That is, the determinant with numbers on the first row, the first column and the diagonal, but all zeros elsewhere, is calculated by multiplying the second column by several times to make the corresponding element of the first column 0, for example, multiplying the second column by a number to make the second element of the first column 0, and multiplying the third column by a number to make the third element of the first column 0. Each column is transformed into a triangular determinant in this way.
Find the determinant Dn, where a 1a2a3 ... one is not equal to 0.
1+a 1... 1
1 1+a2... 1
......
1 1 ... 1+ an
Line 1 is multiplied by-1 and added to other lines.
1+a 1... 1
-a 1 a2...0
......
-a 1 0 ... 1; one
This is the claw determinant.
The calculation method is to use the non-zero elements on the main diagonal from the second column to the nth column to change the elements in the 1 column of the peers into 0.
Column k puts forward that a 1a2a3...an* stands for ak, k= 1, 2, ..., n (note that ai is not equal to 0).
1+ 1/a 1 1/a2... 1/an
- 1 1 ...0
......
- 1 0 ... 1
Add columns 2 to n to column 1 to get an upper triangular determinant.
1+ 1/a 1 1/a2... 1/an
0 1 ...0
......
0 0 ... 1
Determinant = a 1a2a3 ... An (1+1+2/a2+...+1/an) = ∏ ai (1+)
Extended data:
N-order determinant
set up
Is made up of n arranged in an n-order square matrix? The number aiji (I, j = 1, 2, ..., n), whose value is n! Sum of terms
Where k 1, k2, ..., kn are exchanged by the sequence 1, 2, ..., n represents k times, and σ represents k 1, k2, ..., kn. The form is as follows
The sum of the items of, where a 13a2 1a34a42 corresponds to k=3, that is, the symbol before the item should be
(- 1)3.
If the n-order square matrix A=(aij), then the determinant D corresponding to A is recorded as
D=|A|=detA=det(aij)
If the determinant d of matrix A is equal to 0, it is called singular matrix, otherwise it is called nonsingular matrix.
Tag set: any k elements I 1, I 2, ..., n are satisfied in the sequence i 1, i2, ..., ik.
1≤I 1 & lt； i2 & lt...& ltik≤n( 1)
I 1, i2, ..., ik has k {1, 2, ..., n} and {1, 2, ..., n} denoted as C.
σ={i 1, i2, ..., ik} is {1, 2, ..., n}. Let τ = {j 1, J2, ..., JK}
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