How Far Is That Outlier, Really?

An anomaly detector scores a $2$-feature data point by its squared Mahalanobis distance from the training mean:

$d^2 = (x - \mu)^{\top}\,\Sigma^{-1}\,(x - \mu)$

where $\Sigma$ is the covariance matrix of the training data. Unlike plain Euclidean distance, this measure stretches and tilts with the shape of the data — which is exactly why the covariance matrix has to be inverted.

For this detector,

$\Sigma = \begin{pmatrix} 4 & 2 \\ 2 & 5 \end{pmatrix}, \qquad x - \mu = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$

What is $d^2$ for this point?