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Find a matrix A that is not invertible and b such that Ax equals b has a unique solution?
A matrix with a row or a column of zeros cannot have an inverse.Proof:Let A denote a matrix which has an entire row or column of zeros. If B is any matrix, then AB has an entire rows of zeros, or BA has an entire column of zeros. Thus, neither AB nor BA can be the identity matrix, so A cannot have an inverse, or A cannot be invertible.Since A is not invertible, then Ax = b has not a unique solution.
Which of the smartArt graphic type show relationships of parts to a whole?
matrix
What is the relation in which each element in the domain is mapped to exactly one element the range?
It is an invertible function.
What type of matrix is a vector?
Vector matrix has both size and direction. There are different types of matrix namely the scalar matrix, the symmetric matrix, the square matrix and the column matrix.
Is matrix polynomial and polynomial matrix same?
No. A matrix polynomial is an algebraic expression in which the variable is a matrix. A polynomial matrix is a matrix in which each element is a polynomial.
Let A be a 6by4 matrix and B a 4by6 matrix show that the 6by6 matrix AB can not be invertible?
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.
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 an invertible idempotent matrix the identity matrix?
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
Does every square matrix have a determinant?
Yes, every square matrix has a determinant. The determinant is a scalar value that can be computed from the elements of the matrix and provides important information about the matrix, such as whether it is invertible. For an ( n \times n ) matrix, the determinant can be calculated using various methods, including cofactor expansion or row reduction. However, the determinant may be zero, indicating that the matrix is singular and not invertible.
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 ).
How can you tell if a matrix is invertible?
An easy exclusion criterion is a matrix that is not nxn. Only a square matrices are invertible (have an inverse). For the matrix to be invertible, the vectors (as columns) must be linearly independent. In other words, you have to check that for an nxn matrix given by {v1 v2 v3 ••• vn} with n vectors with n components, there are not constants (a, b, c, etc) not all zero such that av1 + bv2 + cv3 + ••• + kvn = 0 (meaning only the trivial solution of a=b=c=k=0 works).So all you're doing is making sure that the vectors of your matrix are linearly independent. The matrix is invertible if and only if the vectors are linearly independent. Making sure the only solution is the trivial case can be quite involved, and you don't want to do this for large matrices. Therefore, an alternative method is to just make sure the determinant is not 0. Remember that the vectors of a matrix "A" are linearly independent if and only if detA�0, and by the same token, a matrix "A" is invertible if and only if detA�0.
Find a matrix A that is not invertible and b such that Ax equals b has a unique solution?
A matrix with a row or a column of zeros cannot have an inverse.Proof:Let A denote a matrix which has an entire row or column of zeros. If B is any matrix, then AB has an entire rows of zeros, or BA has an entire column of zeros. Thus, neither AB nor BA can be the identity matrix, so A cannot have an inverse, or A cannot be invertible.Since A is not invertible, then Ax = b has not a unique solution.
How do you figure out whether a matrix has a determinant?
Any n x n (square) matrix have a determinate. If it's not a square matrix, we don't have a determinate, or rather we don't care about the determinate since it can't be invertible.
Is a singular matrix an indempotent matrix?
A singular matrix is a matrix that is not invertible. If a matrix is not invertible, then:• The determinant of the matrix is 0.• Any matrix multiplied by that matrix doesn't give the identity matrix.There are a lot of examples in which a singular matrix is an idempotent matrix. For instance:M =[1 1][0 0]Take the product of two M's to get the same M, the given!M x M = MSo yes, SOME singular matrices are idempotent matrices! How? Let's take a 2 by 2 identity matrix for instance.I =[1 0][0 1]I x I = I obviously.Then, that nonsingular matrix is also idempotent!Hope this helps!
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.
Prove that the inverse of an invertible mapping is invertible?
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.
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.
