Summary of the calculation method of determinant

Summary of the calculation method of determinant

There is no diagonal rule for determinants of order 2 and 3, and there is no diagonal rule for determinants of order 4 and above!
There are several methods to solve the higher order determinant
Use properties of upper (lower) triangle, upper (lower) oblique triangle, arrow
Expansion theorem by row and column
Laplace expansion theorem
Bordering method
Recursive relation method
Induction
Special determinant (such as Vandermonde determinant)
Ha ha, just think of these
What is the calculation method of determinant?
It is very important to make full use of the characteristics of determinant to simplify the determinant. According to the characteristics of determinant, using the properties of determinant, a row (column) is transformed into only one non-zero element, and then expanded according to the row (column)
How is the determinant calculated?
1. Second order determinant, third-order determinant calculation, the landlord should have learned. But can not be used for fourth-order, fifth-order, second-order, fourth-order or fourth-order determinant calculation, generally speaking, there are two methods. The first is according to any row or any column expansion: A, any row or any column of all elements multiplied by delete the element
Let m and p be two nonempty sets, define the difference set of M and P as M-P = {x | x ∈ m and X &; P}, and prove m - (M-P) = m ∩ P
The answer is divided into two situations: when m ∩ P = & amp; 8709; when m ∩ P ≠ & amp; 8709; but it says that when m ∩ P ≠ & amp; 8709; when M-P = M
When m ∩ P = &
Since any x ∈ m has X &; P, M-P = M
So m - (M-P) = &;
When M-P ≠ & # 8709
M-P denotes elements in m but not in P
M - (M-P) denotes elements in m but not in m-p
Because the elements in M-P are not in P, the elements in M - (M-P) are in P, so the elements in M - (M-P) are in M ∩ P
On the other hand, the element of P-M is also consistent with the definition of p-m
So m - (M-P) = m ∩ P
In order to make the solution of the equation 5x-2m = 3x-6m + 1 about X between - 3 and 2, find the integer value suitable for M
5x-2m=3x-6m+1
2x=-4m+1
x=-2m+1/2
-3
Solution:
2x = -4m + 1
x = -2m + 0.5
-3 < -2m + 0.5 < 2
-3/4 < m < 7/4
m = { 0 ,1}
How to calculate the formula of mathematical rounding up ten?
check
If the equations of X and y have the same solution, find the value of ab
I am now grade two, the method is not too difficult, I will not understand
It is known that 1.4x-y = 5 ① ax + by = - 1 ② 2. Bracket: 3x + y = 9 ① 3ax-4by = 18 ②
1. Multiply 2 in brackets by 4
We get 4ax + 4BY = - 4 ③
1. ② + ③ in brackets
7ax=14
ax=2
1 in brackets + 2 in brackets
7x=14
X=2
Substitute x = 2 for Ax = 2
A=1
Substitute x = 2 for 4x-y = 5
8-y=5
-y=-3
Y=3
Put y = 3, a = 1, x = 2 in brackets
2+b*3=-1
2+3b=-1
3b=-3
b=-1
So a = 1, B = - 1
The image of a linear function passing through y = KX + B (k is not equal to 0) has two empty spaces
Point (0, b) and point (- B / K, 0)
Don't make the title so mysterious, OK?
Empty you always mark out ah, the whole so hard
Across (0, b) (- B / k.o)
Let m and p be two nonempty sets, and define M-P = {x x ∈ m and X &; P}, if M = {x 1 ≤ x ≤ 2011}
Let m and p be two nonempty sets, and define: M-P = {x x ∈ m and X &; P}. If M = {x 1 ≤ x ≤ 2011, X ∈ n}, P = {y 2 ≤ y ≤ 2012, y ∈ n}, then P-M = {x 1 ≤ x ≤ 2011, X ∈ n}=
P-M= {1}
The result is a set containing only element 1
It is known that the solution of the equation 5x-2m = 3x-6m + 1 about X satisfies - 3 < x ≤ 3, and the integer value of M satisfying the condition is obtained
-5/4