Find the minimum value of $\frac{1}{2}(x^{2}+2bxy+9y^{2})-y$ Find the minimum value of $f=\frac{1}{2}(x^{2}+2bxy+9y^{2})-y$
I know that there's a theorem which states $P(x)=\frac{1}{2}x^{T}Ax-x^{T}b$ reaches min at the point $Ax=b$ at $P_{min}=-\frac{1}{2}b^{T}A^{-1}b$.
I first tried to write the function $f$ in the form of $P(x)=\frac{1}{2}x^{T}Ax-x^{T}b$, so
$\frac{1}{2}(\begin{bmatrix}x &y \end{bmatrix}
\begin{bmatrix}1&2b\\2b&9\end{bmatrix}
\begin{bmatrix}x\\y\end{bmatrix})
-\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}$
And $P_{min}=-\frac{1}{2}b^{T}A^{-1}b$
$=-\frac{1}{2}\begin{bmatrix}0&1\end{bmatrix}\frac{1}{9-4b^{2}}\begin{bmatrix}9&-2b\\-2b&1\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}$
$=-\frac{1}{2(9-4b^{2})}$
However the answer key says that the min is $-\frac{1}{2(9-b^{2})}$, where did I go wrong?

Hint: Your matrix A is not right. It should be:
1    b
b    9.
In other words, a(11) = 1, a(12) = b = a(21), and a(22) = 9.
