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

Question

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

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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

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     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]

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

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     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]

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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.

Answer 1

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 $0$.

Answer 2

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.

Answer 3

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.