The invertible matrix equation tells us that the following statements are equivalent, for any square matrix A:(adsbygoogle = window.adsbygoogle || []).push({});

1) A is invertible

2) Ax=0 has only the trivial solution

3) Ax=b has a unique solution for any column vector b

My question:

Suppose you know that A is a singular matrix. Then can you conclude that for every column vector b, either Ax=b has no solution or Ax=b has infinitely many solutions? Or is there a vector b for which Ax=b has a unique solution, despite the fact that A is singular?

Also, given that A is singular, how can you tell whether Ax=b has no solutions or has infinitely many solutions?

BiP

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# The nonhomogenous system Ax=b

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