How Do You Solve for X in a Matrix Equation?

[Cpt Qwark]

Homework Statement

Given the matrices A, B, C, D, X are invertible such that
(AX+BD)C=CA
Find an expression for X.

Homework Equations

N/A
Answer is $$A^{-1}CAC^{-1}-A^{-1}BD$$

The Attempt at a Solution

I know you can't do normal algebra for matrices.
So this means A≠(AX+BD)?

 

[HallsofIvy]

Your last question puzzles me $A\ne AX+ BD$, unless X= 1 and B or D= 0, for numbers, much less matrices! (Oh, I see- no, you cannot just "cancel" C.)

You can do "normal algebra" for matrices as long as you remember that matrix multiplication is not commutative, that some matrices do not have multiplicative inverses, and we say "multiply by A^-1" not "divide by A". Here, we are told that every matrix is invertible.

From (AX+ BD)C= CA, we "unpeel" X just as we would for numbers. The quantity on the left of the equation is multiplied by, on the right, by C. So start by multiplying both sides of the equation, on the right, by C^-1. Continue from there.

 

[SteamKing]

Homework Statement

Given the matrices A, B, C, D, X are invertible such that
(AX+BD)C=CA
Find an expression for X.

Homework Equations

N/A
Answer is $$A^{-1}CAC^{-1}-A^{-1}BD$$

The Attempt at a Solution

I know you can't do normal algebra for matrices.
So this means A≠(AX+BD)?

Instead of worrying about whether A = (AX + BD), why don't you use the rules of matrix algebra (which you know, I assume) to find X?

Start off by expanding the original matrix equation.