The correct answer is $\boxed{\frac{{{{\text{A}}^{ – 1}}{\text{b}}}}{2}}$.
Let $X$ be a vector of dimension $n \times 1$. The function $f(X)$ is defined as follows:
$$f(X) = XTAX + bTX + c$$
where $A$ is a symmetric positive definite matrix with dimension $n \times n$; $b$ and $x$ are vectors of dimension $n \times 1$.
We can write the function $f(X)$ as follows:
$$f(X) = X^T(A^T + A)X + b^TX + c$$
Since $A$ is a symmetric positive definite matrix, it is invertible. Therefore, we can write the function $f(X)$ as follows:
$$f(X) = (X^TA^TX + b^TX + c)(A^{-1})^T$$
We can minimize the function $f(X)$ by setting its derivative to zero. The derivative of $f(X)$ with respect to $X$ is given by:
$$\frac{\partial f(X)}{\partial X} = 2(A^T + A)X + b^T = 2AX + b^T$$
Setting the derivative to zero and solving for $X$, we get:
$$X = -(A^T + A)^{-1}b^T = -A^{-1}b^T$$
Therefore, the minimum value of $f(X)$ will occur when $X = -A^{-1}b^T$.
The other options are incorrect because they do not minimize the function $f(X)$.