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

Let  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  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:

 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}.<br />
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  det B =/= 0 then the equation F=0 defines  y =(y_{1},y_{2},y_{3}) implicitly as a function of  x=(x_{1},x_{2}) near a (that is the function f:R^2--->R^3, y=f(x)).