# For the following set of vectors, determine whether it is linearly independent or linearly dependent. If it is linearly dependent, find all its maximum linearly independent subsets? {(1,2,-1),(2,4,6),(0,0,-8)}

## from my ERO, I found out that it is linearly dependent but I dont know how to find its linearly independent subsets

Jul 9, 2018

#### Explanation:

The vectors are

$\left\{\begin{matrix}1 & 2 & - 1 \\ 2 & 4 & 6 \\ 0 & 0 & - 8\end{matrix}\right\}$

The vectors are independent if

$\alpha \left(\begin{matrix}1 \\ 2 \\ - 1\end{matrix}\right) + \beta \left(\begin{matrix}2 \\ 4 \\ 6\end{matrix}\right) + \gamma \left(\begin{matrix}0 \\ 0 \\ - 8\end{matrix}\right) = \left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)$

Where $\alpha , \beta , \gamma \in {\mathbb{R}}^{3}$

has only the trivial solution

$\alpha = \beta = \gamma = 0$

Perform a row reduction on the augmented matrix

$A = \left(\begin{matrix}1 & 2 & 0 & | & 0 \\ 2 & 4 & 0 & | & 0 \\ - 1 & 6 & - 8 & | & 0\end{matrix}\right)$

$\iff$, $R 2 \leftarrow R 2 - 2 R 1$

$\left(\begin{matrix}1 & 2 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ - 1 & 6 & - 8 & | & 0\end{matrix}\right)$

$\iff$, $R 2 \leftrightarrow R 3$

$\left(\begin{matrix}1 & 2 & 0 & | & 0 \\ - 1 & 6 & - 8 & | & 0 \\ 0 & 0 & 0 & | & 0\end{matrix}\right)$

$\iff$, $R 2 \leftarrow R 2 + R 1$

$\left(\begin{matrix}1 & 2 & 0 & | & 0 \\ 0 & 8 & - 8 & | & 0 \\ 0 & 0 & 0 & | & 0\end{matrix}\right)$

$\iff$, $R 2 \leftarrow \frac{R 2}{8}$

$\left(\begin{matrix}1 & 2 & 0 & | & 0 \\ 0 & 1 & - 1 & | & 0 \\ 0 & 0 & 0 & | & 0\end{matrix}\right)$

$\iff$, $R 1 \leftarrow R 1 - 2 R 2$

$\left(\begin{matrix}1 & 0 & 2 & | & 0 \\ 0 & 1 & - 1 & | & 0 \\ 0 & 0 & 0 & | & 0\end{matrix}\right)$

Since there is a free variable, the system is not linearly independant

The determinant of the matrix $= 0$

Therefore,

$\left\{\begin{matrix}\alpha = - 2 \gamma \\ \beta = \gamma \\ \gamma = \text{ free}\end{matrix}\right.$

So,

$- 2 \gamma \left(\begin{matrix}1 \\ 2 \\ - 1\end{matrix}\right) + \gamma \left(\begin{matrix}2 \\ 4 \\ 6\end{matrix}\right) + \gamma \left(\begin{matrix}0 \\ 0 \\ - 8\end{matrix}\right) = \left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)$

Hope that this will help!!!