Given an n x n matrix [math]A=\left[\mathbf{a}_1 \ \mathbf{a}_2 \ \cdots \mathbf{a}_n\right]=\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ [br]\vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}[/math], we can generalize the definition of the signed volume of parallelepiped to the "n-dimensional signed volume" of the so-called [b]parallelotope[/b] formed by the column vectors [math]\mathbf{a}_1, \mathbf{a}_2 \ldots, \mathbf{a}_n[/math] in [math]\mathbb{R}^n[/math] and use it as the definition of [math]\det(A)[/math]. Similar to what we did in the previous section, we write each column vector as a linear combination of the standard basis in [math]\mathbb{R}^n[/math] and then expand [math]\det\left(\left[\mathbf{a}_1 \ \mathbf{a}_2 \ \cdots \mathbf{a}_n\right]\right)[/math] by linearity. Without getting into detail computation, which is quite involved, we write down the following result:[br][br][math]\det(A)=\sum_{\sigma} \text{sgn}(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}\cdots a_{n\sigma(n)}[/math][br][br]where the sum runs through all the permutation [math]\sigma[/math] of the indices [math]\left\{1,2,\ldots,n\right\}[/math] and [math]\text{sgn}(\sigma)[/math], the signature of [math]\sigma[/math], is either [math]1[/math] or [math]-1[/math], depending whether the permutation [math]\sigma[/math] can be obtained by successively interchanging two indices an even or odd number of times. For example, when [math]n=4[/math] and [math]\sigma(1)=2,\sigma(2)=3,\sigma(3)=4,\sigma(4)=1[/math], [math]\text{sgn}(\sigma)=-1[/math] because [math]\sigma[/math] can be obtained by interchanging 1 and 2, then 1 and 3 and finally 1 and 4 ([math]1234\rightarrow2134\rightarrow2314\rightarrow2341[/math]) i.e. interchanging odd number of times.[br][br]This is the [b]Leibniz formula[/b] for the determinant of an n x n matrix.[br][br]However, this formula is seldom used for actual computation because it is too complicated to use. Fortunately, we have some other ways to compute determinants.

We can express the determinant of an n x n matrix [math]A[/math] in terms of the determinants of [b]submatrices[/b] of [math]A[/math] (matrices obtained by deleting one or several rows and/or columns of [math]A[/math]). Let [math]A_{ij}[/math] be the n-1 x n-1 submatrix of [math]A[/math] formed by deleting the i[sup]th[/sup] row and j[sup]th[/sup] column of [math]A[/math]. Then we have the following formula:[br][br][math]\begin{eqnarray}\det(A) & = & a_{11}\det(A_{11})-a_{12}\det(A_{12})+\cdots+(-1)^{1+n}a_{1n}\det(A_{1n})\\ & = & \sum_{j=1}^n (-1)^{i+j}a_{ij}\det(A_{ij}) \ \ \ \text{for any} \ i=1,2, \ldots, n \end{eqnarray}[/math][br][br]The term [math](-1)^{i+j}\det(A_{ij})[/math] is called the [b][math](i,j)[/math]-cofactor[/b] of [math]A[/math], denoted by [math]C_{ij}[/math]. Hence, for [math]i=1,2, \ldots,n[/math],[br][br][math]\det(A)=a_{i1}C_{i1}+a_{i2}C_{i2}+\cdots+a_{in}C_{in}[/math][br][br]is called a [b]cofactor expansion across the i[sup]th[/sup] row[/b] of A.[br][br]In fact, we can also compute the determinant by a [b]cofactor expansion across the j[sup]th[/sup] column[/b] of A:[br][br][math]\det(A)=a_{1j}C_{1j}+a_{2j}C_{2j}+\cdots+a_{nj}C_{nj}[/math][br][br]The proofs of these formulas are lengthy and are thus omitted.[br][br][u]Example[/u]: Evaluate [math]\det(A)[/math] where [math]A= \begin{pmatrix}2 & -1 & 9 & 8 & 4 \\ 0 & 3 & -1 & 4 & 6 \\ 0 & 0 & 1 & -6 & 0 \\ 0 & 0 & 2 & -5 & 0 \\ 0 & 0 & 0 & 2 & 3\end{pmatrix}[/math].[br]We can choose the first column to do the cofactor expansion because most of its entries are zeros. We have[br][br][math] \det(A)=2\begin{vmatrix} 3 & -1 & 4 & 6 \\ 0 & 1 & -6 & 0 \\ 0 & 2 & -5 & 0 \\ 0 & 0 & 2 & 3\end{vmatrix}[/math][br][br]Then we choose the first column of the 4 x 4 matrix to do the cofactor expansion. We have[br][br][math] \det(A)=2\cdot 3\begin{vmatrix}1 & -6 & 0 \\2 & -5 & 0\\0 & 2 & 3\end{vmatrix}[/math][br][br]Then we choose the third column of the 3 x 3 matrix to do the factor expansion. We have[br][br][math] \det(A)=2\cdot 3 \cdot 3 \begin{vmatrix} 1 & -6\\2 & -5\end{vmatrix}=2\cdot 3\cdot 3 \cdot 7=126[/math][br]