[mathjax]

A 2 × 2 determinant is expressed as follows.

$$\left[\begin{array}{cc}a_{11}&a_{12}\\ a_{21}&a_{22}\\ \end{array}\right]=a_{11}a_{22}-a_{21}a_{12}$$

Then a 3 × 3 determinant is expressed as follows.

$$\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\\ \end{array}\right]\\=a_{11}a_{22}a_{33}+a_{21}a_{32}a_{13}+a_{31}a_{23}a_{12}\\-a_{31}a_{22}a_{13}-a_{21}a_{12}a_{33}-a_{11}a_{23}a_{32}$$

So far you can memorize easily, but it is hard to memorize from 4 × 4. In this article, I’ll show you how to find a 4×4 determinant by looking at the elements of the first row.

## How to find a 4×4 determinant

$$\left[\begin{array}{cccc}a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44}\\ \end{array}\right]\\=a_{11}\left[\begin{array}{ccc}a_{22}&a_{23}&a_{24}\\a_{32}&a_{33}&a_{34}\\a_{42}&a_{43}&a_{44}\\\end{array}\right]-a_{12}\left[\begin{array}{ccc}a_{21}&a_{23}&a_{24}\\a_{31}&a_{33}&a_{34}\\a_{41}&a_{43}&a_{44}\\\end{array}\right]\\+a_{13}\left[\begin{array}{ccc}a_{21}&a_{22}&a_{24}\\a_{31}&a_{32}&a_{34}\\a_{41}&a_{42}&a_{44}\\\end{array}\right]-a_{14}\left[\begin{array}{ccc}a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\a_{41}&a_{42}&a_{43}\\\end{array}\right]$$

With respect to the signs of the 3×3 determinants on the right hand, they are positive if the sum of the numbers of row and column of $a$ is even, and they are negative if the sum is odd. For example, the coefficient of  the first term on the right side is $a_{11}$. Since the sum of the numbers of row and column is 1 + 1 = 2 and is an even number. Therefore, the sign of the first term is positive. On the other hand, the coefficient of the second term is $a_{12}$. Since the sum of the numbers of row and column is 1 + 2 = 3 and is an odd number. Therefore, the sign of the second term is negative.

This method is also useful for finding a determinant with 5 × 5 determinant.