Determine whether the set S spans R2.  If

Let's determine whether each given set spans ${ℝ}^2$. If a set does not span ${ℝ}^2$, we will provide a geometric description of the subspace it does span.
1. Set $S=\{(2,1),(-1,2)\}$:
To check if $S$ spans ${ℝ}^2$, we need to see if every vector in ${ℝ}^2$ can be written as a linear combination of the vectors in $S$. Let's express an arbitrary vector $(x,y)$ in ${ℝ}^2$ as a linear combination of the vectors in $S$:
$(x,y)=a(2,1)+b(-1,2)$,
where $a$ and $b$ are scalars.
Expanding this equation gives us a system of equations:
$2a-b=x$,
$a+2b=y$.
To solve this system, we can use various methods such as substitution or elimination. Let's use elimination to eliminate $a$:
$2a-b=x$,
$2a+4b=2y$.
Subtracting the first equation from the second equation, we get:
$5b=2y-x$.
This equation tells us that $b$ is uniquely determined by the values of $x$ and $y$. Therefore, $S$ spans ${ℝ}^2$ since any vector $(x,y)$ can be expressed as a linear combination of the vectors in $S$.
2. Set $S=\{(1,3),(-2,-6),(4,12)\}$:
Similarly, we express an arbitrary vector $(x,y)$ in ${ℝ}^2$ as a linear combination of the vectors in $S$:
$(x,y)=a(1,3)+b(-2,-6)+c(4,12)$.
Expanding this equation gives us a system of equations:
$a-2b+4c=x$,
$3a-6b+12c=y$.
To solve this system, we can use elimination. Subtracting three times the first equation from the second equation, we obtain:
$-12b=y-3x$.
This equation tells us that $b$ is uniquely determined by the values of $x$ and $y$. Therefore, $S$ spans ${ℝ}^2$.
3. Set $S=\{(-3,5)\}$:
To check if $S$ spans ${ℝ}^2$, we express an arbitrary vector $(x,y)$ in ${ℝ}^2$ as a linear combination of the vector in $S$:
$(x,y)=a(-3,5)$.
This equation gives us the system of equations:
$-3a=x$,
$5a=y$.
From the second equation, we find that $a=\frac{y}{5}$. Substituting this into the first equation, we have $-3\left(\frac{y}{5}\right)=x$, which simplifies to $y=-\frac{5}{3}x$.
The set $S$ does not span ${ℝ}^2$. Geometrically, the set $S$ represents a line in ${ℝ}^2$ passing through the point $(-3,5)$ with a slope of $-\frac{5}{3}$. This line is a subspace of ${ℝ}^2$.
4. Set $S=\{(5,0),(5,-4)\}$:
We express an arbitrary vector $(x,y)$ in ${ℝ}^2$ as a linear combination of the vectors in $S$:
$(x,y)=a(5,0)+b(5,-4)$.
Expanding this equation gives us a system of equations:
$5a+5b=x$,
$-4b=y$.
From the second equation, we find that $b=-\frac{y}{4}$. Substituting this into the first equation, we have $5a+5(-\frac{y}{4})=x$, which simplifies to $20a-5y=4x$.
This equation tells us that $a$ is uniquely determined by the values of $x$ and $y$. Therefore, $S$ spans ${ℝ}^2$.
5. Set $S=\{(-1,2),(2,-4)\}$:
We express an arbitrary vector $(x,y)$ in ${ℝ}^2$ as a linear combination of the vectors in $S$:
$(x,y)=a(-1,2)+b(2,-4)$.
Expanding this equation gives us a system of equations:
$-a+2b=x$,
$2a-4b=y$.
To solve this system, we can use elimination. Multiplying the first equation by 2 and adding it to the second equation, we obtain:
$0=2x+y$.
This equation tells us that $b$ is not uniquely determined by the values of $x$ and $y$. Therefore, $S$ does not span ${ℝ}^2$. Geometrically, the set $S$ represents a line passing through the origin with a slope of $\frac{1}{2}$.