Dual of a Linear Program \begin{align}
\min_{x} c^Tx \\
s.t.~Ax=b
\end{align}
Note that here $x$ is unrestricted. I need to prove that the dual of this program is given by
\begin{align}
\max_{\lambda} \lambda^Tb \\
s.t.~\lambda^TA\leq c^T
\end{align}
But in the constraint, I always get an equality (using what I learnt)
\begin{align}
\max_{\lambda} \lambda^Tb \\
s.t.~\lambda^TA = c^T
\end{align}
Please give some explanation also. 
 A: You wrote the dual problem correctly.  Perhaps whoever wrote your assignment forgot to include the constraint $x \geq 0$ in the primal problem.
Edit: here's how I derived the dual problem.  The Lagrangian is
\begin{align}
L(x,\nu) &= \langle c, x \rangle + \langle \nu, b - Ax \rangle \\
&= \langle c, x \rangle - \langle \nu, Ax \rangle + \langle \nu, b \rangle \\
&= \langle c, x \rangle - \langle A^T \nu, x \rangle + \langle \nu, b \rangle \\
&= \langle c - A^T \nu, x \rangle + \langle \nu, b \rangle.
\end{align}
The dual function is
\begin{equation}
g(\nu) = 
\begin{cases}
\langle \nu,b \rangle & \quad \text{if } A^T \nu = c \\
-\infty & \quad \text{otherwise}.
\end{cases}
\end{equation}
So the dual problem is
\begin{align}
\text{maximize} & \quad \langle \nu, b \rangle \\
\text{subject to} & \quad A^T \nu = c.
\end{align}
A: \begin{align}
\min_{x} c^Tx \\
s.t.~Ax=b
\end{align}
Is the same as:
\begin{align}
\min_{x} c^T(x^+-x^-) \\
s.t.~A(x^+-x^-)=b\\
x^+,x^-\geq 0
\end{align}
Is the same as:
\begin{align}
\min_{x} [c^T|-c^T]z \\
s.t.~[A|-A]z=b\\
z\geq 0
\end{align}
$$z=[x^T|-x^T]^T$$
Dual of this is :
\begin{align}
\max \quad b^Tp\\
s.t. [A|-A]^Tp\leq [cT|-c^T]^T\\
\implies Ap=c
\end{align}
I think your answer is correct.
A: Given 
$\begin{align}
\min_{x} c^Tx \\
s.t.~Ax=b
\end{align}$
I think the dual of the problem would be
$\begin{align}
\max_{x} b^Tw \\
s.t.~A^Tw (*)c
\end{align}$
where $(*)$ actually depends on the restrictions of the variables in the primal problem. If the restriction is $x_i \geq 0$ then the type of constraint will follow the same inequality. But if $x_i$'s are unrestricted, then the constraints will become $=$
A: \begin{align}
\min_{x} c^Tx \\
s.t.~Ax=b \\
Unrestricted 
\end{align}
Take $x=x_1-x_2$
\begin{align}
\min c^T(x_1-x_2) \\
s.t.~A(x_1-x_2)=b \\
~ x_1, x_2 \ge 0
\end{align}
This is can be written as
\begin{align}
\min {\begin {pmatrix}c \\ -c \\ \end  {pmatrix} }^T  \begin {pmatrix} x_1 \\ x_2 \\ \end  {pmatrix} \\
s.t.~\begin {pmatrix} A, & -A  \end  {pmatrix} \begin {pmatrix} x_1 \\ x_2 \\ \end  {pmatrix}=b \\
~ x_1, x_2 \ge 0
\end{align}
Now,  this is in the standard form of the linear program. Therefore, the dual can be written as
\begin{align}
\max b^T y \\
s.t.~ \begin {pmatrix}c \\ -c \\ \end  {pmatrix}  - 
\begin {pmatrix} A^T\\ -A^T \\ \end  {pmatrix}  y \ge 0\\ 
~ y \ge 0
\end{align}
This can be simplify as
\begin{align}
\max b^T y \\
s.t.~ c-A^T y \ge 0\\ 
~ -c+A^T y \ge 0\\ 
~ y \ge 0
\end{align}
\begin{align}
\max b^T y \\
s.t.~ A^T y \le c\\ 
~ A^T y \ge c\\ 
~ y \ge 0
\end{align}
This is equivalent to 
\begin{align}
\max b^T y \\
s.t.~ A^T y = c\\ 
~ y \ge 0
\end{align}
