Assertion: matrix's determinant | /r/HasProperty | invertible
Facets:
value: in the ring
type: LOCATION
value: in r.
type: LOCATION
value: in r
type: LOCATION
Source sentences:
text: If the ring R is commutative, that is, its multiplication is commutative, then M(n, R) is a unitary noncommutative (unless n = 1) associative algebra over R. The determinant of square matrices over a commutative ring R can still be defined using the Leibniz formula; such a matrix is invertible if and only if its determinant is invertible in R, generalising the situation over a field F, where every nonzero element is invertible.
text: This definition can also be extended where K is a commutative ring R, in which case a matrix is invertible if and only if its determinant is an invertible element in R.
text: However, in the case of the ring being commutative, the condition for a square matrix to be invertible is that its determinant is invertible in the ring, which in general is a stricter requirement than being nonzero.