# Non-linear algebra with Cramer's Rule?

I am asked to solve the following system of equations using Cramer's Rule

$$A^2 + B = x\\ AB = 1-x^2$$

where $A$ and $B$ are 2 unknowns that are to be solved for some given $x$. ie I'm expecting the solution to be $A = a(x)$ and $B = b(x)$ where $a(x)$ and $b(x)$ is some function of $x$.

Is it possible to do so? Alternatively, can I differentiate both equations explicitly w.r.t. $x$ and apply Cramer's rule? I have a feeling this is the right way since this involves linear algebra techniques.

$$2AA' + B' = 1\\ A'B + B'A = -2x$$

### UPDATE

Alright, guys...the answer is actually to differentiate the equations w.r.t. $x$. I do not know if this is meaningful to any one. Basically, both $A$ and $B$ are implicitly defined in $x$. So after applying the differentiation, which is above, the next step is to construct the matrix:

$$\begin{pmatrix} 2A & 1 \\ B & A \end{pmatrix}\begin{pmatrix} A'\\B' \end{pmatrix} = \begin{pmatrix} 1\\-2x \end{pmatrix}$$

Applying Cramer's Rule,

\begin{align*} A' &= \frac{A + 2B}{2A^2 - B}\\ \\ B' &= \frac{4Ax + B}{B - 2A^2} \end{align*}

-
So $A = A(x)$?.. – gnometorule Feb 15 '13 at 7:19
Well, you could have saved a lot of confusion by stating right away that you don't really need to solve $A$ and $B$ but rather $A'$ and $B'$ in terms of $A$ and $B$. – Jyrki Lahtonen Feb 15 '13 at 8:09

Cramer's rule is for linear system of equations only. So, no, you can't apply it for the system above.

## Update

Also, what's $x$ in your equations? Because if it's actually a variable, then your system is a functional equations. If it's just a number, never mind then.

-
I updated my question to include that $A$ and $B$ is some function of some value of $x$. – bryansis2010 Feb 15 '13 at 7:24
So, basically, you have two functions $A(x)$ and $B(x)$ that satisfy the given system. It has nothing to do with Cramer's rule and I don't understand why you stick with it. As for the method you chose, it's not good idea to differentiate equations, since you don't have initial conditions, and those equations are implicit! – Kaster Feb 15 '13 at 7:52
Hello Kaster, I know exactly what you are talking about - but my teacher says I need to solve it using Cramer's rule. Since the original equations cannot be solved with Cramer's rule as they are not linear, the next best bet is to differentiate w.r.t $x$ and solve with Cramer's rule the differentiated equations. – bryansis2010 Feb 15 '13 at 7:55
Anyway, Kaster, take a look at valtron's answer...$A$ is eventually a complex number...I guess the question was indeed intentionally done this way. – bryansis2010 Feb 15 '13 at 7:57
Well, in this case you'll get non linear system of ODEs and there's no method to solve it analytically due to the nonlinearity. – Kaster Feb 15 '13 at 7:57

You could write that system of equations as

$$\left( \begin{matrix}A & 1 \\ 0 & A \end{matrix} \right) \left( \begin{matrix}A \\ B \end{matrix} \right) = \left( \begin{matrix}x \\ 1-x^2 \end{matrix} \right)$$

and then solve, but then you get a solution for $A$ and $B$ in terms of $x$ and $A$; it's not clear that such a thing would be useful.

Or, you could write it as a system

$$\left( \begin{matrix}1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right) \left( \begin{matrix}A^2 \\ B \\ AB \end{matrix} \right) = \left( \begin{matrix}x \\ 1-x^2 \end{matrix} \right)$$

and get infinitely many putative solutions for $(A^2, B, AB)$, and you have the problem of figuring out which triples can be solved for $A$ and $B$. (e.g. the solution $(x,0,1-x^2)$ is a solution to the linear system, but you can't solve the system $A^2=x$, $B=0$, $AB=1-x^2$ for $A$ and $B$)

Your approach lets you write the equation as

$$\left( \begin{matrix}2A & 1 \\ B & A \end{matrix} \right) \left( \begin{matrix}A' \\ B' \end{matrix} \right) = \left( \begin{matrix}1 \\ -2x \end{matrix} \right)$$

So gain you could solve, but the answer is a solution for $(A',B')$ in terms of $A$ and $B$ -- i.e. a system of differential equation in two functions, so again it's not clear the result will be useful.

But then, it doesn't hurt to try any of these approaches to see if the result suggests what to do next.

-

Multiply both sides of the first equation to get $A^3+AB=Ax$. Plug in the second equation and rearrange the terms to get $A^3-Ax+(1-x^2)=0$. Solve this cubic either by hand or using WolfamAlpha. You can then solve $B$ by substituting the solution for $A$ in $B=x-A^2$.

-