My book presents the following algebraic construction:
LP: $z_{min}=cx$
$s.t. $ $Ax=b$
$x \ge 0$
Where A is a m*n matrix and x is a vector of variables.
Beyond the meaning of the simplex algorithm the book report the following algebraic steps:
$1.Ax=b$
$2.Bx_B+Nx_N=b$ Here A is divided in two sub-matrices A & B and x is divided in $x_B$ & $x_N$
$3.$ From the book:
"Multiplying the last equation by $B^{-1}$ and rearranging the terms, we get $x_B=B^{-1}b - B^{-1}Nx_N$"
I don't get this step, it's seems that $B^{-1}$ is used like $1/B$.
I explain my self:
Back to the $Bx_B+Nx_N=b$
If I "divide" all by $B$, I get:
$\frac{Bx_B}{B}+\frac{Nx_N}{B}=\frac{b}{B} => x_B+\frac{Nx_N}{B}=\frac{b}{B}$
If I express $\frac{1}{B}$ as $B^{-1}$, I can write: $x_B+B^{-1}Nx_N=B^{-1}b => x_B=B^{-1}b - B^{-1}Nx_N$ exactly the same expression of the book in point 3.
Is my explanation of the expression $x_B=B^{-1}b - B^{-1}Nx_N$ valid in linear algebra and why?
Yes, $B^{-1}$ is equivalent to $\frac{1}{B}$ if $B \in \mathbb{R}$. Or to be more exact, the $x^{-1}$ operator generalizes the inversion $\frac{1}{x}$.
Go back to your definitions to make things more clear. $B^{-1}$ is the matrix such that $$ B \times B^{-1} = I $$ where $\times$ refers to matrix multiplication and $I$ is the identity matrix (with only $1$s on the diagonal). Now if $B$ is a scalar, $I$ equals $1$. And solving the equation $B \times B^{-1} = 1$ yields $B^{-1}=1/B$, indeed.