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The problem is given below:
In the following $X, A, B$ and $C$ are all $n \times n$ matrices. Solve for $X$ and account for any assumptions made for each of the following.
(a) $AX + B = C$
(b) $AXA - B = C$
(c) $AX + B = X$
(d) $A^2 X + BX + C = A$
I solved (a) by $$ X = A^{-1} (C-B). $$
I'm confused how to solve the others, can someone explain the proces?
For completeness, let me spell out what I've said in the comments.
For part a), we have $AX + B = C$, so to solve for $X$, we move $B$ over: $AX=C-B$, and then multiply by the inverse of $A$, i.e. $A^{-1}AX = A^{-1}(C-B) \Rightarrow X = A^{-1}(C-B)$. To do this, we implicitly assume $A^{-1}$ is a well-defined object -- that is, that $A$ is invertible.
In part b), we use the same property. $AXA-B=C \Rightarrow AXA = C+B \Rightarrow X = A^{-1}(C+B)A^{-1}$. Again, for this to make sense, we need $A$ is invertible.
For part c), we do the same procedure, more or less. Isolate $X$: $AX+B=X \Rightarrow B = X-AX$. Now we factor out $X$: $B = X-AX \Rightarrow B = (Id-A)X$.
The $Id$ appears in place of a $1$ as you would do in polynomial factoring -- for example, $x-ax$ as polynomials, we get $(1-a)x$. In a matrix space, the identity matrix $Id$ is the $1$. Further, we must also keep $X$ on the left, because in general, $AX \neq XA$ for any matrices $X$ and $A$. This is another way in which matrix spaces differ from polynomials, with which you might be more familiar.
So then $B = (Id-A)X \Rightarrow (Id-A)^{-1}B = X$. So we use the fact that $Id-A$ is invertible so that $(Id-A)^{-1}$ exists.
Part d) is the same. Isolate everything with $X$ on one side, factor out $X$ (being sure to respect the non-commutativity of the matrix multiplication), and multiply by the inverse of the coefficient on $X$. You must use that the coefficient on $X$ is invertible.
You should also account for assumptions you have made. What assumptions have you made in solving for $X$ in part (a)?
Regarding the other questions, follow the same process. Recall, however, that matrix multiplication is non-commutative in general. That is to say, $AX \neq XA$. This should be of use to you in part (b). Part (c) and (d) make use of the same assumptions and properties.
