The sum of the nth order determinant is 50, and the nth order determinant is summed

As follows:

1) The first line is swapped with the second line, and then the ,.. is exchanged with the third lineSwap with the last row, a total of n-1 row swaps, the first row is swapped to the last row, and the other rows are moved up one line;

2) The first line at this point is swapped with the second row, and then the ,.. is exchanged with the third rowSwap with the penultimate row, a total of n-2 row swaps;

3) The first row is swapped with the following row in turn until it is swapped with the penultimate row, for a total of n-3 swaps;

n-1) The first line is exchanged with the second line, and 1 line is exchanged;

After performing the above n-1 steps, the determinant.

This becomes a determinant with all the main diagonals of 1, and the result is 1, and the total number of row exchanges is: (n-1) + (n-2) +2+1=n(n-1) 2 times, each line exchanged, multiplied by a minus sign.

Therefore, the result is (-1) [n(n-1) 2].

Brief introduction. A determinant, in mathematics, is a function that defines a domain.

is the matrix a of det, and the value is a scalar.

Polynomial theory, still in calculus (for example, in commutation integral), determinants as a basic mathematical tool, have important applications.

The determinant can be seen as the concept of directed area or volume in general Euclidean space.

in promotion. Or, in an n-dimensional Euclidean space, the determinant describes a linear transformation.

Effect on Volume.

This is not the main diagonal, you can't do it directly, you have to transform it first.

1) The first line is swapped with the second line, and then the ,.. is exchanged with the third lineSwap with the last row, a total of n-1 row swaps, the first row is swapped to the last row, and the other rows are moved up one line;

2) The first line at this point is swapped with the second row, and then the ,.. is exchanged with the third rowSwap with the penultimate row, a total of n-2 row swaps;

3) The first row is swapped with the following row in turn until it is swapped with the penultimate row, for a total of n-3 swaps;

n-1) The first line is exchanged with the second line, and 1 line is exchanged;

After the above n-1 steps, the determinant becomes a determinant with all the main diagonals of 1, and the result is 1, and the total number of row exchanges is: (n-1)+(n-2)+2+1=n(n-1) 2 times, each line exchanged, multiplied by a minus sign.

Therefore, the result is (-1) [n(n-1) 2].

The sum of the algebraic remainders of line 1 is equal to the determinant obtained by replacing all elements in line 1 of the original determinant with 1, the sum of the algebraic remainders of line 2 is equal to the determinant obtained by replacing all elements in line 2 of the original determinant with 1, and the sum of the algebraic remainders of line n is equal to the determinant obtained by replacing all elements in line 1 of the original determinant.

The sum of all algebraic remainders is the sum of the n new determinants above.

In the nth-order determinant, after the o-row and e-column where the element a i is located are crossed out, the remaining n-1 determinant is called the coremainant of element a i, which is denoted as m, and the coundit m is multiplied by the o+e power of -1 as a, and a is called the algebraic coremainant of element a.

The algebraic remainder of an element a i has nothing to do with the element itself, only with the position of that element.

What does determinant summing mean? There's no such thing.

1²+2²+3²+.n = n(n+1)(2n+1) 6 proof: n+1) = n +3n +3n+1

n+1)³-n³=3n²+3n+1

n -(n-1) =3(n-1) +3(n-1)+1 are added to each other.

n+1)³-1³=3*(1²+2²+.n²)+3(1+2+..n)+1*n

n³+3n²+3n)-3n(n+1)/2-n=3sn3sn=n(2n²+3n+1)/2=n(n+1)(2n+1)/2sn=n(n+1)(2n+1)/6

Van der Mon determinant, as shown in the figure below:

The first line is to the power of 0 to the 3rd power of 1, the second line is to the power of 0 to the 3rd power of 2, the third line is to the power of 0 to the 3rd power of 3, and the first line is to the power of 0 to the 3rd power of the 4.

Conforms to the form of the van der Mon determinant, which is evaluated using the formula.

The standard form of the van der Mon determinant is the bridge only: the n-order van der Mon determinant is equal to the product of all possible differences of this number. According to the characteristics of the Van der Mon determinant, Pi Ye can turn the given determinant into the Van der Monde determinant, and then use the result to calculate.

The basic formula is:

Common formula: 1)1 [n(n+1)]=1 n)- 1 (n+1)]2)1 [(2n-1)(2n+1)]=1 2[1 (2n-1)-1 (2n+1)].

3）1/[n(n+1)(n+2)]=1/24）1/(√a+√b)=[1/(a-b)](a-√b)5） n·n!=(n+1)!-n!

6）1/[n(n+k)]=1/k[1/n-1/(n+k)]7）1/（√n+√n+1）=√n+1）-√n8）1/（√n+√n+k）=（1/k）·[n+k）-√n]

1. The sum of all preceding it is an inverse ordinal number.

It is defined as: the nth order determinant (definition 1) has n 2 numbers, arranged in n rows and n columns, and makes the product of n numbers in different rows and columns in the table, and is crowned with the symbol (-1) t, in the form of the following terms, where the natural numbers 1, 2 ,..n is an arrangement, and t is the inverse ordinal number of this arrangement.

Since there is a total of n!a, this n!The algebraic sum of terms is called nth-order determinant.

Think of it this way: here j1, j2 ,......The jn status is all equivalent, each representing a number from 1 to n, and the two do not repeat each other. So j1j2 ......jn represents a full permutation from 1 to n.

Because the full permutation under the summation symbol does not specify which one it is, this summing requires that all possible permutations be combined.

As an example, if n = 3, there are 6 terms on the right side of the equation. Among these 6 items, J1, J2, and J3 correspond to 1 2 3, 1 3 2, 2 1 3, 2 3 1, 3 1 2, 3 2 1, respectively.

Thus the right side of the equation = (-1) to the power of (1,2,3) multiplied by a11a12a13 + (-1) to the power (1,3,2) to the power of a11a13a12 + ......

Using the property of the determinant to simplify, step by step, it is simple to turn it into the lower triangle determinant, and the result is 2 2

Let the value of the determinant be dn

Add row n to line 1 and follow line 1 to get the recursive formula, dn=2d(n-1).