Find a linear dependence relation between the following vectors:
x1 = (1, -1, 2)
x2 = (-3, 2, 1)
x3= (1, 2, -3)
x4= (2, 3, 1)
I've already created a matrix and reduced and I know how to tell whether it is linearly independent or not, but I don't understand how to find an actual relation. Thanks
$\endgroup$ 62 Answers
$\begingroup$We are tasked with finding a nonzero solution $(a_1,a_2,a_3,a_4)$ to the equation:
$$a_1x_1+a_2x_2+a_3x_3+a_4x_4=\textbf{0}$$
This is the same as finding a nonzero solution $(a_1,a_2,a_3,a_4)$ to the system
$$\begin{cases}1a_1-3a_2+1a_3+2a_2=0\\-1a_1+2a_2+2a_3+3a_4=0\\2a_1+1a_2-3a_3+1a_4=0\end{cases}$$
This is the same as finding a solution $[a_1,~a_2,~a_3,~a_4]^T$ to the matrix equation
$$\begin{bmatrix}1&-3&1&2\\-1&2&2&3\\2&1&-3&1\end{bmatrix}\begin{bmatrix}a_1\\a_2\\a_3\\a_4\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$
In other words, we are trying to find the kernel of the matrix.
Row reducing $$rref\left(\begin{bmatrix}1&-3&1&2\\-1&2&2&3\\2&1&-3&1\end{bmatrix}\right) = \begin{bmatrix}1&0&0&3\\0&1&0&1\\0&0&1&2\end{bmatrix}$$
Remembering that row-reducing a system of equations retains the solution-set of the system and reinterpreting this as a system of equations this is the system:
$$\begin{cases} 1a_1&&&+3a_4&=0\\&1a_2&&+1a_4&=0\\&&1a_3&+2a_4&=0\end{cases}$$
This tells us that for $(a_1,a_2,a_3,a_4)$ to be a solution that $a_1=-3a_4, a_2=-a_4$ and $a_3=-2a_4$
Picking any number (other than zero) for $a_4$, we can then write a linear combination of $x_1,\dots,x_4$ resulting in zero.
$\endgroup$ $\begingroup$$-3x_1-x_2-2x_3+x_4=0$
(I don't have enough reputation to post a comment, so posting it as here) one way to find the dependency is track your row reduction steps. For example if these are the steps that reduced the matrix
R1 = R2 - R1 ,
R3 = R3 - R1,
R3 = R3 + 2R2
Now R3 = R3 + 2R2 replace R3 and R2 from first two steps.
So (R3 - R1) + 2(R2 - R1) = -3R1 + 2R2 + R3 = 0 (this would be the dependency between the between the rows.
