We already know that for any matrix [math]A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}[/math], we define [math]\det(A)[/math] to be the signed area of the parallelogram formed by the column vectors of [math]A[/math]. Now we will show that [math]\det(A)=ad-bc[/math]. The proof is based on the following important properties of determinant: Let [math]u, v, w[/math] be column vectors in [math]\mathbb{R}^2[/math]. Then[br][br][list=1][*][math]\det\left(\left[\hat{\mathbf{i}} \ \hat{\mathbf{j}}\right]\right)=1[/math][br][/*][*][math]\det\left(\left[u \ v \right]\right)=-\det\left(\left[v \ u \right]\right)[/math] [br][/*][*][math]\det\left(\left[k u \ v \right]\right)=k\det\left(\left[u \ v \right]\right)[/math] for any real number [math]k[/math][br][/*][*][math]\det\left(\left[u+w \ v \right]\right)=\det\left(\left[u \ v \right]\right)+\det\left(\left[w \ v \right]\right)[/math] [/*][/list][br](1) is obviously true because the parallelogram formed by the standard basis is a unit square.[br][br](2) is the convention we use to assign the sign to the area of the parallelogram formed by the column vectors of the matrix.[br][br](3) is obviously true when [math]k\ge0[/math] because scaling the length of a side of a parallelogram will scale the area of the parallelogram by the same factor. When [math]k<0[/math], the area will be scaled by factor [math]\left|k\right|[/math] and the sign of the area will be changed because the vectors [math]u[/math] and [math]ku[/math] are pointing in the opposite direction. Therefore, the result in (3) still holds.[br][br]Try the following applet to see why (4) holds. [br][br][u]Remarks[/u]: [br][list][*]From (2), we can easily see that [math]\det\left(\left[u \ u\right]\right)=0[/math] for any vector [math]u[/math].[br][/*][*](3) and (4) together imply that any determinant is "linear" in its first column vector. Using (2), we can deduce that any determinant is also "linear" in its second column vector.[/*][*]Other commonly-used notations for determinant: [math]\left|A\right|[/math] or [math]\begin{vmatrix} a & b\\ c & d \end{vmatrix}[/math].[br][/*][/list]

Now, we are ready to prove the formula for the determinant of 2 x 2 matrices:[br][br][math]\begin{eqnarray}\begin{vmatrix} a & b \\ c & d\end{vmatrix} & = & \det\left(\left[a\hat{\mathbf{i}}+c\hat{\mathbf{j}} \ \ b\hat{\mathbf{i}}+d\hat{\mathbf{j}}\right]\right) \\ [br]& = & a\det\left(\left[\hat{\mathbf{i}} \ \ b\hat{\mathbf{i}}+d\hat{\mathbf{j}}\right]\right)+c\det\left(\left[\hat{\mathbf{j}} \ \ b\hat{\mathbf{i}}+d\hat{\mathbf{j}}\right]\right) \\[br]& = & a\left(b\det\left(\left[\hat{\mathbf{i}} \ \ \hat{\mathbf{i}}\right]\right)+d\det\left(\left[\hat{\mathbf{i}} \ \ \hat{\mathbf{j}}\right]\right)\right)+c\left(b\det\left(\left[\hat{\mathbf{j}} \ \ \hat{\mathbf{i}}\right]\right)+d\det\left(\left[\hat{\mathbf{j}} \ \ \hat{\mathbf{j}}\right]\right)\right)\\[br]& = & ad-bc\end{eqnarray}[/math][br][br]This completes the proof.

We can use the same idea to derive a formula for the determinant of any 3 x 3 matrix [math]A=\left[\mathbf{a}_1 \ \mathbf{a}_2 \ \mathbf{a}_3\right]=\begin{pmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{pmatrix}[/math]. We define [math]\det(A)[/math] to be the signed volume of the parallelepiped formed by the column vectors [math]\mathbf{a}_1, \mathbf{a}_2[/math] and [math]\mathbf{a}_3[/math]. Then it has the following properties:[br][br][list=1][*][math]\det\left(\left[\hat{\mathbf{i}} \ \hat{\mathbf{j}} \ \hat{\mathbf{k}}\right]\right)=1[/math][br][/*][*][math]\det\left(\left[u \ v \ w\right]\right)= -\det\left(\left[v \ u \ w \right]\right)= -\det\left(\left[u \ w \ v\right]\right)=-\det\left(\left[w \ v \ u\right]\right)[/math][br][/*][*][math]\det\left(\left[ku \ v \ w\right]\right)= k\det\left(\left[u \ v \ w \right]\right)[/math] for any real number [math]k[/math][br][/*][*][math]\det\left(\left[u+p \ v \ w\right]\right)= \det\left(\left[u \ v \ w \right]\right)+\det\left(\left[p \ v \ w\right]\right)[/math] [br][/*][/list][br]Therefore, we can compute the determinant as follows:[br][br][math]\begin{eqnarray}\begin{vmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{vmatrix} & = & \det\left(\left[(a_{11}\hat{\mathbf{i}}+a_{21}\hat{\mathbf{j}}+a_{31}\hat{\mathbf{k}}) \ \ (a_{12}\hat{\mathbf{i}}+a_{22}\hat{\mathbf{j}}+a_{32}\hat{\mathbf{k}}) \ \ (a_{13}\hat{\mathbf{i}}+a_{23}\hat{\mathbf{j}}+a_{33}\hat{\mathbf{k}})\right]\right) \\& = & \text{Expansion by the linearity of the determinant} \\ & = & a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31}\end{eqnarray}[/math][br][br]This formula looks a bit complicated. Let's rewrite it into the one that is easier to memorize.[br][br][math]\begin{eqnarray}\begin{vmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{vmatrix} & = & a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31}) \\ & = & a_{11}\begin{vmatrix}a_{22} & a_{23}\\a_{32} & a_{33}\end{vmatrix}-a_{12}\begin{vmatrix}a_{21} & a_{23}\\a_{31} & a_{33}\end{vmatrix}+a_{13}\begin{vmatrix}a_{21} & a_{22}\\a_{31} & a_{32}\end{vmatrix}\end{eqnarray}[/math][br][br][br][br]