I can't get the right answer to this question I think I have a small mistake somewhere but I can't find it...
a) for which numbers $b$ is the matrix $A = \begin{bmatrix}1 & b \\ b & 9\end{bmatrix}$ positive definite?
b) Factor $A = LDL^T$ when $b$ is in the range for positive definiteness
c) Find the minimum value of $F = \frac{1}{2}(x^2 + 2bxy + 9y^2) - y$ for $b$ in this range
d) what is the minimum if $b = 3$
My answer is as follows
a) when $-3 < b < 3$ because $a > 0 \text{ and } ac > b^2$ must be true
b)
$$ \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} $$
c) This is where I am stuck with the wrong answer... I should check that $\frac{\partial}{\partial x} = \frac{\partial}{\partial y} = 0$ when x is in the right range (lets say x = 2)...
$$ \begin{aligned} \frac{\partial F}{\partial x} &= x + 2y \\ \frac{\partial F}{\partial y} &= 2x + 9y - 1 \\ \end{aligned} $$
the first one is equal to 0 at $(0, 0)$ and the second one is equal to 0 at $(\frac{1}{2}, 0) \text{ and } (0, \frac{1}{9})$ which means they don't equal 0 at the same spot??
c) You need to find the partial derivatives parametrically: $$F_x=\left(\frac{1}{2}(x^2 + 2bxy + 9y^2) - y\right)'_x=x+by=0 \Rightarrow x=-by;\\ F_y=\left(\frac{1}{2}(x^2 + 2bxy + 9y^2) - y\right)'_y=bx+9y-1=0 \Rightarrow -b^2y+9y-1=0 \Rightarrow \\ y=\frac1{9-b^2}, x=\frac{b}{b^2-9}\\ $$ The second order test: $$F_{xx}=1>0,F_{yy}=9>0,F_{xy}=b,\\ \Delta=F_{xx}F_{yy}-F_{xy}^2=9-b^2>0 \ \ \text{for minimum, which is the range of $b$ from a)}$$ d) if $b=3$, there is no stationary point as the system of equations $F_x=0,F_y=0$ does not have a solution.