The system has a unique solution if and only if $w^{T}A^{-1}v\ne 0$. I have the following problem: 
$$\left\{
    \begin{array}{ll}
        Ax+\lambda v=b \\
        w^{T}x=\rho
    \end{array}
\right.$$
where $A\in Gl_n(\Bbb{R})$, $u,v$ are column vector in $\Bbb{R}^n$ and $\lambda\in \Bbb{R}$, $x\in \Bbb{R}^n$ are the unknown values. 

Question: Show that the system has a unique solution if and only if $w^{T}A^{-1}v\ne 0$.

I rewrite the system as $$
 \left[ {\begin{array}{cc}
   A & v \\
   w^{T} & 0 \\
  \end{array} } \right]\left[ {\begin{array}{cc}
   x  \\
   \lambda \\
  \end{array} } \right]=\left[ {\begin{array}{cc}
   b  \\
   \rho \\
  \end{array} } \right].$$
Now I have to compute the determinant of $$\begin{pmatrix} * &*&*& v_1\\\vdots &\vdots&\vdots & \vdots\\ *&*&*& v_n\\ w_1 &\cdots & w_n & 0\end{pmatrix}$$ where the stars are $A$. The determinant looks complicated to compute and this is the first question a problem so I think I miss something..

Your idea to rewrite the system as a larger matrix is excellent.  Our goal, then, is to determine when the larger matrix is singular.  Note that
$$
\pmatrix{A^{-1}\\&1}
\pmatrix{
A & v \\
w^{T} & 0} = 
\pmatrix{I & A^{-1}v\\w^T & 0}
$$
Now, when is this second matrix singular?

Consider the matrix
$$
\pmatrix{
I & x\\
y^T & 0} = 
\pmatrix{
I & \pmatrix{x_1\\ \vdots \\ x_n}\\
\pmatrix{y_1 & \cdots & y_n} & 0}
$$
with $x,y \in \Bbb R$.  Calculate
$$
\det
\pmatrix{
I & x\\
y^T & 0} = x_1 \det \pmatrix{0&I\\y_1& *} - 
x_2 \det \pmatrix{*} + \cdots = \sum_{i=1}^n x_iy_i
$$
