0
More precisely, I think you're asking whether the set of n X n matrices forms an abelian group under multiplication. The answer is no (assuming n>1). For example
(1 0)(0 1) = (0 1)
(0 0)(0 0) (0 0),
but
(0 1)(1 0) = (0 0)
(0 0)(0 0) (0 0). However, the set of n x n diagonalmatrices does form an Abelian set. This is true regardless of the direction of the diagonality, right-to-left or left-to-right. Note that the resulting matrix will also be diagonal, but always right-to-left.
Wiki User
∙ 14y ago
Add your answer:
Earn +20 pts
Is the set of irrational numbers a group under the operation of multiplication?
No. It is not even closed. sqrt(3)*sqrt(3) = 3 - which is rational.
Is w-10 closed under multiplication?
No.
Is set of integers is a group under multiplication?
No. One of the group axioms is that each element must have an inverse element. This is not the case with integers. In other words, you can't solve an equation like: 5 times "n" = 1 in the set of integers.
What is closed under multiplication means?
A set is closed under multiplication if for any two elements, x and y, in the set, their product, x*y, is also a member of the set.
Is the set of rational numbers closed under multiplication?
yes
Example of group is an abelian group?
The set of integers, under addition.
What does S R followed by a number mean?
If you mean what does something like SL(3, R) mean, it is the group of all 3X3 matrices with determinant 1, with real entries, under matrix multiplication.
How are the inverse matrix and identity matrix related?
If an identity matrix is the answer to a problem under matrix multiplication, then each of the two matrices is an inverse matrix of the other.
Is the group of all real numbers except 0 under multiplication is an infinite group?
s
Does the set of even integers form a group under the operation of multiplication?
No. The inverses do not belong to the group.
Is the set of all integers a group under multiplication?
No. The set does not include inverses.
Is the set of all 2x2 invertible matrices a subspace of all 2x2 matrices?
I assume since you're asking if 2x2 invertible matrices are a "subspace" that you are considering the set of all 2x2 matrices as a vector space (which it certainly is). In order for the set of 2x2 invertible matrices to be a subspace of the set of all 2x2 matrices, it must be closed under addition and scalar multiplication. A 2x2 matrix is invertible if and only if its determinant is nonzero. When multiplied by a scalar (let's call it c), the determinant of a 2x2 matrix will be multiplied by c^2 since the determinant is linear in each row (two rows -> two factors of c). If the determinant was nonzero to begin with c^2 times the determinant will be nonzero, so an invertible matrix multiplied by a scalar will remain invertible. Therefore the set of all 2x2 invertible matrices is closed under scalar multiplication. However, this set is not closed under addition. Consider the matrices {[1 0], [0 1]} and {[-1 0], [0 -1]}. Both are invertible (in this case, they are both their own inverses). However, their sum is {[0 0], [0 0]}, which is not invertible because its determinant is 0. In conclusion, the set of invertible 2x2 matrices is not a subspace of the set of all 2x2 matrices because it is not closed under addition.
Why are the rational numbers under the operation of multiplication not a group?
I believe it is because 0 does not have an inverse element.
Is the set of irrational numbers a group under the operation of multiplication?
No. It is not even closed. sqrt(3)*sqrt(3) = 3 - which is rational.
Is the set of composite number closed under multiplication?
Is { 0, 20 } closed under multiplication
Is 018 closed under multiplication?
No.
What property does the equation 19 plus 11 equals 11 plus 19 settle in?
The Abelian (commutative) property of integers under addition.
