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I have matrix

$$A = \left( \begin{array}{ccc} -1 & 1 & 1 \\ 2 & 3 & -1 \\ -2 & 3 & 3 \end{array} \right)$$

and matrix

$$B = \left( \begin{array}{ccc} -1 & 2 & -1 \\ 1 & 2 & -2 \\ 2 & 1 & -1 \end{array} \right)$$

And I want find a matrix $X$, for which the equation

$$ XA-B=2X$$ holds, but I don't know how to do it.

I tried

Solve[XmatA - matB - 2 X = 0],

but it gives:

Set::write: "Tag Plus in -2\ X+XmatA+{{1,-2,1},{-1,-2,2},{-2,-1,1}} is Protected."

Solve::naqs: 0 is not a quantified system of equations and inequalities.
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closed as off topic by Fabian, Hagen von Eitzen, Cameron Buie, Nameless, Davide Giraudo Jan 1 '13 at 18:52

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2  
This question is more relevant for Mathematica SE. –  Fabian Jan 1 '13 at 16:33

1 Answer 1

up vote 4 down vote accepted

It says right there - 0 is not a system of equations. It's looking at 0 as the real number, not as the zero matrix. So you need to find a way to reference the matrix $$M = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ and see if there are any other problems.

However, to solve this by hand, you write the equation as $$XA - 2X = B$$ and since $$2X = 2IX = X(2I)$$you get that $$X(A-2I) = B$$ and if you (can) find $(A-2I)^{-1}$ (i.e. if it exists Spoiler alert: it does), then you may multiply by this inverse to get $$X = B(A-2I)^{-1}$$

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1  
Problem is of different nature. In Mathematica = is used for assignement and == for equality check. –  Fabian Jan 1 '13 at 16:35
    
Oh okay that would make sense –  andybenji Jan 1 '13 at 16:40

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