# Matrix equation - solve for X

#### mpdancow

Hi everybody! I need some help with this matrix equation:

$$\pmatrix{6 & 1 \cr -3 & -8} -X \pmatrix{1 & 6 \cr -1 & 5}=I$$

So this is how I would solve it:

$$A-X\cdot B=I \Leftrightarrow -X=I\cdot B^{-1}-A$$

Is this correct? Also, is -X different from X when it comes to matrices? Should I just see it as -1 times the matrix X, so then I would multiply all the numbers in the matrix with -1? I appreciate any help I can get with this!

#### Plato

MHF Helper
matrix equation:
$$\pmatrix{6 & 1 \cr -3 & -8} -X \pmatrix{1 & 6 \cr -1 & 5}=I$$
$$\displaystyle X = \left( {\begin{array}{*{20}{c}}5&1\\{ - 3}&{ - 9}\end{array}} \right){\left( {\begin{array}{*{20}{c}}1&6\\{ - 1}&5\end{array}} \right)^{ - 1}}$$

#### Debsta

MHF Helper
$$\displaystyle a - x . B = i$$
$$\displaystyle a - i = x . B$$
$$\displaystyle (a - i) . B^ -1 = x$$

Sorry, not too good at LaTex. They should all be capitals.

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#### HallsofIvy

MHF Helper
Hi everybody! I need some help with this matrix equation:

$$\pmatrix{6 & 1 \cr -3 & -8} -X \pmatrix{1 & 6 \cr -1 & 5}=I$$

So this is how I would solve it:

$$A-X\cdot B=I \Leftrightarrow -X=I\cdot B^{-1}-A$$

Is this correct?
No, it's not. In fact, it would not be correct even if these were numbers. From a- xb= 1, multiplying by $$\displaystyle b^{-1}$$ would give $$\displaystyle ab^{-1}- x=b^{-1}$$ so that $$\displaystyle x= ab^{-1}- b^{-1}$$. With A, B, X matrices, it is exactly the same with $$\displaystyle X= AB^{-1}- B^{-1}= (A- I)B^{-1}$$.

Also, is -X different from X when it comes to matrices? Should I just see it as -1 times the matrix X, so then I would multiply all the numbers in the matrix with -1? I appreciate any help I can get with this!
-X is different from X in any algebraic system as long as X is not 0! Yes, you can interpret -X as -1 times X which would be all entries of X multiplied by -1. It is also the same as "X subtracted from the 0 matrix" as well as the diagonal matrix with all diagonal entries "-1" times X.

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