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I have 3 known matrices A B and C and wish to find X. My question here is why AXB = C ⇔ XB = A^-1 *C and not AXB = C ⇔ XB = C*A^-1

The A^-1 matrix is of size 3x3 and C is 3x2 is it the size that decided to multiply A^-1 by C and not the other way around?

Kind regards

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2 Answers

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Like any other kind of equation, an equation of matrices have two basic permitted manipulations:

1. Swap some term for something else with exactly the same value
2. Do something to the left-hand side, then do exactly the same thing to the right-hand side.

In this case it's a manipulation of type 2 that we're interested in, and the operation what we want to do is "multiply by $A^{-1}$. However, unlike normal equations of numbers and functions, it's not enough to specify that we multiply by $A^{-1}$, we have to say from which direction we're multiplying by $A^{-1}$. In this case, in order to exploit manipulations of type 1 to make $A$ eventually disapepar from the left-hand side, $A$ and $A^{-1}$ have to be adjacent. To make that happen, we must have $A^{-1}$ enter by multiplication from the left. Thus: $$ AXB = C\\ A^{-1}\times (AXB) = A^{-1}\times C\\ (A^{-1}A) (XB) = A^{-1}C\\ I(XB) = A^{-1}C\\ XB = A^{-1}C $$ where the operations we do from line to line are the following (in order):

• Multiply from the left by $A^{-1}$ (type 2)
• Rearrange parenthesises (type 1, justified because matrix multiplication is associative)
• Swap out $A^{-1}A$ with $I$ (type 1, by definition of inverse)
• Remove $I$ from the product (type 1, by general property of the identity matrix)

In this case, an expression like $CA^{-1}$ wouldn't make sense since $C$ is $3\times 2$ and $A^{-1}$ is $3\times 3$: the dimensions don't match up. But it is important to note that this is not what makes $AXB = C \implies XB = CA^{-1}$ wrong.

There is an easy way to remember which way the dimensions of a product of matrices $AB$ should match up, and it's the following: If $A$ is $a_1\times a_2$ and $B$ is $b_1 \times b_2$, then write it up like this: $$ (\color{blue}{a_1}\times \color{red}{a_2)(b_1} \times \color{blue}{b_2}) $$ and the dimensions match up iff $\color{red}{a_2 = b_1}$. Also, the resulting product matrix $AB$ has dimensions $\color{blue}{a_1\times b_2}$.

This generalises easily to products of three or more matrices. Just write out the dimensions of every one of them in order; the product makes sense if all adjacent dimensions are equal (the second dimensino of one matrix to the first dimension of the next), and the resulting product matrix has dimensions equal to the first dimension of the first matrix by the last dimension of the last matrix.

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When you have $AXB=C$ and multiply with $A^{-1}$ from the left you get $$ A^{-1}AXB = A^{-1}C $$ where the left-hand side simplifies to $XB$.

If you multiply with $A^{-1}$ from the right (assuming that the dimensions even match) you get $$ AXBA^{-1} = CA^{-1} $$ which doesn't simplify anywhere.

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