Please check whether the following is correct:

For $A \in M_{n\times m}$

1) If $n \neq m$, $A$ is not invertible

2)If $n = m$,

2-1) If maximal number of independent row or column = n(or =m) it's invertible

2-1-1) sequential application of "only one kind of elementary operation" - row or column would be allowed to check the indpendency among rows and columns.

2-1-2) else, not invertible

2-2) $Det(A) \neq 0$ it's invertible

2-2-1) else, it's not invertible.

Is this reasoning correct?

  • $\begingroup$ I don't understand what is the "reasoning" $\endgroup$ – x4rkz Jun 5 '17 at 11:28
  • $\begingroup$ If I read this as a programming code with if-else statements, I would say: No it's not, because if you have n independent rows (2-1) you'll never reach and (2-1-2) nor at (2-2)&(2-2-1), the latter being essentially the same as (2-1) and (2-1-1)... $\endgroup$ – draks ... Jun 7 '17 at 5:15

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