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(i+X).(X.A+B)+C=0

i is an integer

A, B, C are known matrices of n*m integer elements

X is unknown matrix of n*n elements

EDIT: More standard notation would be:

(I+X).(X.A+B)+C=0

I is a matrix of n*n elements each of integer value i

A, B, C are known matrices of n*m integer elements

X is unknown matrix of n*n elements

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    What does $i+X$ mean when $i$ is an integer and $X$ is a matrix? For example, what is $$ 1 + \pmatrix{1&0\\0&0}? $$ – 2017-02-28
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    I am not sure about standard notation. But you can replace i with I matrix that has n*n elements each of value i. I wrote it this way as a concise version. – 2017-02-28
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    So, is it correct to say that $$ 1 + \pmatrix{1&0\\0&0} = \pmatrix{2&1\\1&1}? $$ – 2017-02-28
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    Yes, sorry for inconvenience. – 2017-02-28
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    See if this link provides any help: http://math.stackexchange.com/questions/649121/solving-the-quadratic-equation-for-matrices – 2017-02-28
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    @Mengchun Zhang Thank you, I can convert my equation to the form in your link for i=0, but for other values of i I am not sure whether it is possible. – 2017-02-28
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    @azerbajdzan You are welcome :) and could you write this – 2017-02-28
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    @azerbajdzan Sorry to mess up the comment above. I think you could try this: let $T$ be the matrix whose entries are all equal $i$, and let $Y=X+T$, then the equality becomes $Y.[(Y-T).A+B]+C=0$, which is $(Y^2).A+Y.(B-T.A)+C=0$ – 2017-02-28
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    @Mengchun Zhang What a simple "trick". :-D I have overcomplicated it and overlooked this substitution. Thank you a lot again. So now I can proceed with studying your link in more detail. :-) – 2017-02-28
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    @azerbajdzan No worries :) but I am not sure whether this equation will work because all the "coefficient matrices" are multiplied after the unknown matrix $Y$, whereas in the link all the "coefficient matrices" are before the unknown matrix. – 2017-02-28
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    @Mengchun Zhang In the link they require the matrices to be commuting, so it does not matter for them on which side the unknown is. But my matrices are not commuting so their method is not applicable to my problem. – 2017-03-01
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    @azerbajdzan Yes you are right, sorry about that. I should check it before sending the link to you. – 2017-03-01

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