Number of terms in the expansion of general determinant of order n is
Answers
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Concept
For a square matrix, the determinant can be defined in many ways.
The first and simplest way is to formulate the determinant by considering the elements of the upper row and the corresponding minor entries. Take the first element of the top row and multiply it by the minor element, then subtract the product of the second element and its minor element. Continue to alternately add and subtract the product of each element of the top row with its corresponding minor element until all the elements of the top row have been accounted for.
If In is an identity matrix of order nxn, then det(I) = 1
If the matrix MT is the transpose of the matrix M, then det (MT) = det (M)
If the matrix M-1 is the inverse of the matrix M, then det (M-1) =
= det (M)-1
If two square matrices M and N have the same size, then det (MN) = det (M) det (N)
If the matrix M has ax size and C is a constant, then det (CM) = Ca det (M)
If X, Y and Z are three positive semidefinite matrices of equal size, then the following holds together with the corollary det (X+Y) ≥ det(X) + det (Y) for X,Y, Z ≥ 0 det (X+Y+ Z) + det C ≥ det (X+Y) + det (Y+Z)
In a triangular matrix, the determinant is equal to the product of the diagonal elements.
The determinant of a matrix is zero if all elements of the matrix are zero.
Given
It is given that there is a determinant of order n
Find
We need to find the number of terms in the expansion of general determinant of order n
Solution
The maximum number of terms is the extension
|A|(n×n)
= n!
Hence the number of terms in the expansion of general determinant of order n is n!
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