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What else can I help you with?
Is the invertible to be one to one function?
Yes, a function needs to be one-to-one in order to have an inverse.
What is the range of the inverse of relation (1 7) (2 -4) (5 6) (2 8)?
The original relationship is one-to-many. It is therefore not an invertible relationship.
What is the answer of opposite or inverse?
Two operators are opposites or inverses if their combined mapping is the identity mapping. Less technically, one mapping must reverse the effect of the other. There are problems, though, when dealing with even fairly common functions. Squaring is a function from the real numbers to the non-negative real numbers, but there is not a single inverse operation. [+sqrt and -sqrt are the two inverse functions over the range.]
Why is zero not an invert number?
Zero is not an invertible number because it does not have a multiplicative inverse. An invertible number has a reciprocal that, when multiplied by the original number, results in one. However, multiplying zero by any number always yields zero, so there is no number that can be multiplied by zero to produce one. Thus, zero cannot be inverted.
How do you prove if the determinant of A is not equal to zero then the matrix A is invertible?
To prove that a matrix ( A ) is invertible if its determinant ( \det(A) \neq 0 ), we can use the property of determinants related to linear transformations. If ( \det(A) \neq 0 ), it implies that the linear transformation represented by ( A ) is bijective, meaning it maps ( \mathbb{R}^n ) onto itself without collapsing any dimensions. Consequently, there exists a matrix ( B ) such that ( AB = I ) (the identity matrix), confirming that ( A ) is invertible. Thus, the non-zero determinant serves as a necessary and sufficient condition for the invertibility of the matrix ( A ).
If matrix a is invertible and a b is invertible and a 2b a 3b and a 4b are all invertible how can you prove that a 5b is also invertible?
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?
Is the invertible to be one to one function?
Yes, a function needs to be one-to-one in order to have an inverse.
What are applications of determinants?
If you think of a matrix as a mapping of one vector to another, by either rotation or stretching, then the determinant tells you what size one unit volume is mapped to. This also can tell you if a matrix has an inverse as at least one dimension in a non-invertible matrix will be mapped to zero, making the determinant zero.
Can an invertible function have more than one x-intercept?
No. If the function has more than one x-intercept then there are more than one values of x for which y = 0. This means that, for the inverse function, y = 0 should be mapped onto more than one x values. That is, the inverse function would be many-to-one. But a function cannot be many-to-one. So the "inverse" is not a function. And tat means the original function is not invertible.
What is the range of the inverse of relation (1 7) (2 -4) (5 6) (2 8)?
The original relationship is one-to-many. It is therefore not an invertible relationship.
What is the answer of opposite or inverse?
Two operators are opposites or inverses if their combined mapping is the identity mapping. Less technically, one mapping must reverse the effect of the other. There are problems, though, when dealing with even fairly common functions. Squaring is a function from the real numbers to the non-negative real numbers, but there is not a single inverse operation. [+sqrt and -sqrt are the two inverse functions over the range.]
What is an input or output relation that has exactly one output for each input?
A one-to-one function, a.k.a. an injective function.
Why is zero not an invert number?
Zero is not an invertible number because it does not have a multiplicative inverse. An invertible number has a reciprocal that, when multiplied by the original number, results in one. However, multiplying zero by any number always yields zero, so there is no number that can be multiplied by zero to produce one. Thus, zero cannot be inverted.
How do you prove if the determinant of A is not equal to zero then the matrix A is invertible?
To prove that a matrix ( A ) is invertible if its determinant ( \det(A) \neq 0 ), we can use the property of determinants related to linear transformations. If ( \det(A) \neq 0 ), it implies that the linear transformation represented by ( A ) is bijective, meaning it maps ( \mathbb{R}^n ) onto itself without collapsing any dimensions. Consequently, there exists a matrix ( B ) such that ( AB = I ) (the identity matrix), confirming that ( A ) is invertible. Thus, the non-zero determinant serves as a necessary and sufficient condition for the invertibility of the matrix ( A ).
Is the inverse of the function y x2 still a function?
Y = 1/X2 ==============Can it pass the line test? * * * * * That is not the inverse, but the reciprocal. Not the same thing! The inverse is y = sqrt(x). Onless the range is resticted, the mapping is one-to-many and so not a function.
What is a projection in mathematics?
Generally speaking, in mathematics, a projection is a mapping of a set (or of a mathematical structure) which is idempotent, which means that a projection is equal to its composition with itself. A projection may also refer to a mapping which has a left inverse.
Why order doesn't matter when you find inverse of the matrix specificly?
When finding the inverse of a matrix, order doesn't matter because the operation of taking the inverse is inherently defined for square matrices. Specifically, if ( A ) is an invertible matrix, then its inverse ( A^{-1} ) satisfies the property ( A A^{-1} = I ) and ( A^{-1} A = I ), where ( I ) is the identity matrix. This means that multiplying ( A ) by its inverse will always yield the identity matrix, regardless of the order in which the matrices are multiplied. However, note that the order does matter when multiplying different matrices together; it's only the specific case of a matrix and its inverse that ensures commutativity in this regard.
