Matrix equations, simplify them.

Question

I have some equations and I don't know am I doing the simplification right. Can someone check it?

For example, we have an equation $AX = 4X + B$, where $A$ and $B$ are matrices. So, what I have done with it is

1. $AX - 4X = B $
2. $X(A - 4) = B$
3. $X = B(A-4)^{-1}$

And the second example. $(A+4X)^{-1} = B$

1. $A^{-1} + \frac{1}{4}X^{-1} = B$
2. $\frac{1}{4}X^{-1} = B - A^{-1}$
3. $X^{-1} = 4(B - A^{-1})$
4. $X = (4(B - A^{-1}))^{-1}$

I didn't find a lot of information about that, so maybe someone can share any links? Or just help me here pls.

Answer 1

Both examples contain some errors. For the first one, you have the right idea subtracting to get $$AX-4X=B$$ however, you have to be careful about factoring matrices. Observe that $$AX-4X=(A-4I)X$$ where $I$ is the identity matrix. Keep in mind that "$4$" does not denote a matrix (unless we are talking about $1\times1$ matrices). To complete this, we have $$X=(A-4I)^{-1}B$$

For the second example, your first step is wrong. In general, whether for real numbers or matrices $a$ and $b$ and integers $n$, it is not true that $$(a+b)^{n}=a^{n}+b^{n}$$

What we can do is the following: $$(A+4X)^{-1}=B$$ $$I=B(A+4X)=BA+4BX$$ $$I-BA=4BX$$ $$\frac{1}{4}(I-BA)=BX$$ $$X=\frac{1}{4}B^{-1}(I-BA)$$

I should mention that with all of the above manipulations, things only make sense if the corresponding matrices are invertible. For example, if $(A-4I)$ is not invertible then we cannot simplify the first example as shown.

Answer 2

Your first example is almost correct. You just change your $4$ to $ 4I$

Your second example needs serious adjustments.

First take inverse of both sides , then try to isolate $X$ as you did in the first example.