Assertion: determinant | /r/HasProperty | equal
Facets:
value: to zero
type: OTHER
value: to the product of determinants
type: OTHER
value: thus,
type: DEGREE
value: briefly
type: DEGREE
value: to the product of its eigenvalues
type: OTHER
value: by definition
type: OTHER
value: always
type: DEGREE
value: to 0
type: OTHER
value: to one
type: OTHER
value: as a consequence
type: OTHER
value: to the entry in that matrix
type: OTHER
value: namely
type: DEGREE
Source sentences:
text: One of them is multiplicativity, namely that the determinant of a product of matrices is equal to the product of determinants.
text: When the determinant is equal to one, the linear mapping defined by the matrix is equi-areal and orientation-preserving.
text: A determinant is equal to zero if the elements in two of its rows or columns are proportional; thus, for example,
text: The determinant of a matrix with two identical rows is equal to 0.
text: Note that the determinant of 1 by 1 matrix is equal to the entry in that matrix.
text: A determinant is equal to zero if the elements in two of its rows or columns are proportional; thus, for example,
text: A determinant is equal to zero if the elements in two of its rows or columns are proportional; thus, for example,
text: Then, the determinant of is equal to the product of its diagonal entries:
text: Therefore, it is triangular and its determinant is equal to the product of its diagonal entries.
text: As a consequence, the determinant of is equal to .
text: The next proposition shows that the determinant of a product of two matrices is equal to the product of their determinants.
text: If in a given matrix, we have all zero elements in a particular row or column then determinant of such a matrix is equal to zero.
text: Therefore, we can notice that determinant of such a matrix is equal to zero.
text: Various other theorems can be proved as well, including that the determinant of a product of matrices is always equal to the product of determinants; and, the determinant of a Hermitian matrix is always real.
text: If $n = m$, then the determinants are equal and both will be non-zero iff $A$
text: If all the determinants of the submatrices of order 3 were equal to zero, we would have to continue with the submatrices of order 2, until we found a determinant that was different from 0 and in that case, the range of the matrix would be 2.
text: Its determinant is equal to zero.
text: $\vert \Sigma \vert$ captures their product, because by definition the determinant of a matrix is equal to the product of its eigenvalues.
text: (This result is painfully obvious since the determinant is just equal to the one number of $A$, but we discuss the one-dimensional case just to give intuition on higher dimensions.)
text: Briefly, the determinant of matrix A is equal to the absolute value of matrix A, which in this case is a.
text: Since the determinants of a matrix and its transpose are equal, you can use column operations to simplify a determinant if you wish: performing a column operation on a matrix has the same effect as performing the corresponding row operation on its transpose.
text: The determinant of the product of matrices is equal to the product of determinants of those matrices, so it may be beneficial to decompose a matrix into simpler matrices, calculate the individual determinants, then multiply the results.
text: b)Will I be wrong to assume that, in a case when determinant is equal to zero,there are infinitely many solutions
text: If, we have any matrix with two identical rows or columns then its determinant is equal to zero.
text: So why is the determinant of Q equal one or minus one?