An invertible animals is any animal that is born and lived without having a vertebra also known as a back bone .
an animal lacking a backbone
ask your self
subtropical animals are animals that live in the subtropics. subtropical animals are animals that live in the subtropics.
Invertebrate animals are animals with out Backbones.
There are a variety of different lifestyle's of different animals like some animals are domestic or house animals and some animals are live stock animals some animals are wild as well.
Aborel animals are animals that lives in trees.
What is "a 3b"? Is it a3b? or a+3b? 3ab? I think "a3b" is the following: A is an invertible matrix as is B, we also have that the matrices AB, A2B, A3B and A4B are all invertible, prove A5B is invertible. The problem is the sum of invertible matrices may not be invertible. Consider using the characteristic poly?
Assume that f:S->T is invertible with inverse g:T->S, then by definition of invertible mappings f*g=i(S) and g*f=i(T), which defines f as the inverse of g. So g is invertible.
Indeterminate. How? a4b5 is the expression instead of equality. Since we are not given the equality of two variables, there is no way to determine whether it's invertible or not. Otherwise, if you are referring "a" and "b" as invertible matrices, then yes it's invertible. This all depends on the details.
Invertible counterpoint The contrapuntal design of two or more voices in a polyphonic texture so that any of them may serve as an upper voice or as the bass. Invertible counterpoint involving two (three, four) voices is called double (triple, quadruple) counterpoint. http://www.answers.com/topic/invertible-counterpoint-music
ask your self
The assertion is true. Let A be an idempotent matrix. Then we have A.A=A. Since A is invertible, multiplying A-1 to both sides of the equality, we get A = I. Q. E. D
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.
Noninvertible debenture
It is not possible to show that since it is not necessarily true.There is absolutely nothing in the information that is given in the question which implies that AB is not invertible.
For small matrices the simplest way is to show that its determinant is not zero.
Theorem 5.12- A rotation is a composition of two reflections, and hence is an invertible isometry.
It is an invertible function.