The proof of determinant property 2 is incomprehensible

I think you may have misunderstood.

If you think about it, in fact, d itself should be.

1)^t*a(1,p1)a(2,p2)……a(i,pj),…a(j,pi),…a(n,pn)

Otherwise, if you exchange the two lines of d, you will get nothing in b.

b(i,p)=a(j,p).

Obviously, it's already been said, p1 p2 ......pi……pj……The reverse order of pn is t

p1 p2 ……pj……pi……The reverse order of pn is t1

Obviously, taking advantage of the property of reverse order, after exchanging two numbers, -1) t=-(-1) t1

Then in the last line.

d=(-1)^t1*a(1,p1)a(2,p2)……a(i,pj),…a(j,pi),…a(n,pn)

At this point, the subscript of A has become P1 P2 ......pj……pi……pn, and t1 also represents the reverse order of this arrangement, combined with the definition of the determinant, this is not equal to the original value of d

d itself should be.

1)^t*a(1,p1)a(2,p2)……a(i,pj),…a(j,pi),…a(n,pn)

Otherwise, if you swap the two lines of d, you will get b, and there will be no relationship between b(i,p)=a(j,p).

Obviously, it's already been said, p1 p2 ......pi……pj……The reverse order of PN is TP1 P2 ......pj……pi……The reverse order of pn is t1Obviously, using the property of reverse order, after exchanging two numbers, 1) t=-(-1) t1

Then in the last line.

d=(-1)^t1*a(1,p1)a(2,p2)……a(i,pj),…a(j,pi),…a(n,pn)

At this point, the subscript of A has become P1 P2 ......pj……pi……pn, and t1 also denotes the reverse order under this permutation, combined with the definition of the determinant.

You can look at it with a concrete example, so it's easy to understand.

If you really don't understand it, you can admit that it's forgotten, and it's all tested in practice anyway.

The determinant nature book is very clear, if you can't understand it, then who can say that you can understand it?

Are you sick? After learning the points, I came here to do the determinant???

Am you mistaken?? Take this point to fool people??

It's math, so many letters even think it's a foreign language

Q: How do you do math the same as a foreign language?

There is a suspicion of earning clicks.

The second column is multiplied by -a and added to the first column, and the elements in the first column are all 0, so the original determinant = 0.

The second and first columns are split sequentially and then calculated according to the nature of the determinant.

It's too much to prove. To correct you, you only have one narrative complete.

Swap two rows of the determinant, and the determinant only changes the symbol, which should be changed to: swap the two rows or two columns of the determinant, and the value of the determinant only changes the sign;

If a row of the determinant is added to an infinite number of other rows, the determinant should be changed to: the value of the determinant should be changed to the following: a row of the determinant is added to a number of other rows, and the value of the determinant remains unchanged;

If two rows in the determinant are identical, the value of the determinant is zero, which should be: if two rows in the determinant are identical, then the value of the determinant is zero;

If the determinant has two rows of corresponding elements proportional, then this determinant is equal to zero, which should be: if the determinant has two rows of corresponding elements proportional, then this determinant is equal to zero;

6.If a is reversible and the determinant is 0, it should be changed to: 6If a is reversible, then the value of the determinant of a is not 0;

7.If the determinant is 0 and a is irreversible, it should be changed to: 7If the value of the determinant ad is 0, then a is irreversible.

The proof of this property depends on another spin-off property.

Consider adding k times from line j to line i. Remember that this determinant is the property of d1 by the determinant, and divide the determinant d1 into the sum of the two determinants in line i:

One of them is the original determinant, and the elements of the ith row of the other determinant are k times the elements of the j-row, i.e., the two rows are proportional, so they are 0

So d1 = d, i.e. the value of the determinant does not change.

First take the 2nd, 3rd, and 4th columns and subtract the first column respectively, subtract a 2 d 2 from the third column, then subtract the second column 2 and remove a d, get the second column and each row is 2, the 4th column minus the second column 3 and remove a d, get the third column, each row is 6, and finally the 4th column minus the third column 3, so that the fourth column is all 0, so the value of the determinant is 0

It is divided into two items, and simplified separately with the determinant property, and finally canceled out, and the determinant is equal to 0.

Add column 3 to column 1 and subtract column 2*2 Subtract column 1 from column 2 A 2 2a+1 2 b 2 2b+1 2 c 2 2c+1 2 Subtract column 3 from column 2 and extract the factor 2 of column 2 to get the Van der Mon determinant a 2 a 1 b 2 b 1 c 2 c 1 So the result: 4(a-b)(a-c)(b-c).

Column n * 1) is added to the first and second columns, respectively;

The resulting new first column is all identical; The new second column is also all the same;

Extract the common factors separately;

We get that the first column and the second column are both 1;

So the determinant is 0

I am like an ostrich ...

A b means that matrix A corresponds to each element in matrix b, and the left part of the equation is not (a b), but only the addition of a and a row of elements in b, so the book is correct.

In the above question, the matrix (a b) should be equal to:

a1 a2 b1 b2 c1 c22l 2m 2n2x 2y 2z instead of the left side of the question:

A1 A2 B1 B2 C1 C2L m Nx y Z got it?