A geometrical explanation of the determinant formula for $2 \times 2$ matrices

The determinant of the linear function that maps the two-dimensional vectors $\pmat{1 \\ 0}$ and $\pmat{0 \\ 1}$ to $\vec{a} = \pmat{a_1 \\ a_2}$ and $\vec{b} = \pmat{b_1 \\ b_2}$ is \begin{equation} \label{eq:determinant} D = \vmat{a_1 & b_1 \\ a_2 & b_2} = a_1 b_2 - a_2 b_1\,. \tag{$*$}\end{equation}

It can also be viewed as the area of the parallelogram that is spanned by the vectors $\vec a$ and $\vec b$:

\begin{mypicture} \draw (x1) node[below right] {$x_1$}; \draw (x2) node[left] {$x_2$}; \path[fill=middlegray] (0, 0) -- (a) -- (sum) -- (b) -- cycle; \draw[thick, ->] (0,0) -- (a) node[midway, below right] {$\vec a$}; \draw[thick, ->] (0,0) -- (b) node[midway, above left] {$\vec b$}; \draw[thick, -{>[right]}] (a) -- (sum); \draw[thick, -{>[left]}] (b) -- (sum); \draw[densely dashed] (a) -- (a |- 0,0) (b) -- (b -| 0,0); \end{mypicture}

With this definition, the formula $(*)$ can easily be proved geometrically. This is because the area of the grey parallelogram is the difference of the two areas \begin{equation*} A_1 = \begin{mypicture} \path[fill=weakgray] (0, 0) -- (b) -- (sum) -- (corner) -- cycle; \draw (sum) -- (corner); \draw (b) -- (b -| corner) node[midway, below] {$a_1$}; \draw (b) -- (b |- corner) node[midway, right] {$b_2$}; \path (sum) -- (b -| corner) node[midway, right] {$a_2$}; \path (0,0) -- (b |- corner) node[midway, below] {$b_1$}; \node[anchor=base] at ($(0,0)!.5!(b)!.5!(b |- corner)$) {$T_1$}; \node[anchor=base] at ($(b)!.5!(sum)!.5!(b -| sum)$) {$T_2$}; \draw[thick, ->] (0,0) -- (b) node[midway, above left] {$\vec b$}; \draw[thick, ->] (b) -- (sum) node[midway, above left] {$\vec a$}; \end{mypicture} \qquad\text{and}\quad A_2 = \begin{mypicture} \path[fill=weakgray] (0, 0) -- (a) -- (sum) -- (corner) -- cycle; \draw (sum) -- (corner); \draw (a) -- (a -| corner) node[midway, below] {$b_1$}; \draw (a) -- (a |- corner) node[midway, right] {$a_2$}; \path (sum) -- (a -| corner) node[midway, right] {$b_2$}; \path (0,0) -- (a |- corner) node[midway, below] {$a_1$}; \node[anchor=base] at ($(0,0)!.5!(a)!.5!(a |- corner)$) {$T_2$}; \node[anchor=base] at ($(a)!.5!(sum)!.5!(a -| sum)$) {$T_1$}; \draw[thick, ->] (0,0) -- (a) node[midway, above left] {$\vec a$}; \draw[thick, ->] (a) -- (sum) node[midway, above left] {$\vec b$}; \end{mypicture}\end{equation*} CThe triangles $T_1$ and $T_2$ occur in both diagrams, therefore the value of the determinant is $A_1 - A_2 = a_1 b_2 - b_1 a_2$.

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