A determinant of order $n$ can be calculated by expanding it in terms of determinants of order $n-1$. Let $A=(a_{ij})$ be an $n\times n$ matrix and let us denote by $A_{ij}$ the $(n-1)\times (n-1)$ matrix obtained by deleting the $i$-th row and the $j$-th column from $A$:

Then $\det A$ is given by the *Laplace expansion*

\begin{align*}

\det A&=(-1)^{i+1}a_{i1}\det A_{i1}+\cdots+(-1)^{i+n}a_{in}\det A_{in}\\

&=(-1)^{1+j}a_{1j}\det A_{1j}+\cdots+(-1)^{n+j}a_{nj}\det A_{nj}.

\end{align*}

All the properties of the determinants of order 2 we studied here still hold in general for the determinants of order $n$. In particular,

*Theorem*. Let $A^1,\cdots,A^n$ be column vectors of dimension $n$. They are linearly dependent if and only if

$$\det(A^1,\cdots,A^n)=0.$$

*Example*. Let us calculate the determinant of the following $3\times 3$ matrix

$$A=\begin{pmatrix}

a_{11} & a_{12} & a_{13}\\

a_{21} & a_{22} & a_{23}\\

a_{31} & a_{32} & a_{33}

\end{pmatrix}.$$

You may use any column or row to calculate $\det A$ using the Laplace expansion. In this example, we use the first row to calculate $\det A$. By the Laplace expansion,

\begin{align*}

\det A&=a_{11}\det A_{11}-a_{12}\det A_{12}+a_{13}\det A_{13}\\

&=a_{11}\left|\begin{array}{cc}

a_{22} & a_{23}\\

a_{32} & a_{33}

\end{array}\right|-a_{12}\left|\begin{array}{cc}

a_{21} & a_{23}\\

a_{31} & a_{33}

\end{array}\right|+a_{13}\left|\begin{array}{cc}

a_{21} & a_{22}\\

a_{31} & a_{32}

\end{array}\right|.

\end{align*}

Replay this with

$$A=\begin{pmatrix}

2 & 1 & 0\\

1 & 1 & 4\\

-3 & 2 & 5

\end{pmatrix}.$$

Since the first row or the third column contains 0, you may want to use the first row or the third column to do the Laplace expansion.

For $3\times 3$ matrices, there is a quicker way to calculate the determinant as shown in the following figure. You multiply three entries along each indicated arrow. When you multiply three entries along each red arrow, you also multiply by $-1$. This is called the *Rule of Sarrus* named after a French mathematician Pierre Frédéric Sarrus. Please be warned that the rule of Sarrus works only for $3\times 3$ matrices.

*Example*. [Cross Product] Let $v=v_1E_1+v_2E_2+v_3E_3$ and $w=w_1E_1+w_2E_2+w_3E_3$ be two vectors in Euclidean 3-space $\mathbb{R}^3$. The *cross product* is defined by

$$v\times w=\left|\begin{array}{ccc}

E_1 & E_2 & E_3\\

v_1 & v_2 & v_3\\

w_1 & w_2 & w_3

\end{array}\right|.$$

Note that the cross product is perpendicular to both $v$ and $w$.

Clearly, if there are many 0 entries in a given determinant, it would be easier to calculate the determinant since you will have a lesser than usual number of terms that you actually have to calculate in the Laplace expansion. For any given determinant, we can indeed make it happen. Recall the theorem we studied here:

*Theorem*. If one adds a scalar multiple of one column (row) to another column (row), then the value of the determinant does not change.

Using the particular column (row) operation in the Theorem, we can turn a given determinant into one with more 0 entries.

*Example*. Find

$$\left|\begin{array}{cccc}

1 & 3 & 1 & 1\\

2 & 1 & 5 & 2\\

1 & -1 & 2 & 3\\

4 & 1 & -3 & 7

\end{array}\right|.$$

*Solution*. By the above Theorem,

\begin{align*}

\left|\begin{array}{cccc}

1 & 3 & 1 & 1\\

2 & 1 & 5 & 2\\

1 & -1 & 2 & 3\\

4 & 1 & -3 & 7

\end{array}\right|&=\left|\begin{array}{cccc}

0 & 4 & -1 & -2\\

2 & 1 & 5 & 2\\

1 & -1 & 2 & 3\\

4 & 1 & -3 & 7

\end{array}\right|\ (\mbox{add $-1$ times row 3 to row 1})\\

&=\left|\begin{array}{cccc}

0 & 4 & -1 & -2\\

0 & 3 & 1 & -4\\

1 & -1 & 2 & 3\\

4 & 1 & -3 & 7

\end{array}\right|\ (\mbox{add $-2$ times row 3 to row 2})\\

&=\left|\begin{array}{cccc}

0 & 4 & -1 & -2\\

0 & 3 & 1 & -4\\

1 & -1 & 2 & 3\\

0 & 5 & -11 & -5

\end{array}\right|\ (\mbox{add $-4$ times row 3 to row 4})\\

&=\left|\begin{array}{ccc}

4 & -1 & -2\\

3 & 1 & -4\\

5 & -11 & -5

\end{array}\right|\ (\mbox{Laplace expansion along column 1})

\end{align*}

You may compute the resulting determinant of order 3 using the rule of Sarrus or you may further simpify it. For instance, you may do:

\begin{align*}

\left|\begin{array}{ccc}

4 & -1 & -2\\

3 & 1 & -4\\

5 & -11 & -5

\end{array}\right|&=\left|\begin{array}{ccc}

7 & 0 & -6\\

3 & 1 & -4\\

5 & -11 & -5

\end{array}\right|\ (\mbox{add row 2 to row 1})\\

&=\left|\begin{array}{ccc}

7 & 0 & -6\\

3 & 1 & -4\\

38 & 0 & -49

\end{array}\right|\ (\mbox{add 11 times row 2 to row 3})\\

&=\left|\begin{array}{cc}

7 & -6\\

38 & -49

\end{array}\right|\ (\mbox{Laplace expansion along column 2})\\

&=-115.

\end{align*}