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 November 9th, 2010, 07:34 AM #1 Member   Joined: Apr 2010 Posts: 91 Thanks: 0 Solving for X (inverses) Let $\mathbf{A}(2\mathbf{B} + 3\mathbf{XC})= D$, where $\mathbf{A}, \mathbf{B}, \mathbf{C}$ and $\mathbf{D}$ are invertible n x n matrices. What does X equal to? Attempt: $A(2B + 3XC)= D$ $A2B + A3XC= D$ $A^{-1}A3XC= A^{-1}(D - A2B)$ $3XCC^{-1}= A^{-1}(D - A2B)C^{-1}$ $3X= A^{-1}(D - A2B)C^{-1}$ $X= \frac{1}{3}A^{-1}(D - A2B)C^{-1}$ but I dont think this is right.
November 9th, 2010, 01:47 PM   #2
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Re: Solving for X (inverses)

Quote:
 Originally Posted by TsAmE Let $\mathbf{A}(2\mathbf{B} + 3\mathbf{XC})= D$, where $\mathbf{A}, \mathbf{B}, \mathbf{C}$ and $\mathbf{D}$ are invertible n x n matrices. What does X equal to? Attempt: $A(2B + 3XC)= D$ $A2B + A3XC= D$ $A^{-1}A3XC= A^{-1}(D - A2B)$ $3XCC^{-1}= A^{-1}(D - A2B)C^{-1}$ $3X= A^{-1}(D - A2B)C^{-1}$ $X= \frac{1}{3}A^{-1}(D - A2B)C^{-1}$ but I dont think this is right.
Looks good to me.

 November 9th, 2010, 02:42 PM #3 Senior Member   Joined: Oct 2009 Posts: 105 Thanks: 0 Re: Solving for X (inverses) What makes you think it's not correct?
 November 9th, 2010, 11:28 PM #4 Member   Joined: Apr 2010 Posts: 91 Thanks: 0 Re: Solving for X (inverses) It was a multiple choice question, but I am not getting the right form. Here are the options: A) $\frac{1}{3}C^{-1}(A^{-1}D - 2B)$ B) $(\frac{1}{3})^nC^{-1}(A^{-1}D - 2B)$ C) $(\frac{1}{3}A^{-1}D - \frac{1}{2}B^{-1})C^{-1}$ D) $3(A^{-1}D - 2B)C^{-1}$ E) $\frac{1}{3}(A^{-1}D - 2B)C^{-1}$
 November 10th, 2010, 04:07 AM #5 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Solving for X (inverses) What happens when you multiply the $A^{-1}$ into the terms in parentheses on its right hand side?
 November 10th, 2010, 04:21 AM #6 Member   Joined: Apr 2010 Posts: 91 Thanks: 0 Re: Solving for X (inverses) You get the answer E). Thanks

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