Showing a set is a basis for $R^2$ I am reading this text:

How did they find the determinant of both systems? Is there a shortcut I'm missing? How does the determinant help you answer this question?
 A: Yes, there is a shortcut, the determinant of a 2x2 matrix A is given by
$$\det A= \begin{vmatrix}a_{11}&a_{12}\\ a_{21} &a_{22} \end{vmatrix}=a_{11}a_{22}-a_{21}a_{21}$$
A nxn non-homogeneous system of linear equations has a unique non-trivial solution iff $\det A\neq0$. 
On the other hand, if $\det A=0$, then the system has either no nontrivial solutions or an infinite number of solutions. 
Edit: for a nxn homogeneous system there is a unique (trivial) solution $x_1=x_2=x_n=0$ iff $\det A\neq 0$. In the case $\det A=0$ the homogeneous system  has an infinite number of solutions. 
A: Let me give you first a more detailed outline of the approach. They have considered the system of equations
$$c_1+c_2=0\\c_1-c_2=0$$
as a matrix equation $Ac=b$ where $A$ is the so called coefficient matrix, i.e.
$$\begin{pmatrix}1 &1\\1 &-1\end{pmatrix}\begin{pmatrix}c_1\\c_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
For $2\times 2$-matrices, you may evaluate the determinant by a simple formula according to
$$\det\begin{pmatrix}a &b\\c &d\end{pmatrix}=ad-cb$$
Applying this to our coefficient matrix, we derive
$$\det\begin{pmatrix}1 &1\\1 &-1\end{pmatrix}=-1-1=-2$$
As the determinant of this matrix is non-zero, we have that the corresponding homogeneous system just has the trivial solution.

First, you may derive this short formula for two-dimensional determinants by the Leibniz formula or a technique for evaluating determinants called expansion. Let me give you a glimpse on the approach via the Leibniz formula. I don't go into detail too far, so you may come back to this after having learned about the Leibniz formula in some more detail. Let $$A=\begin{pmatrix}a_{11} &a_{12}\\a_{21} &a_{22}\end{pmatrix}$$ be a $2\times 2$-matrix, then
$$\det A=\sum_{\sigma\in S_2}\mathrm{sign}(\sigma)a_{\sigma(1)1}a_{\sigma(2)2}$$
by the Leibniz formula where $S_2$ is the set of permutations(bijections) on the set $\{1,2\}$ and $\mathrm{sign}(\sigma)$ is the sign of the permutation $\sigma$. Now, $S_2$ only contains $\mathrm{id}$ and the map $\sigma_0:\begin{cases}1\mapsto 2\\2\mapsto 1\end{cases}$. Now
$$\sum_{\sigma\in S_2}\mathrm{sign}(\sigma)a_{\sigma(1)1}a_{\sigma(2)2}=\mathrm{sign}(\mathrm{id})a_{11}a_{22}+\mathrm{sign}(\sigma_0)a_{21}a_{12}=a_{11}a_{22}-a_{21}a_{12}$$
In the last step, I've used that $\mathrm{sign}(\mathrm{id})=1$ and $\mathrm{sign}(\sigma_0)=-1$.

On why the determinant is helpful to answer the question, note the following intimate relationship between determinants and linear independence:

Lemma: Let $\{a_1,\dots a_n\}\subseteq\mathbb{R}^n$ and $A=(a_1,\dots,a_n)$. We have $$\det A=0\text{ iff }a_1,\dots,a_n\text{ are linearly dependent iff }Ax=\mathbf{0}\text{ has a solution with }x\neq\mathbf{0}$$.

The right iff is just definition of linear dependence/independence. The left one follows from two properties of the determinant called anti-symmetry and multi-linearity:

Anti-symmetry: Let $A=(a_1,\dots,a_n)$. If $a_i=a_j$ for $i\neg j$, then $\det A=0$.

