Have a positive-definite QP ($\varLambda$ is positive-definite, $A$ is full row rank ):
$$\begin{equation}\begin{aligned} &min \quad f(x)=c^{T} \varLambda c + k^{T}c \\ &s.t. \quad Ac-b=0 \end{aligned}\end{equation}$$
Use Lagrangean multiplier
$$\begin{equation}\begin{aligned} &L(c,\alpha) = c^{T} \varLambda c + \alpha^{T} (Ac-b) \\ &\bigtriangledown_{c}L= \varLambda c + k + A^{T}\alpha = 0 \\ &\bigtriangledown_{\alpha}L = Ac-b=0 \end{aligned}\end{equation}$$
Then the $c_{min}$ can represent
$$\begin{equation}\begin{aligned}
&\alpha= (-A \varLambda^{-1}A^{T})^{-1}(A \varLambda^{-1}k + b)\\
&c_{min} = -\varLambda^{-1}(k+A^{T} \alpha)
\end{aligned}\end{equation}$$
This soluton $c_{min}$ should be the minial solution of QP but I feel confused when i use Moore–Penrose inverse to solve $ Ac=b$ and put the solution ($c_{m-p}$) to the object function. Somtime the $f(c_{m-p})<f(c_{min})$ is happend. Why?
ex: $$\begin{equation}\begin{aligned} &min \quad f(x)=c^{T} \begin{pmatrix} 0.54534832 & 0\\ 0 & 0.15573046\\ \end{pmatrix} c + \begin{pmatrix} 1.86544158 & 6.19765616 \end{pmatrix} c \\ &s.t. \quad \begin{pmatrix} 8.53061926 & 2.34567083 \end{pmatrix} c- 3.84608369 = 0\\ & c_{min} = \begin{pmatrix} 8.29260734 \\ -28.51849082 \end{pmatrix} \quad f(c_{min}) =2.879981 \\ & c_{m-p} = \begin{pmatrix} 0.41916381 \\ 0.11525779 \end{pmatrix} \quad f(c_{m-p}) =1.594139 \\ &\color{red}{f(c_{m-p})<f(c_{min})} \end{aligned}\end{equation}$$
There is a mistake in your computation:
$$\bigtriangledown_c L = \color{red}2\Lambda c + k + A^T\alpha=0$$