Question

If they are linearly dependent, find scalars that are not all zero such that the equation below is true. If they are linearly independent, find the only scalars that will make the equation below true :
[-5] [2] [-17]
[4] [-1] [16]
[5] [5] [45]
Are the vactors linerly independent.

Answer 1

I bet you're talking about column vectors,

$\begin{bmatrix}-5\\4\\5\end{bmatrix}, \begin{bmatrix}2\\-1\\5\end{bmatrix},\text{ and }\begin{bmatrix}-17\\16\\45\end{bmatrix}$

If these vectors are linearly independent, this means that

$c_1 \begin{bmatrix}-5\\4\\5\end{bmatrix} + c_2 \begin{bmatrix}2\\-1\\5\end{bmatrix} + c_3 \begin{bmatrix}-17\\16\\45\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}$

only if $c_1=c_2=c_3=0$.

In matrix form, we can write this as a system of linear equations,

$\begin{bmatrix}-5&2&-17\\4&-1&16\\5&5&45\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}$

This system has a potentially non-zero solution so long as the coefficient matrix is not singular.

Compute the determinant:

$\begin{vmatrix}-5&2&-17\\4&-1&16\\5&5&45\end{vmatrix} = -5 \begin{vmatrix}-1&16\\5&45\end{vmatrix} - 4 \begin{vmatrix}2&-17\\5&45\end{vmatrix} + 5 \begin{vmatrix}2&-17\\-1&16\end{vmatrix} \\\\ \begin{vmatrix}-5&2&-17\\4&-1&16\\5&5&45\end{vmatrix} = -5(-45-80) - 4(90+85) + 5(32-17) \\\\ \begin{vmatrix}-5&2&-17\\4&-1&16\\5&5&45\end{vmatrix} = 0$

The coefficient matrix is singular, so there is no non-zero solution to the system, and in turn the given vectors are linearly dependent.