I'm having some trouble with a question, I've been staring at it for a day or two now and it's not getting any clearer.
We have $ f(\mathbf{x}) = (\mathbf{x - b})^TA(\mathbf{x-b}) +\mathbf{(x-c)^T (x-c)} + \mathbf{b^T c} $, and I wish to find $ \nabla f $, and hence solve $\nabla f = \mathbf{0}$. We have $a$ and $b$ both constant collumn vectors, $A$ a square symmetric matrix, and $x$ our argument as a collumn vector.
My attempts so far have involved recognising the first term as a quadratic form in $(\mathbf{x-b}) $and thus contributing $2A(\mathbf{x-b})$ to the gradient vector, and the second term contributing $2(\mathbf{x-c})$ to the gradient.
So my gradient function is $\nabla f = 2A(\mathbf{x-b}) + 2(\mathbf{x-c})$. When solving equal to zero I get $ \mathbf{x}=(A+I)^{-1}(\mathbf{c}+A\mathbf{b})$ which when plugging in values is not correct.
Thanks!