Find a linearly independent set of vectors that spans the same substance of $\mathbb{R}^3$ as that spanned by the vectors $\begin{bmatrix}2\\2\\-1\end{bmatrix}, \begin{bmatrix}-8\\-2\\5\end{bmatrix}, \begin{bmatrix}-3\\0\\2\end{bmatrix}$
So I put this matrix into RREF to get:
$\begin{bmatrix}1&0&1/2\\0&1&1/2\\0&0&0\end{bmatrix}$ , but how do I know if this is a linearly independent set of vectors? And also I tried to put in this for the answer and it said it was incorrect:
$\begin{bmatrix}1/2\\1\\0\end{bmatrix}, \begin{bmatrix}1/2\\0\\1\end{bmatrix}$
Best Answer
I personally feel that automatically putting things into "matrix form" and "row reducing" is too often a substitute for understanding what you are doing! A set of vectors, $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$, is defined to be "linearly independent" if and only if the only solution to $a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0$ is $a_1= a_2= \cdot\cdot\cdot= a_n= 0$.
The three vectors here are $(2, 2, -1)$, $(-8, -2, 5)$, $(-3, 0, 2)$ so the equation is $a(2, 2, -1)+ b(-8, -2, 5)+ c(-3, 0, 2)= (2a- 8b- 3c, 2a- 2b, -a+ 5b+ 2c)= (0, 0, 0)$ so we must have 2a- 8b- 3c= 0, 2a- 2b= 0, -a+ 5b+ 2c= 0. From 2a- 2b= 0, a= b. Putting that into the first equation, 2a- 8a- 3c= -6a- 3c= 0 so c= -2a. Putting b= a and c= -2a into the third equation -a+ 5a- 4a= 0a= 0. Then a(2, 2, -1)+ a(-8, -2, 5)- 2a(-3, 0, 2)= 0 so these vectors are linearly dependent. We can write a(2, 2, -1)= -a(-8, -2, 5)+ 2a(-3, 0, 2) and then divide by a: (2, 2, -1)= (-8, -2, 5)+ 2(-3, 0, 2). Since that vector can be written as a linear combination of the other two vectors, and those vectors [b]are[/b] independent (one is not a multiple of the other), the set {(-8, -2, 5), (-3, 0, 2)} is the largest subset of linearly independent vectors.