SOLUTION: Please help me
given that{{{ A }}}is non-degenerate and {{{B}}} is multiplication commutable with{{{ A}}} prove that {{{B }}}and {{{A^-1 }}} also multiplication commutable
Algebra ->
College
-> Linear Algebra
-> SOLUTION: Please help me
given that{{{ A }}}is non-degenerate and {{{B}}} is multiplication commutable with{{{ A}}} prove that {{{B }}}and {{{A^-1 }}} also multiplication commutable
Log On
You can put this solution on YOUR website!
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
------------------------------------------------------------------
Here's one way to do that
A*B = B*A Start with fact #2
A^(-1)*A*B = A^(-1)*B*A Left-Multiply both sides by A^(-1). This step is possible because of fact #1
I*B = A^(-1)*B*A
B = A^(-1)*B*A
B*A^(-1) = A^(-1)*B*A*A^(-1)Right-Multiply both sides by A^(-1)
B*A^(-1) = A^(-1)*B*I
B*A^(-1) = A^(-1)*B
So that shows A^(-1) and B are multiplication commutable
------------------------------------------------------------------
Here's another way
A*B = B*A
A*B*A^(-1) = B*A*A^(-1) Right-Multiply both sides by A^(-1)
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)Left-Multiply both sides by A^(-1)
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.
(