What is the angle between $\vec{x}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $\vec{y}=\begin{bmatrix} -1 \\ 1 \end{bmatrix}$ using the inner product defined as $⟨\vec{x},\vec{y}⟩=\vec{x}^{T}\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}\vec{y}$
My Attempt
$$\Rightarrow ⟨\vec{x},\vec{y}⟩ = \begin{Vmatrix} \vec{x} \end{Vmatrix}\begin{Vmatrix} \vec{y} \end{Vmatrix} \cos \theta$$ $$\Rightarrow \cos \theta = \frac{⟨\vec{x},\vec{y}⟩}{\begin{Vmatrix} \vec{x} \end{Vmatrix}\begin{Vmatrix} \vec{y} \end{Vmatrix}}$$ $$\Rightarrow \cos \theta = \frac{\vec{x}^{T}\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}\vec{y}}{\begin{Vmatrix} \vec{x} \end{Vmatrix}\begin{Vmatrix} \vec{y} \end{Vmatrix}}$$ $$\Rightarrow \cos \theta = \frac{{\begin{bmatrix} 1 \\ 1 \end{bmatrix}}^{T}\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} -1 \\ 1 \end{bmatrix}}{\begin{Vmatrix} 1 \\ 1 \end{Vmatrix}\begin{Vmatrix} -1 \\ 1 \end{Vmatrix}}$$ $$\cos \theta = \frac{-1}{\sqrt{2}\sqrt{2}}$$ $$\theta = cos^{-1}-\frac{1}{2} \approx 2.09 radian \approx 120 degree$$
But the accepted answer is 1.9
What am I missing?
Note The question is from the Week-2 Coursera Course Mathematics for Machine Learning: PCA video Inner product: angles and orthogonality