Let $A=\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}$. Then we define the determinant $\det A$ by
$$\det A=ad-bc.$$
$\det A$ is also denoted by $|A|$ or $\left|\begin{array}{ccc}
a & b\\
c & d
\end{array}\right|$. In terms of the column vectors $A^1,A^2$, the determinant of $A$ may also be written as $\det(A^1,A^2)$.

Example. If $A=\begin{pmatrix}
2 & 1\\
1 & 4
\end{pmatrix}$, then $\det A=8-1=7$.

Property 1. The determinant $\det (A^1,A^2)$ may be considered as a bilinear map of the column vectors. As a bilinear map $\det (A^1,A^2)$ is linear in each slot. For example, if $A^1=C+C’$ then
$$\det(A^1,A^2)=\det(C,A^2)+\det(C’,A^2).$$
If $x$ is a number,
$$\det(xA^1,A^2)=x\det(A^1,A^2).$$

Property 2. If the two columns are equal, the determinant is 0.

Property 3. $\det I=\det (E^1,E^2)=1$.

Combining Properties 1-3, we can show that:

Theorem. If one adds a scalar multiple of one column to another, then the value of the determinant does not change.

Proof. We prove the theorem for a particular case.
\begin{align*}
\det(A^1+xA^2,A^2)&=\det(A^1,A^2)+x\det(A^2,A^2)\\
&=\det(A^1,A^2).
\end{align*}

Theorem. If the two columns are interchanged, the determinant changes by a sign i.e.
$$\det(A^2,A^1)=-\det(A^1,A^2).$$

Proof. \begin{align*}
0&=\det(A^1+A^2,A^1+A^2)\\
&=\det(A^1,A^2)+\det(A^2,A^1).
\end{align*}

Theorem. $\det A=\det {}^tA$.

Proof. This theorem can be proved directly from the definition of $\det A$.

Remark. Because of this theorem, we can also say that if one adds a scalar multiple of one row to another row, then the value of the determinant does not change.

Theorem. The column vectors $A^1,A^2$ are linearly dependent if and only if $\det(A^1,A^2)=0$.

Proof. Suppose that $A^1,A^2$ are linearly dependent. Then there exists numbers $x,y$, not all equal to 0 such that $xA^1+yA^2=0$. Let us say $x\ne 0$. Then $A^1=-\frac{y}{x}A^2$. So,
\begin{align*}
\det(A^1,A^2)&=\det\left(-\frac{y}{x}A^2,A^2\right)\\
&=-\frac{y}{x}\det(A^2,A^2)\\
&=0.
\end{align*}
To prove the converse, suppose that $A^1,A^2$ are linearly independent. Then $E^1,E^2$ can be written as linear combinations of $A^1,A^2$:
\begin{align*}
E^1&=xA^1+yA^2,\\
E^2&=zA^1+wA^2.
\end{align*}
Now,
\begin{align*}
1&=\det(E^1,E^1)\\
&=xw\det(A^1,A^2)+yz\det(A^2,A^1)\\
&=(xw-yz)\det(A^1,A^2).
\end{align*}
Hence, $\det(A^1,A^2)\ne 0$.