Question 1101996
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Given Fact #1: Matrix A is non-degenerate
This means that the inverse of A exists. The inverse is denoted A^(-1)


Given Fact #2: Matrix B is multiplication commutable with matrix A.
In other words, A*B = B*A holds true. The order of multiplication doesn't matter in this specific case (keep in mind that matrix multiplication isn't always commutative)


The goal is to prove that matrix B and the inverse A^(-1) are also multiplication commutable. We need to show the following:
A^(-1)*B = B*A^(-1)
or
B*A^(-1) = A^(-1)*B


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Here's one way to do that


A*B = B*A <font color=blue>Start with fact #2</font>
A^(-1)*A*B = A^(-1)*B*A <font color=blue>Left-Multiply both sides by A^(-1). This step is possible because of fact #1</font>
I*B = A^(-1)*B*A
B = A^(-1)*B*A
B*A^(-1) = A^(-1)*B*A*A^(-1)<font color=blue>Right-Multiply both sides by A^(-1)</font>
B*A^(-1) = A^(-1)*B*I
B*A^(-1) = A^(-1)*B


So that shows A^(-1) and B are multiplication commutable

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Here's another way


A*B = B*A
A*B*A^(-1) = B*A*A^(-1) <font color=blue>Right-Multiply both sides by A^(-1)</font>
A*B*A^(-1) = B*I
A*B*A^(-1) = B
B = A*B*A^(-1)
A^(-1)*B = A^(-1)*A*B*A^(-1)<font color=blue>Left-Multiply both sides by A^(-1)</font>
A^(-1)*B = I*B*A^(-1)
A^(-1)*B = B*A^(-1)


Giving us the same conclusion as before. Keep in mind that if X = Y, then Y = X for any matrices X and Y. 
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