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If $A, B,$ and $C$ are $n \times n$ invertible matrices, does the equation $C^{-1}(A+X) B^{-1}=I_{n}$ have a solution, $X ?$ If so, find it.

$X = C B - A$

Algebra

Chapter 2

Matrix Algebra

Section 2

The Inverse of a Matrix

Introduction to Matrices

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University of Michigan - Ann Arbor

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for this exercise were provided with three matrices A, B and C, and we're told that these matrices are in vertebral and that their size is and buy in. Well, what we're going to do next is consider a matrix equation where the unknown is also a matrix. The equation is see in verse times a plus. X times be in verse equals the end by N identity matrix, and the unknown is X. But here X is going to be considered a matrix. The reason it must be a matrix is because of the addition here. That means X must be a matrix of the same size and buy in now to solve an equation like this. One thing we can do is pre multiply by a matrix C. Then it will multiply the left side of the equation by sea in verse, times a plus X times be in verse and the right side of the equation will be see times the identity matrix. Now the next step here we know that there is a cancellation with sea by sea inverse. It results in the identity matrix, but there is no difference when we multiply here with identity we still get a plus X times be in verse for the left side, on the right hand site we get See by the same logic. Now recall. We're trying to isolate X here, so the next problem is that be inverse matrix. But we post multiply by. Be on the same side of that both sides of the equation we'll have in another elimination. Just as before, this turns into the identity matrix and eight times x times. The identity matrix is still a plus X, so the right hand side is see. Time's be left hand side is a plus X, the less operation we can take to solve for the Matrix A is subtract a from both sides, so we next have that X is equal to see times B minus A, and this is the solution to our matrix equation. Let's indicate what we've shown. We've shown that if this matrix equation has a solution than the solution must be see time's B minus A to really prove without shadow of a doubt that we know that this is the solution. What you need to do is take this solution substituted here and show that we get a true statement that the left hand side would be thy Danny Matrix

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