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Inverse matrices question. 
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#1
Nov1512, 05:33 PM

P: 818

Hello,
Would it be correct to say that if for every two different vectors x and y, A*x ≠ A*y (where A is a symmetrical matrix), then A is NOT necessarily invertible? In other words, albeit for any two different vectors x and y symmetrical matrix A times one of the vectors is not equal to A times the other, A is not necessarily invertible? 


#2
Nov1512, 05:44 PM

P: 818

Correction, in case A is a square matrix (of order nXn)



#3
Nov1612, 06:38 AM

P: 345

If A is an n x n  matrix such that Ax ≠ Ay for all pairs of distinct nvectors x and y, then A is invertible.



#4
Nov1612, 06:51 AM

P: 818

Inverse matrices question.
Let us examine the following singular matrix:
2 0 4 1 1 3 0 1 1 For any two different vectors I claim that that matrix multiplied by the first vector will never be equal to the multiplication of that same matrix by the second vector. Hence, the matrix does not necessarily have to be singular for the proposition to be valid and hold. Wouldn't you agree? 


#5
Nov1612, 06:58 AM

P: 345




#6
Nov1612, 07:31 AM

P: 818

Okay, so the proposition does not hold in case A is singular, but does that per se guarantee that it holds, for EVERY two different vectors, if A were not singular, i.e. invertible?



#7
Nov1612, 08:18 AM

P: 345

The system Ax=b has a unique solution, x=A^{1}b, if x is invertible. Otherwise, it has either no solution or infinitely many solutions. 


#8
Nov1612, 09:00 AM

P: 818

Thank you very much! :)



#9
Nov1612, 03:57 PM

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P: 7,160

The easy way to see this is false is consider the special case of x ≠ 0 and y = 0.
Ay = 0, so for every x ≠ 0, Ax ≠ 0. If A is singular, there is a vector x ≠ 0 such that Ax = 0, which is a contradiction. Therefore A is nonsingular. 


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