Showing linear dependence in a matrix Let 
$v_1=\begin{bmatrix}
1\\
-3\\
2\\
\end{bmatrix}
$
Let $V_2=\begin{bmatrix}
-3\\
9\\
-6\\
\end{bmatrix}
$
Let $V_3=\begin{bmatrix}
5\\
-7\\
h\\
\end{bmatrix}
$
For what value of h is the span {v1,v2,v3} linear dependent.
This be my work.
Let $A= \begin{bmatrix}
1 & -3 & 5&  0 \\
-3&9& -7&   0 \\
2& -6 &   h&   0 \\
\end{bmatrix}
$
then doing 3r1+r3 and -2r1+r2
Let $A= \begin{bmatrix}
1 & -3 & 5&  0 \\
0&0& 8&   0 \\
0&0 &   h-10&   0 \\
\end{bmatrix}
$
 A: Hint: 
for what values of $h$, is the determinant $\det{(v_1,v_2,v_3)}$ exactly $0$? What if it is always $0$ no matter what the value of $h$ is?
A: Adding the last zero column is irrelevant; with Gaussian elimination, you can not only decide if a set of vectors is linearly independent, but also tell what vectors are linear combinations of the remaining ones.
Your elimination is good, the reduced matrix is
\begin{bmatrix}
1 & -3 & 5\\
0&0& 8\\
0&0 &   h-10
\end{bmatrix}
and this shows the three vectors are linearly dependent for all values of $h$.
The elimination shows, in particular, that the second vector is equal to the first one multiplied by $-3$ (linear relations among columns holding in the reduced matrix hold the same in the original matrix).
Since the last column is not a multiple of the first one, a basis of the subspace generated by $v_1$, $v_2$ and $v_3$ is $\{v_1,v_3\}$.
A: You must have zeros in last row, so $h+30=0$ and $h=-30$.
A: One way of doing that is to look at $\det A$ where $A$ is the matrix formed by $v_1,v_2,v_3.$ As long as determinant is non zero the matrix is invertible so these three vectors span $\mathbb{R}^3$.
