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Problem:enter image description here

In this problem, I multiplied the left side of the AX=B equation by the inverse of A, but what I am confused about is why then I must multiply B by the inverse of A in this order: (inverse of A) * (B). When I do (B) * (inverse of A), it is wrong, but why?

54284User
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  • Remember that matrix multiplication is not commutative. If you have $AX=B$. Then it follows that by left-multiplying both sides of the expression by $C$ we have $C\color{blue}{(AX)}=C\color{blue}{(B)}$ since the parenthetical expressions in blue are equal. Using $A^{-1}$ (assuming it exists) in place of $C$ we have then $AX=B\implies A^{-1}(AX)=A^{-1}(B)\implies X=A^{-1}B$ which is potentially unequal to $BA^{-1}$ – JMoravitz Jul 04 '18 at 21:23
  • @54284User Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Aug 06 '18 at 21:49

2 Answers2

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There's no need to multiply by anything at all. An $m\times n$ matrix multiplied by an $n\times k$ matrix will be an $m\times k$ matrix. We cannot multiply an $m\times n$ matrix by a $p\times k$ matrix unless $n=p.$ Use this to find the size of $X,$ instead.

Cameron Buie
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We have that

$$AX=B$$

and by left multiplication by $A^{-1}$ and associativity we obtain

$$A^{-1}(AX)=A^{-1}B\implies (A^{-1}A)X=A^{-1}B\implies X=A^{-1}B$$

Recall also that matrix multiplication is not commutative and moreover in that case the multiplication $BA^{-1}$ is not well defined since the dimensions of the two matrices, for a multiplication in that order, are not compatible.

José Carlos Santos
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user
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  • What happens then if I have the problem BAC=D and I need to solve for BC. On what side of D would I multiply the inverse of A? – 54284User Jul 04 '18 at 21:32
  • @54284User What are the dimensions involved? – user Jul 04 '18 at 21:33
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    @54284User there is not enough information to get $BC$ by itself with that. Assuming each matrix involved is invertible, you could learn from $BAC=D$ that $C=A^{-1}B^{-1}D$, and so $BC=BA^{-1}B^{-1}D$, but you cannot from that information alone get $BC$ written solely in terms of $A$ and $D$ – JMoravitz Jul 04 '18 at 21:35
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    @54284User The only legal locations that you can multiply matrices is on the far left as is the case for $A=B\implies CA=CB$, or on the far right as is the case for $A=B\implies AC=BC$. Multiplying the middle of an expression by $A^{-1}$ as you seem to be trying to do is simply not allowed and does not make sense to do. – JMoravitz Jul 04 '18 at 21:40