Section determinants almost finished.

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

@@ -81,7 +81,31 @@ $$
 
 ??? note "*Proof*:"
 
-    Will be added later.
+    Let $A$ be a $n \times n$ triagular matrix with $n \in \mathbb{N}$ given by
+
+    $$
+        A = \begin{pmatrix} a_{11} & \cdots &a_{1n}\\ & \ddots & \vdots \\ & & a_{nn} \end{pmatrix}.
+    $$
+
+    We claim that $\det(A) = a_{11} \cdot a_{22} \cdots a_{nn}$. We first check the claim for $n=1$ which is given by $\det(A) = a_{11}$. 
+
+    Now suppose for some $k \in \mathbb{N}$, the determinant of a $k \times k$ triangular $A_{k}$ is given by
+
+    $$
+        \det(A_k) = a_1{11} \cdot a_{22} \cdots a_{kk}
+    $$
+
+    then by assumption 
+
+    $$
+        \det(A_{k+1}) = \begin{pmatrix} A_k & a_{(k+1)1}\\& \vdots\\ 0 \cdots 0 & a_{(k+1)(k+1)}\end{pmatrix} = a_{(k+1)(k+1)} \det(A_k) + 0 = a_{11}a_1{11} \cdot a_{22} \cdots a_{kk} \cdot a_{(k+1)(k+1)}. 
+    $$
+
+    Hence if the claim holds for some $k \in \mathbb{N}$ then it also holds for $k+1$. The principle of natural induction implies now that for all $n \in \mathbb{N}$ we have
+
+    $$
+        \det(A) = a_{11} \cdot a_{22} \cdots a_{nn}.
+    $$
 
 > *Theorem*: let $A$ be an $n \times n$ matrix
 >
@@ -92,3 +116,86 @@ $$
 
     Will be added later.
 
+> *Lemma*: let $A$ be an $n \times n$ matrix with $n \in \mathbb{N}$. If $A_{jk}$ denotes the cofactor of $a_{jk}$ for $k \in \mathbb{N}$ then
+>
+> $$
+>   a_{i1} A_{j1} + a_{i2} A_{j2} + \dots + a_{in} A_{jn} = \begin{cases} \det(A) &\text{ if } i = j,\\ 0 &\text{ if } i \neq j.\end{cases}
+> $$
+
+??? note "*Proof*:"
+
+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 $A$
+
+$$
+    A^* = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} \begin{array}{ll} j\text{th row}\\ \\ \\ \\\end{array}
+$$
+
+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
+
+$$
+\begin{align*}
+    0 &= \det(A^*) = a_{i1} A_{j1}^* + a_{i2} A_{j2}^* + \dots + a_{in} A_{jn}^*, \\
+      &= a_{i1} A_{j1} + a_{i2} A_{j2} + \dots + a_{in} A_{jn}.
+\end{align*}
+$$
+
+> *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
+>
+> $$
+>   \det(E A) = \det(E) \det(A),
+> $$
+>
+> where
+>
+> $$
+>   \det(E) = \begin{cases} -1 &\text{ if $E$ is of type I},\\ \alpha \in \mathbb{R}\backslash \{0\} &\text{ if $E$ is of type II},\\ 1 &\text{ if $E$ is of type III}. \end{cases}
+> $$
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+Similar results hold for column operations, since for the elementary matrix $E$, $E^T$ is also an elementary matrix and $\det(A E) = \det((AE)^T) = \det(E^T A^T) = \det(E^T) \det(A^T) = \det(E) \det(A)$. 
+
+> *Theorem*: an $n \times n$ matrix A with $n \in \mathbb{N}$ is singular if and only if
+>
+> $$
+>   \det(A) = 0
+> $$
+
+??? note "*Proof*:"
+
+    Let $A$ be an $n \times n$ matrixwith $n \in \mathbb{N}$. Matrix $A$ can be reduced to row echelon form with a finite number of row operations obtaining
+
+    $$
+        U = E_k E_{k-1} \cdots E_1 A,
+    $$
+
+    where $U$ is in $n \times n$ row echelon form and $E_i$ are $n \times n$ elementary matrices for $i \in \{1, \dots, k\}$. It follows then that 
+
+    $$
+    \begin{align*}
+        \det(U) &= \det(E_k E_{k-1} \cdots E_1 A), \\
+                &= \det(E_k) \det(E_{k-1}) \cdots \det(E_1) \det(A). 
+    \end{align*}
+    $$
+
+    Since the determinants of the elementary matrices are all nonzero, it follows that $\det(A) = 0$ if and only if $\det(U) = 0$. If $A$ is singular then $U$ has a row consisting entirely of zeros and hence $\det(U) = 0$. If $A$ is nonsingular then $U$ is triangular with 1's along the diagonal and hence $\det(U) = 1$.
+
+From this theorem we may pose a method for computing $\det(A)$ by taking 
+
+$$
+\det(A) = \Big(\det(E_k) \det(E_{k-1} \cdots \det(E_1)\Big)^{-1}. 
+$$
+
+> *Theorem*: let $A$ and $B$ be $n \times n$ matrices with $n \in \mathbb{N}$ then
+>
+> $$
+>   \det(AB) = \det(A) \det(B)
+> $$
+
+??? note "*Proof*:"
+
+    Will be added later.