c=0\\ b=3c \\ c\in\R\end{cases}


{a=3b2c=11cb=3ccR\begin{cases} a=-3b-2c=-11c\\ b=3c \\ c\in\R\end{cases}


Let c=1,c=1, then b=3, a=11.b=3,\ a=-11. We have that 11(1,3,1,4)+3(3,8,5,7)+(2,9,4,23)=(0,0,0,0),-11(1,3,-1,4)+3(3,8,-5,7)+(2,9,4,23)=(0,0,0,0), and hence we conclude that this vectors are linearly dependent.


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