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Although an explicit inverse is not necessary to estimate the vector of unknowns, it is the easiest way to estimate their accuracy, found in the diagonal of a matrix inverse (the posterior covariance matrix of the vector of unknowns). However, faster algorithms to compute only the diagonal entries of a matrix inverse are known in many cases. [19]
[1] [2] [3] That is, given an invertible matrix and the outer product of vectors and , the formula cheaply computes an updated matrix inverse (+)). The Sherman–Morrison formula is a special case of the Woodbury formula .
A matrix with entries in a field is invertible precisely if its determinant is nonzero. This follows from the multiplicativity of the determinant and the formula for the inverse involving the adjugate matrix mentioned below. In this event, the determinant of the inverse matrix is given by
The sum of the entries along the main diagonal (the trace), plus one, equals 4 − 4(x 2 + y 2 + z 2), which is 4w 2. Thus we can write the trace itself as 2w 2 + 2w 2 − 1; and from the previous version of the matrix we see that the diagonal entries themselves have the same form: 2x 2 + 2w 2 − 1, 2y 2 + 2w 2 − 1, and 2z 2 + 2w 2 − 1. So ...
If p and q are both 1 (i.e., A, B, C and D are all scalars), we get the familiar formula for the inverse of a 2-by-2 matrix: = [] provided that AD − BC is non-zero.. In general, if A is invertible, then
In mathematics, and in particular linear algebra, the Moore–Penrose inverse + of a matrix , often called the pseudoinverse, is the most widely known generalization of the inverse matrix. [1] It was independently described by E. H. Moore in 1920, [2] Arne Bjerhammar in 1951, [3] and Roger Penrose in 1955. [4]
2.1 Determinant and inverse matrix. ... the formula for the coefficients c i is given in terms of complete exponential Bell ... B 2 (x 1, x 2) = x 2 1 + x 2, and B 3 ...
This gives a formula for the inverse of A, provided det(A) ≠ 0. In fact, this formula works whenever F is a commutative ring , provided that det( A ) is a unit . If det( A ) is not a unit, then A is not invertible over the ring (it may be invertible over a larger ring in which some non-unit elements of F may be invertible).