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
- $AX - 4X = B $
- $X(A - 4) = B$
- $X = B(A-4)^{-1}$
And the second example. $(A+4X)^{-1} = B$
- $A^{-1} + \frac{1}{4}X^{-1} = B$
- $\frac{1}{4}X^{-1} = B - A^{-1}$
- $X^{-1} = 4(B - A^{-1})$
- $X = (4(B - A^{-1}))^{-1}$
I didn't find a lot of information about that, so maybe someone can share any links? Or just help me here pls.
$\endgroup$ 22 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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.
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