TITLE: Matrix Equation with factoring
QUESTION [1 upvotes]: I've been working through some practice problems for my upcoming linear algebra exam and have a matrix equation that has me questioning properties of matrix equations.
Problem:
Given the two matrices 
A = \begin{bmatrix}1&2\\-1&0\end{bmatrix} 
and
B = \begin{bmatrix}2&0\\3&4\end{bmatrix}, 
find the solution X to the matrix equation 3(A - B + X) = AX + B.
My Attempt:
Carry out the A - B to create a new matrix C, then we are left with 
3(C + X) = AX + B.
Distribute the 3 so now we have 
3C + 3X = AX + B
Swap the terms to the correct sides
3X - AX = B - 3C
Pull out the X
(A + 3)X = B - 3C
Then this is where things get fuzzy. My gut instinct is to do
X = $(A+3)^-1$(B - 3C)
but how would that work with the 3? Any help would be appreciated.

REPLY [0 votes]: The number 3 can be represented as the matrix with 3s down the middle. However be careful, there is a mistake in your algebra, when you swapped terms around, AX should have become negative.
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