# How can we memorize the formula for the determinant of a $4\times4$ matrix?

This is the formula for the determinant of a $4\times4$ matrix.

.

0,0 | 1,0 | 2,0 | 3,0
0,1 | 1,1 | 2,1 | 3,1
0,2 | 1,2 | 2,2 | 3,2
0,3 | 1,3 | 2,3 | 3,3

.

m[0,3] * m[1,2] * m[2,1] * m[3,0] - m[0,2] * m[1,3] * m[2,1] * m[3,0] -
m[0,3] * m[1,1] * m[2,2] * m[3,0] + m[0,1] * m[1,3] * m[2,2] * m[3,0] +
m[0,2] * m[1,1] * m[2,3] * m[3,0] - m[0,1] * m[1,2] * m[2,3] * m[3,0] -
m[0,3] * m[1,2] * m[2,0] * m[3,1] + m[0,2] * m[1,3] * m[2,0] * m[3,1] +
m[0,3] * m[1,0] * m[2,2] * m[3,1] - m[0,0] * m[1,3] * m[2,2] * m[3,1] -
m[0,2] * m[1,0] * m[2,3] * m[3,1] + m[0,0] * m[1,2] * m[2,3] * m[3,1] +
m[0,3] * m[1,1] * m[2,0] * m[3,2] - m[0,1] * m[1,3] * m[2,0] * m[3,2] -
m[0,3] * m[1,0] * m[2,1] * m[3,2] + m[0,0] * m[1,3] * m[2,1] * m[3,2] +
m[0,1] * m[1,0] * m[2,3] * m[3,2] - m[0,0] * m[1,1] * m[2,3] * m[3,2] -
m[0,2] * m[1,1] * m[2,0] * m[3,3] + m[0,1] * m[1,2] * m[2,0] * m[3,3] +
m[0,2] * m[1,0] * m[2,1] * m[3,3] - m[0,0] * m[1,2] * m[2,1] * m[3,3] -
m[0,1] * m[1,0] * m[2,2] * m[3,3] + m[0,0] * m[1,1] * m[2,2] * m[3,3]

.

0, 1, 2, 3,
4, 5, 6, 7,
8, 9, 10, 11,
12, 13, 14, 15

.

m[12] * m[9]  * m[6]  * m[3]   -  m[8] * m[13] * m[6]  * m[3]   -
m[12] * m[5]  * m[10] * m[3]   +  m[4] * m[13] * m[10] * m[3]   +
m[8]  * m[5]  * m[14] * m[3]   -  m[4] * m[9]  * m[14] * m[3]   -
m[12] * m[9]  * m[2]  * m[7]   +  m[8] * m[13] * m[2]  * m[7]   +
m[12] * m[1]  * m[10] * m[7]   -  m[0] * m[13] * m[10] * m[7]   -
m[8]  * m[1]  * m[14] * m[7]   +  m[0] * m[9]  * m[14] * m[7]   +
m[12] * m[5]  * m[2]  * m[11]  -  m[4] * m[13] * m[2]  * m[11]  -
m[12] * m[1]  * m[6]  * m[11]  +  m[0] * m[13] * m[6]  * m[11]  +
m[4]  * m[1]  * m[14] * m[11]  -  m[0] * m[5]  * m[14] * m[11]  -
m[8]  * m[5]  * m[2]  * m[15]  +  m[4] * m[9]  * m[2]  * m[15]  +
m[8]  * m[1]  * m[6]  * m[15]  -  m[0] * m[9]  * m[6]  * m[15]  -
m[4]  * m[1]  * m[10] * m[15]  +  m[0] * m[5]  * m[10] * m[15]

.

There seems to be some symmetrical patterns in the formula, but most numbers seem to be too much random and aleatory, so memorizing the 96 numbers would seem to be pretty hard...

Is there any mnemotechnical way of making this easier to memorize?

Note: I know there are other ways to calculate the determinant of a $4\times4$ matrix, but this question is only about the brute force approach which consists in memorizing the entire formula.

• To get the formula for a 4x4 determinant, we traverse along one row or column, calculating four 3x3 determinants. If you know the 3x3 determinant formula, you know the 4x4 one. Apr 25, 2015 at 18:38
• What Andrey suggest is known as determinant expansion. Then you can look for rows and columns with a lot of zeros and expand along those, which makes the sum much shorter. Apr 25, 2015 at 18:48
• Memorizing the formula for the 4x4 determinant is only less absurd than memorizing the formula for the 5x5 determinant. Apr 25, 2015 at 19:49
• But why do you want to memorize it? If it is a question truly about memorizing to make it easier then this will be a LOT harder then just learning it Apr 25, 2015 at 19:54

## 3 Answers

Laplace expansion:

Let $A$ be an $n \times n$ matrix. Denote $a_{ij}$ be the entry of row $i$ and column $j$. Denote $A_{ij}$ be the $(n-1) \times (n-1)$ matrix by deleting row $i$ and column $j$. Then $\det A$ is given by:

$$\det A = \sum^n_{j=1}(-1)^{i+j}a_{ij}\det A_{ij}$$

Picture:

Intuitively, the signs are alternating in this pattern: $$\begin{pmatrix} + & - & + & \dots \\ - & + & - & \dots \\ + & - & + & \dots \\ \vdots & \vdots & \vdots &\ddots \end{pmatrix}$$ You pick the first row (say), do this:

$$\begin{pmatrix} + & {\rm delete} \\ {\rm delete} & \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}\end{pmatrix}$$ Compute the determinant of the matrix "left" by deleting row $i$ and column $j$, multiply by $a_{ij}$. You sum similar terms like this for a whole row. (Thanks Ivo Terek for helping TEXing)

Notice that you can do any row, any column, which provides you convenience. For example, if one row or column has many $0$'s, choose this row/column.

This can be helped further by Gaussian elimination because of this rule:

Adding multiple of a row to another row does not change determinant.

Therefore after Gaussian elimination, usually there are many $0$'s, you can compute determinant much faster. If one row has all entries $0$, then the determinant is $0$.

• I tried making the drawing better. If you want, you can rollback the edit to what it was before :) Apr 25, 2015 at 18:56
• @Ivo Terek Thank you very much! With your help I can put more matrices by copying and pasting. Actually I want to draw more but I'm too lazy:( Apr 25, 2015 at 19:02

Instead of memorizing it, we can express the determinant of a $n \times n$ matrix by: $$\det(A) = \sum_{\sigma \in S_n} {\rm sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma(i)}.$$ The above is often used as the definition of the determinant. For the case $n=2$, for example, $$\sigma_1 = {\rm id},\, {\rm sgn}(\sigma_1)=1, \quad \sigma_2 = (1\, 2), \, {\rm sgn}(\sigma_2) = -1,$$so that: \begin{align} \det(A) &= {\rm sgn}(\sigma_1) a_{1,\sigma_1(1)}a_{2,\sigma_1(2)} + {\rm sgn}(\sigma_2)a_{1,\sigma_2(1)}a_{2,\sigma_2(2)} \\ &= a_{11}a_{22} - a_{12}a_{21}, \end{align} which agree with what we know.

You can do the same computations for the $4 \times 4$ case, if you want. You can organize the permutations in $S_4$ in the way that works better for you. For example, think first of the identity, then of the permutations that fix only one number, then the permutations that fix two numbers, etc.

Write down all 24 permutations of $\{1,2,3,4\}$, adding next to them the signature: $$1\ 2\ 3 \ 4\ (+)\\ 1\ 2\ 4 \ 3\ (-)\\ 1\ 3\ 2 \ 4\ (-)\\ \dots$$ and then consider all expressions $$+a_{11}a_{22}a_{33}a_{44}\\ -a_{11}a_{22}a_{34}a_{43}\\ -a_{11}a_{23}a_{32}a_{44}\\ \dots$$ where the first index always grows from $1$ to $4$ and the second index is determined by the current permutation from the list. Put the signature in front and sum up.

It's of course easier for $3\times 3$ matrices: $$1\ 2\ 3\ (+)\\ 1\ 3\ 2\ (-)\\ 2\ 1\ 3\ (-)\\ 2\ 3\ 1\ (+)\\ 3\ 1\ 2\ (+)\\ 3\ 2\ 1\ (-)$$ so we get $$+a_{11}a_{22}a_{33}\\ -a_{11}a_{23}a_{32}\\ -a_{12}a_{21}a_{33}\\ +a_{12}a_{23}a_{31}\\ +a_{13}a_{21}a_{32}\\ -a_{13}a_{22}a_{31}$$

And no, you don't memorize it. Use a Laplace development, instead.