- 17 Jan '10 19:33First 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 - 17 Jan '10 21:15
*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. - 18 Jan '10 10:17

Why do you say that?*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.**

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. - 18 Jan '10 11:57
*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.