Verify if a span of vectors is $\mathbb{R}^2$ if $H = \mathrm{Span}\{v_1, v_2\} $
and 
$v_1 = (2,3)$
$v_2 = (-1,4)$
Is knowing $v_1$ and $v_2$ are linearly independent since $\det \begin{bmatrix} 2 & -1 \\ 3 & 4\end{bmatrix} = 11$ enough to say that $ H = \mathbb{R}^2$
 A: That is a fine answer, as long as you know two results:


*

*$n$ elements of $\mathbb R^n$ are linearly independent if the determinant of the matrix composed of the rows (or columns) of those  vectors is non-zero.

*If $n$ vectors in $\mathbb R^n$ are linearly independent, they span the entire space. 


The direct approach would be to show explicitly that every $v=(a,b)$ can be written as a linear combination of $v_1,v_2$:
$$(a,b)=\frac{a}{11}(4v_1-3v_2)+\frac{b}{11}(v_1+2v_2)=\frac{4a+b}{11}v_1+\frac{-3a+2b}{11}v_2$$
That is just the equivalent of finding the inverse of the matrix, of course.
Alternatively, you could show that $(1,0)$ and $(0,1)$ are in the span.
The above two results could be summarized as "the matrix composed of the rows is invertible if and only if the determinant is zero," since the explicit expression of $(a,b)$ in terms of $v_1,v_2$ is essentially the explicit inverse of that matrix. 
A: Also if don't want to use a theorem and can prove this directly like this:
Let $x \in \mathbb{R}^2$ such that $x=(x_1,x_2)$ and $x=av_1+bv_2$
$(x_1,x_2)=a(2,3)+b(-1,4) \Longleftrightarrow (x_1,x_2)=(2a-b,3a+4b)$
Thus  $$2a-b=x_1$$ $$3a+4b=x_2$$
So we have by solving this system: $$b=\frac{2x_2-3x_1}{11}$$ $$a=\frac{8x_1+2x_2}{22}$$
Thus for any $x=(x_1,x_2)$ you can find $a,b \in \mathbb{R}$ such that $x=av_1+bv_2$
