I tried to solve this, but this was slightly different.
I set,
$A =\begin{equation} \begin{bmatrix} 1 & 4 & 2 & -1 & 0\\ 3 & 2 & 0 & 0 & -1\\ \end{bmatrix} \end{equation}$
$C = \begin{equation} \begin{bmatrix} 4M-2 & 6M-3 & 2M-2 & 2M-2 & M\\ \end{bmatrix} \end{equation}$
$b = \begin{equation} \begin{bmatrix} 8 \\ 6\\ \end{bmatrix} \end{equation}$
and by final simplex tableau,
$C_bB^{-1} = \begin{equation} \begin{bmatrix} M-0.5 & M-0.5\\ \end{bmatrix} \end{equation}$
$B^{-1} = \begin{equation} \begin{bmatrix} 0.3 & -0.1\\ -0.2 & 0.4\\ \end{bmatrix} \end{equation}$
And I got $C_BB^{-1}b = 14M-7$ which is $Z$ by multiplying $c_BB^{-1}$ and $b$
But this solution is different of mine which is solved by tableau form. And that was just $-7$
Why the solutions are different in different two forms?