I'm having some problems with the second part of this problem.

Let $\displaystyle F(x_{1},x_{2},y_{1},y_{2},y_{3}) = (2x_{1}+x_{2}+y_{1}+y_{3}-1 ,x_{1}x_{2}^3+x_{2}^2y_{2}^2 - y_{2}y_{3}, x_{2}y_{1}y_{3} +x_{1}y_{1}^2 + y_{2}y_{3}^2)$ and let $\displaystyle a = (0,1,-1,1,1) $

So the first part says to Evaluate DF(a) and write it as DF(a)=[A|B] (a partitioned matrix) where A is a 3x2 matrix and B is a 3x3 matrix.

I got:

$\displaystyle DF = \begin{bmatrix}2&1&1&0&1\\ x_{2}^3&3x_{1}x_{2}^2+2x_{2}y_{2}^2&0&2x_{2}^2y_{2 }-y_{3}&-y_{2}\\ y_{1}^2&y_{1}y_{3}&2x_{1}y_{1}+x_{2}y_{3}&y_{3}^2& 2y_{2}y_{3}+x_{2}y_{1}\end{bmatrix}.

$

and

$\displaystyle DF(a) = \begin{bmatrix}2&1&1&0&1\\1&2&0&1&-1\\1&-1&1&1&1\end{bmatrix}.$

I need to show that if $\displaystyle det B =/= 0$ then the equation F=0 defines $\displaystyle y =(y_{1},y_{2},y_{3})$ implicitly as a function of $\displaystyle x=(x_{1},x_{2})$ near a (that is the function f:R^2--->R^3, y=f(x)).