one more backtick block

Gram-matrix-parameterization.md

@@ -64,12 +64,12 @@ It follows that
 Since the subspace $C$ is transpose-invariant, we also have
 \[ \mathcal{P}(X^\top) = \mathcal{P}(X)^\top. \]
 We can now see that
-\[
+```math
 \begin{align*}
-d\Delta & = -\mathcal{P}(dA^\top Q A + A^\top Q\,dA) \\
-& = -\big[\mathcal{P}(A^\top Q\,dA)^\top + \mathcal{P}(A^\top Q\,dA)\big].
+   d\Delta & = -\mathcal{P}(dA^\top Q A + A^\top Q\,dA) \\
+                 & = -\big[\mathcal{P}(A^\top Q\,dA)^\top + \mathcal{P}(A^\top Q\,dA)\big].
 \end{align*}
-\]
+```
 Plugging this into our formula for $df$, and recalling that $\Delta$ is symmetric, we get
 ```math
 \begin{align*}