Best Answer

For small matrices the simplest way is to show that its determinant is not zero.

User Avatar

Wiki User

โˆ™ 2015-12-28 20:13:56
This answer is:
User Avatar
Study guides


20 cards

A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

See all cards
1793 Reviews

Add your answer:

Earn +20 pts
Q: How do you show a matrix is invertible?
Write your answer...
Still have questions?
magnify glass
Related questions

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

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.

What is the definition of hills matrix?

A Hills matrix is an encryption tool. For a language comprising 26 characters, you would use an invertible n*n matrix where each element is modulo 26. That is a Hill matrix, H.To encode a plain text message,encode the plain-text message using a = 0, b = 1, ..., z = 25.chop the message into strings of n characters.premultiply the column vector so formed by H.convert the resulting n*1 matrix to modulo 26.that is the encrypted decrypt, premultiply the encrypted version by H^-1, modulo 26 .The matrix H, is selected from a set of n*n matrices which are invertible and, for a 26-letter language, the determinant should not be divisible by a factor of 26 (2 or 13).

Is a to power of 4 multiplied by b to power of 5 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.

People also asked