If there is a vector named \( \vec v \) on this page, then \(v_x\), \(v_y\) and \(v_z\) denote the x, y and z components of that vector.
The above animation shows a geometric way to understand a \( 2 \times 2 \) determinant. Draw the vectors \(\vec a\) and \(\vec b\) corresponding to the first and second row of the determinant. Then draw a parallelogram with the vectors as sides. Now the determinant is plus or minus the area of the parallelogram.
To figure out the sign of the determinant, draw an arc-shaped arrow from the vector representing the top row of the determinant to the other vector. Choose the direction of the arrow so that the arrow arc is as short as possible. Then the determinant is
Starting with this geometric definition, it's possible to derive the usual formula $$ \det\left[ \begin{matrix} a & b \\ c & d \end{matrix}\right] = ad - bc, $$ where \( a,b,c,d \in \mathbb{R} \).
This animation shows a geometric way to understand a \( 3 \times 3 \) determinant. The biggest difference to \( 2 \times 2 \) determinant is that the determinant is plus or minus the volume of a parallelepiped instead of area of a parallelogram.
To figure out the sign of a \( 3 \times 3 \) determinant, first you need to draw the arrow from first vector to second vector, then grab the third vector with your right hand so that your thumb presses the tip of the vector and your fingers are wrapped around the vector. If the fingers go in the same direction as the arrow, the determinant is positive, otherwise it's negative.
From this definition, it's possible to derive the "subdeterminant formula" $$ \det\left[\begin{matrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \\ \end{matrix}\right] = a_x \det\left[\begin{matrix} b_y&b_z \\ c_y&c_z \end{matrix}\right] - a_y \det\left[\begin{matrix} b_x&b_z \\ c_x&c_z \end{matrix}\right] + a_z \det\left[\begin{matrix} b_x&b_y \\ c_x&c_y \end{matrix}\right]. $$