A $2$x$2$ system of heterogeneous equations has a unique solution if and only if the determinant of the matrix of its coefficients be non-zero.
There is a $2$x$2$ system of heterogeneous equations and the variables are $x$ and $z^2$, I decided to convert the $z^2 = y$ and solve the system for $x, y$, Then square the solution to the $y$ to get the $z$.
I would like to know is there any thing that determines whether the $y$ has a positive solution?
Thanks in advance
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I am solving a problem and there is a recursive function $f(x)$ and another function $g(x)$. I need to solve this system of equations:
$$ af(a)-xf(a)-z^2g(a) = 0 $$ $$ bf(b)-xf(b)-z^2g(b) = 0 $$
clearly this system has solution if determinant of
$$ \begin{bmatrix} f(a)&g(a)\\ f(b)&g(b)\\ \end{bmatrix} $$ is non-zero, I set the $z^2 = y$. Since $y$ must be positive, I need other conditions to guarantee that $y \geq 0$ .
You can express the solution in terms of the parameters, using the general formula of the inverse matrix of a $2\times 2$ matrix, see for example: https://www.mathsisfun.com/algebra/matrix-inverse.html
There is no better way, this is the precise formula that provides you with the answer. You just have to check in this concrete case when the resulting $y$ is positive, which means you have to solve an inequality. There is no general recipee for this, it can be any inequality, depending on the parameters.