<s>[INST] Matrices equation with one unknown matrix Let us consider following equation:
$$
2X^2+2X=
\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 &  -3
\end{array} \right). 
$$
I have to show that there is no real matrix which satisfy my equation.
I consider three diffrent way to do this.

*

*Direct method. Let $$X=\left( \begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h & i
\end{array} \right).
$$ Then we can write euqation in equivalent form $X(X+I)=\frac{1}{2}\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 &  -3
\end{array} \right)$, and we obtain system of nonlinear equations
$$
\left\{ \begin{array}{llll}
a^2+a+bd+cg=-\frac{1}{2} \\
ab+be+b+ch=\frac{5}{2}\\
ac+bf+ci+c=\frac{3}{2}\\
da+d+ed+fg=-1\\
db+e^2+e+fh=\frac{1}{2}\\
dc+ef+fi+f=1\\
ga+g+hd+ig=0\\
gb+he+h+ih=-2\\
gc+hf+i^2+i=-\frac{3}{2}
\end{array} \right. 
$$
I was trying to deal with it, but it was impossible to solve for me (I can not use Mathematica, Matlab,etc)

*Determinant method. From the fact, that $X(X+I)=\frac{1}{2}\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 &  -3
\end{array} \right)$, we have that $\det X \det (X+I)=-\frac{11}{8}$, and then using direct method i was traing to write formula for $\det (X+I)$ in terms of $\det X$, but here also appear formulas,to difficult for me to interpret it.

*My last idea to solve is to use formula for inverse matrix $3\times3$  ($X$ is invertible, which is consequence of the fact, that $\det X\neq 0$) and from relation $X+I=X^{-1}\frac{1}{2}\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 &  -3
\end{array} \right)$ we also obtain system of equations, which is much more complicated, that this previous one.

I would be grateful if you give me some hints.
 [/INST] If there were a real matrix $X$ satisfying ...
Scalar-multiply the given equation with $2$ and add the identity matrix to get
$$(2X+\mathbb 1)^2 \;=\; \begin{pmatrix} -1& 10& 6\\
-4& 3& 4\\
0& -8& -5 \end{pmatrix}$$
Apply the determinant which produces $\,-25\,$ on the RHS, hence a contradiction because the resulting square on the LHS cannot get negative if $X$ is a real matrix.
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