Hw 3 problem 2 let a be an mtimes n matrix u an ntimes 1...

Name: hw 3 problem 2 let a be an mtimes n matrix u an ntimes 1 matrix and b an mtimes 1 matrix over f such that aub show that the set of all solutions to the matrix equation axb in x is precisely 07089
Uploaded: 2022-07-04T23:31:08-08:00
Duration: 13 min 48 s
Channel: Oswaldo Jiménez
Description: hw 3 problem 2 let a be an mtimes n matrix u an ntimes 1 matrix and b an mtimes 1 matrix over f such that aub show that the set of all solutions to the matrix equation axb in x is precisely 07089

HW 3 Problem 2. Let A be an m imes n matrix, u an n imes 1 matrix, and b an m imes 1 matrix over F such that Au=b. Show that the set of all solutions to the matrix equation Ax=b in x is precisely u+ker(A):={u+v:vinker(A)} HW 3 Problem 2. Let A be an m x n matrix : u an n x 1 matrix, and . b an m x 1 matrix over F such that Au= b. Show that the set of all solutions to the matrix equation Ax = b in is precisely u+ker(A):={u+v:v E ker(A)}

Transcript

00:01 You're going to show that the following statement is false.

00:05 If the matrix equation a x equal b has solutions, then the equation a transpose x equal c must also have solutions.

00:17 A can be any matrix, a t is transpose, and b and c are any vector of compatible size.

00:26 So you should produce a matrix a such that a x equal b has solutions, that a transpose x equals c does not.

00:38 And because we want to prove that for and not indicated size of matrices we what we should always do is try to do the example in two by two or using two by two matrices.

00:55 That's a common way of doing it.

00:58 And because we are going to show that the statement is false we got to do a counter example of this implication here, that is, we want to produce a matrix a such that ax equal b has solutions, but a transpose x equal c does not.

01:19 Okay, so the first thing we get noticed here is that if a were a square and invariable matrix, then the statement is true.

01:30 That is, if we have the matrix equation, a x equal b we have the other matrix equation a transpose of x equals c if a is invariable then the the solution x will be a inverse times v but we know that the determinant of the matrix a is equal to the determinant of its transpose and because of that if this is true that is a inverse exist then the determinant of a is different from zero, and from this equality, this also will be different from zero.

02:15 And so this system also have a solution.

02:23 What this means is that the only way to give this example here is to choose a matrix that is not invertible, because in the case of an invertible matrix, as we saw here, the system ax equal b, having a solution, in this case a unique solution, implies that the other system will also have an unique solution.

02:49 So our matrix must be a non -invertible matrix.

02:57 And so we are going to select, for example, this matrix here 2x3 -4 -2.

03:09 The determinant of a is 0, so a is not inverteable...

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