Let be $\varphi:\mathbb C^{2\times 2}\to\mathbb C$ with the following properties: $$
It is linear on the columns: \left\{\begin{align} \varphi\left(\left[\begin{matrix}a_1+a_2&b\\c_1+c_2&d\\\end{matrix}\right]\right) = \varphi\left(\left[\begin{matrix}a_1&b\\c_1&d\\\end{matrix}\right]\right) + \varphi\left(\left[\begin{matrix}a_2&b\\c_2&d\\\end{matrix}\right]\right)\\ \varphi\left(\left[\begin{matrix}a&b_1+b_2\\c&d_1+d_2\\\end{matrix}\right]\right) = \varphi\left(\left[\begin{matrix}a&b_1\\c&d_1\\\end{matrix}\right]\right) + \varphi\left(\left[\begin{matrix}a&b_2\\c&d_2\\\end{matrix}\right]\right)\\ \qquad\varphi\left(\left[\begin{matrix}ka&b\\kc&d\\\end{matrix}\right]\right) = \varphi\left(\left[\begin{matrix}a&kb\\c&kd\\\end{matrix}\right]\right) = k\varphi\left(\left[\begin{matrix}a&b\\c&d\\\end{matrix}\right]\right)\\ \end{align}\right.\qquad.
It is anti-symmetric on the columns: \varphi\left(\left[\begin{matrix}b&a\\d&c\\\end{matrix}\right]\right) = -\,\varphi\left(\left[\begin{matrix}a&b\\c&d\\\end{matrix}\right]\right)\quad.
It maps identity to identity: \varphi\left(\left[\begin{matrix}1&0\\0&1\\\end{matrix}\right]\right)=1\quad.
$ I was told that, given these properties, \varphi\small\left(\left[\begin{matrix}a&b\\c&d\\\end{matrix}\right]\right)$ has to be $ad-bc\,$, good old $\det\small\left[\begin{matrix}a&b\\c&d\\\end{matrix}\right]$, and that applies to any $\mathbb C^{n\times n}$. Is this true?