Find $\det X$ if $8GX=XX^T$ I need to find $\det X$ where
$$8GX=XX^T,\quad G=\left(\begin{matrix}5 & 4\\3 & 2\\\end{matrix}\right).$$
My answer is that the determinant of $X$ is $-128$ and that is correct but there is one more value of $\det X$ that can solve the equation.
 A: This question had been heavily downvoted and was once closed, but frankly I don't think this is an easy question and I don't understand the reasons behind the downvotes.
Let $A=8G=\pmatrix{40&32\\ 24&16}$. If $AX=XX^T$, then $\det(A)\det(X)=\det(X)^2$. So, the only possible values for $\det(X)$ are $0$ and $\det(A)=-128$. Clearly, the equation $AX=XX^T$ with $\det(X)=0$ is solvable --- $X=0$ is a trivial solution in the first place, and it is easy to show that all singular solutions are given by $X=0$ or $X=\lambda uu^T$, where $(\lambda,u)$ is an eigenpair of $A$ with $\|u\|=1$.
The real question, though, is whether the determinant $-128$ is attainable, i.e. whether there exists a nonsingular matrix $X$ such that $AX=XX^T$. This is equivalent to $A=XX^TX^{-1}$ and it apparently is a very strong requirement --- not only must $A$ be similar to $X^T$, the change-of-basis matrix must also be equal to $X$. Perhaps I am missing something here, but I honestly don't see any obvious answer to this problem.
Edit: For those who are interested, if we put $Y=AX$, then $AYA^T=AXX^TA^T=YY^T$. So, if $X$ is nonsingular, then $Y$ is symmetric positive definite and $\det(Y)=128^2=d$ (say). So, if we let $t=\operatorname{trace}(Y)$ and write $Y=\pmatrix{x&y\\ y&z}$, then by Cayley-Hamilton theorem, we get $AYA^T=YY^T=Y^2=tY-dI$ and this equation can be rewritten as
$$
\pmatrix{x\\ y\\ z}=
-256\left[\pmatrix{
25&40&16\\
15&22& 8\\
 9&12& 4}
-sI\right]^{-1} \pmatrix{1\\ 0\\ 1},\tag{1}
$$
where $s=t/64$. Since $t=64s=x+z$, if we left-multiply $(1)$ by $(1,0,1)$ on both sides, we will eventually obtain $p(s)=(s-2)(s^3 - 49s^2 - 208s - 216)=0$. It turns out that $p$ has two real roots. The root $s=2$ does not give rise to a valid $X$, but the other root, $s\approx53.00132225047258$ gives the solution
$$
X=\pmatrix{
  43.30336531969982  &26.18058434772915\\
  25.88968225037343  &12.69663468030015
}.
$$
