Removed error in determinants.

docs/en/mathematics/linear-algebra/determinants.md

@@ -124,22 +124,22 @@ $$
 
 ??? note "*Proof*:"
 
-If $i = j$ then we obtain the cofactor expansion of $\det(A)$ along the $i$th row of $A$. 
+    If $i = j$ then we obtain the cofactor expansion of $\det(A)$ along the $i$th row of $A$. 
 
-If $i \neq j$, let $A^*$ be the matrix obtained by replacing the $j$th row of $A$ by the $i$th row of $A$
+    If $i \neq j$, let $A^*$ be the matrix obtained by replacing the $j$th row of $A$ by the $i$th row of $A$
 
-$$
-    A^* = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} \begin{array}{ll} j\text{th row}\\ \\ \\ \\\end{array}
-$$
+    $$
+        A^* = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} \begin{array}{ll} j\text{th row}\\ \\ \\ \\\end{array}
+    $$
 
-since two rows of $A^*$ are the same its determinant must be zero. It follows from the cofactor expansion of $\det(A^*)$ along the $j$th row that
+    since two rows of $A^*$ are the same its determinant must be zero. It follows from the cofactor expansion of $\det(A^*)$ along the $j$th row that
 
-$$
-\begin{align*}
-    0 &= \det(A^*) = a_{i1} A_{j1}^* + a_{i2} A_{j2}^* + \dots + a_{in} A_{jn}^*, \\
-      &= a_{i1} A_{j1} + a_{i2} A_{j2} + \dots + a_{in} A_{jn}.
-\end{align*}
-$$
+    $$
+    \begin{align*}
+        0 &= \det(A^*) = a_{i1} A_{j1}^* + a_{i2} A_{j2}^* + \dots + a_{in} A_{jn}^*, \\
+        &= a_{i1} A_{j1} + a_{i2} A_{j2} + \dots + a_{in} A_{jn}.
+    \end{align*}
+    $$
 
 > *Theorem*: let $E$ be an $n \times n$ elementary matrix and $A$ an $n \times n$ matrix with $n \in \mathbb{N}$ then we have
 >