Matrices equation with one unknown matrix

linear algebramatrices

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.

  1. 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)
  2. 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.
  3. 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.

Best Answer

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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