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I am new to matrix multiplication and trying to understand something. Suppose you have a matrix equation $ A x=b $. I know to solve for $x$ you should left multiply by the inverse of A. But what is the reason you can't solve for $b$ like this: $ A A^{-1} x=b A^{-1} $ so that $ x=b A^{-1} $?

I tried with an example to see that it doesn't work but I don't have a good understanding of the mechanics why not.

What if you had something like this instead:

$ A x = By $

could you solve like this?

$ AA^{-1} x = BA^{-1}y $

and then get the solution $ x = BA^{-1}y $

or do you have to solve like this

$ A^{-1}A x = A^{-1}By $

to get the solution

$ x = A^{-1}By $

Anna
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    $bA^{-1}$ does not make sense - nor it does $(Ax)A^{-1}$ - because $b$ is a $(n\times 1)$ matrix, while $A^{-1}$ is a $(n\times n)$ matrix. Matrix product is not commutative. –  Sep 08 '15 at 09:44
  • OK yes, I was forgetting to think about the dimensions! Mostly I was just trying to understand why multiplication "in the middle" didn't really make sense (even if dimensions was well-defined) but thanks to your answer plus the answer below it is more clear to me now – Anna Sep 08 '15 at 10:54

3 Answers3

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You are asking why $AB=A'B'$ does not imply $AEB = A'EB'$. You noticed that the thing is false. So you should not ask yourself "why is false" (which is nonsense) but you should ask yourself "why I thought it could be true?".

You know that applying the same operation to the same object will held the same result. So if $A=B$ you can right multiply and obtain $AC=BC$ or you can left multiply and obtain $CA=CB$. If multiplication were commutative you could also multiply "in the middle" since the order of factors is not relevant. But when the operation is not commutative this is not admissible. Adding a factor in the middle of a product as in $AB \to AEB$ is not an operation performed on the object $AB$ itself, so there is no reason to think that $AB=A'B'$ would imply $AEB \neq A'EB'$.

Emanuele Paolini
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Emanuele and user228113 have already given very good answers. If I may add, the reason why $$Ax=By \nRightarrow AA^{-1}x = BA^{-1}y$$ is same one behind $$f(x) = g(y) \nRightarrow (f \circ f^{-1})(x) = (g \circ f^{-1})(y).$$

Matrix multiplication is defined so that it works right to left, just like function composition. This allows matrices to represent linear transformations more intuitively. It's also why we conventionally represent vectors as column matrices.

William A. Noble
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Your second explanation is the right one. When multiypling with matricies you have to think of multipling from left $A^{-1} \cdot$ or from the right $ \cdot A^{-1}$. In your first calculation you multiplied from the right $Ax\cdot A^{-1} = By \cdot A^{-1}$. This would not solve your problem, as you cant use commutativity on matricies like $AB\neq BA$.

MrYouMath
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