Linear algebra problem, can anyone help?

Given a system of equations Ax = b, with det(A) = 0, show that there are no solutions unless TRyb = 0, where y is any nontrivial solution of the system TRAy = 0

Since superscript is not available I've used TRA to mean the transpose of matrix A, similarly for TRy.

I barely passed Linear Algebra. 😕

First of all, Ax = b is not a system of equations, but just one equation.
I assume, you are talking about matrices and vectors over a field and not about modules over a ring ( "det A != 0" should be "det A invertible" ).

Ok. Let us have the equation A x = b (det A 0 or not), and lets have an y as solution of ATy = 0 and with yTb != 0 . ( I prefer a postfix "T" as mark for the transposition )

If there was a solution to
Ax = b then multipliate both sides with yT from the left:
yTAx = yTb
(ATy)Tx = yTb
the left side is 0, the right side is != 0, which is a contradiction, so there is no
solution to Ax = b. qed

I did not use det A = 0 anywhere!
But if there is a nontrivial solution of ATy = 0, then det A must be zero.
So det A = 0 is no precondition, but a concolusion.
Your teacher did a bad job when building this homework.
By the way, you shoud do your homework at home and not just pasting it here

Originally posted by afx
First of all, Ax = b is not a system of equations, but just one equation.
I assume, you are talking about matrices and vectors over a field and not about modules over a ring ( "det A != 0" should be "det A invertible" ).

Ok. Let us have the equation A x = b (det A 0 or not), and lets have an y as solution of ATy = 0 and with yTb != 0 . ( I prefer a postfix " ...[text shortened]... his homework.
By the way, you shoud do your homework at home and not just pasting it here

Originally posted by afx

Thanks for your solution. There are some errors that you have made, here are a few:

First of all, Ax = b is not a system of equations, but just one equation.
This is false. A, b and x represent a square matrix, a column vector and a column vector respectively. In the context of the question this represents a system of equations in x1, x2, ...xn.

By the way, you shoud do your homework at home and not just pasting it here
It isn't my homework and I am not a student, so this rather sanctimonious statement is based on a false premise.

But if there is a nontrivial solution of ATy = 0, then det A must be zero.
So det A = 0 is no precondition, but a concolusion.
The question postulates the existence of such a solution, so I think saying the determinant is zero is just like saying x != 0 when considering y = 1/x. It is just tidy.

Originally posted by afx
ATy = 0

Lies! 😠

ATy = 0 => (ATy)T = 0T => yTA = 0T => (yTA)x = 0Tx => yT(Ax) = 0 (the scalar 0) => yTb = 0

Originally posted by Lord Shark
ATy = 0 => (ATy)T = 0T => yTA = 0T => (yTA)x = 0Tx => yT(Ax) = 0 (the scalar 0) => yTb = 0

i think he was joking

Originally posted by randolph
i think he was joking

yeah since ATY is an abbreviation for his name i think he "took offense" to being called a zero! 🙂

Originally posted by Aetherael
yeah since ATY is an abbreviation for his name i think he "took offense" to being called a zero! 🙂

🙂

Sorry, missed that. Linear algebra caused a temporary sense of humour bypass.

Originally posted by Lord Shark
The question postulates the existence of such a solution, so I think saying the determinant is zero is just like saying x != 0 when considering y = 1/x. It is just tidy.

Why do you say that?

I thought saying the det(A)=0 is because, if not, the solution of the system would be in the form x = A^(-1)b and would be unique. Since rows of A will be linearly independent (when det(A)~=0), it must be unique.

However, since some rows of A are linearly dependent, there will be no solution unless it shares with b some properties (like the one described above). If it has such properties, this will imply then that there are infinite solutions.

As afx shows by contradiction, the property that (A'y = 0 & y'b = 0) is a necessary one.

Originally posted by Palynka
Why do you say that?

I thought saying the det(A)=0 is because, if not, the solution of the system would be in the form x = A^(-1)b and would be unique. Since rows of A will be linearly independent (when det(A)~=0), it must be unique.

However, since some rows of A are linearly dependent, there will be no solution unless it shares with b some properties ...[text shortened]... s.

As afx shows by contradiction, the property that (A'y = 0 & y'b = 0) is a necessary one.

Why do you say that?

I thought saying the det(A)=0 is because, if not, the solution of the system would be in the form x = A^(-1)b and would be unique.
Yes that's right. There is a unique solution if and only if det(A) is not equal to zero. With det(A) = 0 there are either no solutions and the equations are inconsistent, or there are infinitely many solutions, giving a complete row of zeros on reduction of the augmented matrix.
So I think det(A) is a precondition for that part of the question, not a conclusion, just as x not = 0 is a precondition for y = 1/x to be defined on its domain. Hence I disagree with afx in that regard, although I could have put it better.

Originally posted by Lord Shark
y = 1/x

I still don't understand where this is coming from...

Originally posted by Palynka
I still don't understand where this is coming from...

Ok, maybe it was a bad analogy, sorry. (They are my specialty 🙂)

Originally posted by Lord Shark
Ok, maybe it was a bad analogy, sorry. (They are my specialty 🙂)

Oops, I thought they were the same x and y of the problem, hence my confusion. 🙂

Originally posted by Palynka
Oops, I thought they were the same x and y of the problem, hence my confusion. 🙂

I should have made it more clear that they weren't, sorry.