2 Laplace expansion along the top row

This is a technique which can be used to evaluate determinants of any order. In principle, this method can use any row or any column as its starting point. We quote one example: using the top row.

Consider $\Delta =\left|\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|$ .

First we introduce the idea of a minor . Each element in this array of numbers has an associated minor formed by removing the column and row in which the element lies and taking the determinant of the remainder. For example consider element $a_{23}=3$ . We strike out the second row and the third column:

to leave $\left|\begin{array}{cc}\hfill 4\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 1\hfill \end{array}\right|=4-3=1$ .

For the element $a_{31}=3$ we strike out the third row and first column:

to leave $\left|\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right|=3-2=1$ .

Task!

What is the minor of the element $a_{22}=2$ ?

Answer

Next we introduce the idea of a cofactor . This is a minor with a sign attached. The appropriate sign comes from the pattern of signs appropriate to a $3\times 3$ array:

$\begin{array}{ccc}\hfill +\hfill & \hfill -\hfill & \hfill +\hfill \\ \hfill -\hfill & \hfill +\hfill & \hfill -\hfill \\ \hfill +\hfill & \hfill -\hfill & \hfill +\hfill \end{array}$

(i.e. positive signs on the leading diagonal and the signs ‘alternate’ everywhere else.)

Each element has a cofactor associated with it. The cofactor of element $a_{11}$ is denoted by $A_{11}$ , that of $a_{23}$ by $A_{23}$ and so on.

To obtain the cofactor of an element of a $3\times 3$ matrix we simply multiply the minor of that element by the corresponding sign from the $3\times 3$ array of signs.

Hence the cofactor corresponding to $a_{23}$ is

$A_{23}=-\left|\begin{array}{cc}\hfill 4\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 1\hfill \end{array}\right|=-1$

and the cofactor corresponding to $a_{31}$ is $A_{31}=+\left|\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right|=1$ .

Task!

What is the cofactor of the element $a_{22}$ ?

Answer

Cofactors are important as it can be shown that the value of the determinant of a $3\times 3$ matrix can be found from the formula

$\Delta =a_{11}{A}_{11}+a_{12}{A}_{12}+a_{13}{A}_{13}.$

In words “the determinant of a $3\times 3$ matrix is obtained by multiplying each element of the first row by its corresponding cofactor and then adding the three together”. (In fact this rule can be extended to apply to any row or any column and to any order square matrix.)

In the case of $\Delta =\left|\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|$ we have $a_{11}=4,a_{12}=1$ , $a_{13}=1,$

$A_{11}=+\left|\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=4-3=1$

$A_{12}=-\left|\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 2\hfill \end{array}\right|=-\left(2-9\right)=7$

$A_{13}=+\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 1\hfill \end{array}\right|=1-6=-5$

Hence $\Delta =4\times 1+1\times 7+1\times -5=6$ .

Alternatively, choosing to expand along the second row:

Task!

Use expansion along the first row to find $\Delta =\left|\begin{array}{ccc}\hfill 1& \hfill -1& \hfill 3\\ \hfill 0& \hfill 2& \hfill 6\\ \hfill -2& \hfill 1& \hfill 5\end{array}\right|$

Answer