What are the methods and steps of calculating the fourth-order determinant

What are the methods and steps of calculating the fourth-order determinant

If it's a pure numeric determinant
In general, the property of determinant is used to simplify the determinant
Choose a row (or a column) of numbers that are relatively simple, use properties to get three zeros, and then use the expansion theorem to expand
If it contains letters, it depends on the specific situation
Note whether there is a special block matrix

How to use algebraic cofactor to solve 4-order determinant
As shown in the picture

A31, A32, A33 and A34 are the congruent algebraic expressions of the elements in the third row of determinant D. where d=
3 1 -1 2
-5 1 3 -4
2 0 1 -1
1 -5 3 -3
In this paper, we construct a new determinant G such that G=
3 1 -1 2
-5 1 3 -4
1 3 -2 2
1 -5 3 -3
The elements of G and D are equal except for the third line
According to the property of determinant, the algebraic cofactor of the third row element of G is equal to that of the third row element of D
That is, G is expanded by the third line
G = A31+ 3*A32 - 2*A33 +2* A34……………………………………………… （*）
[now find the value of determinant g]
First of all, replace the first, third, second and fourth lines of G in turn
1 3 -2 2
1 -5 3 -3
3 1 -1 2
-5 1 3 -4
Then subtract the first line from the second line, subtract three times the first line from the third line, and add five times the first line to the fourth line
1 3 -2 2
0 -8 5 -5
0 -8 5 -4
0 16 -7 6
Then subtract the second line from the third line, and double the second line from the fourth line
1 3 -2 2
0 -8 5 -5
0 0 0 1
0 0 3 -4
Multiply the fourth line by (- 1), and then exchange the third and fourth lines
1 3 -2 2
0 -8 5 -5
0 0 -3 4
0 0 0 1
∴G = 1 * (- 8)* (- 3)* 1 = 24
Substituting (*), we get
A31+ 3*A32 - 2*A33 +2* A34 = 24