diff --git a/docs/en/mathematics/linear-algebra/determinants.md b/docs/en/mathematics/linear-algebra/determinants.md
index ddfb1b8..d49fb5d 100644
--- a/docs/en/mathematics/linear-algebra/determinants.md
+++ b/docs/en/mathematics/linear-algebra/determinants.md
@@ -198,4 +198,106 @@ $$
 
 ??? note "*Proof*:"
 
-    Will be added later.
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+    If $n \times n$ matrix $B$ is singular with $n \in \mathbb{N}$ then it follows that $AB$ is also singular and therefore
+
+    $$
+        \det(AB) = 0 = \det(A) \det(B),
+    $$
+
+    If $B$ is nonsingular, $B$ can be written as a product of elementary matrices. Therefore
+
+    $$
+    \begin{align*}
+        \det(AB) &= \det(A E_k \cdots E_1)
+                &= \det(A)\det(E_k)\cdots\det(E_1)
+                &- \det(A)\det(E_K \cdots E_1)
+                &= \det(A)\det(B).
+    \end{align*}
+    $$
+
+> *Theorem*: let $A$ be a nonsingular $n \times n$ matrix with $n \in \mathbb{N}$,then we have
+>
+> $$
+>   \det(A^{-1}) = \frac{1}{\det(A)}.
+> $$
+
+??? note "*Proof*:"
+
+    Suppose $A$ is a nonsingular $n \times n$ matrix then
+
+    $$
+    A^{-1} A = I,
+    $$
+
+    and taking the determinant on both sides
+
+    $$
+    \det(A^{-1}A) = \det(A^{-1})\det(A) = \det(I) = 1,
+    $$
+
+    therefore
+
+    $$
+        \det(A^{-1}) = \frac{1}{\det(A)}.
+    $$
+
+## The adjoint of a matrix
+
+> *Definition*: let $A$ be an $n \times n$ matrix with $n \in \mathbb{N}$, the adjoint of $A$ is given by
+>
+> $$
+> \mathrm{adj}(A) = \begin{pmatrix} A_{11} & A_{21} & \dots & A_{n1} \\ A_{12} & A_{22} & \dots & A_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ A_{1n} & A_{2n} & \dots & A_{nn}\end{pmatrix}
+> $$
+>
+> with $A_{ij}$ for $(i,j) \in \{1, \dots, n\} \times \{1, \dots, n\}$ the cofactors of $A$. 
+
+The use of the adjoint becomes in the following theorem, that generally saves a lot of time and brain capacity.
+
+> *Theorem*: let $A$ be a nonsingular $n \times n$ matrix with $n \in \mathbb{N}$ then we have
+>
+> $$
+>   A^{-1} = \frac{1}{\det(A)} \text{ adj}(A).
+> $$
+
+??? note "*Proof*:"
+
+    Suppose $A$ is a nonsingular $n \times n$ matrix with $n \in \mathbb{N}$, from the definition and the lemma above it follows that
+
+    $$
+        \text{adj}(A) A= \det(A) I,
+    $$
+
+    this may be rewritten into
+
+    $$
+        A^{-1} = \frac{1}{\det(A)} \text{ adj}(A).
+    $$
+
+## Cramer's rule
+
+> *Theorem*: let $A$ be an $n \times n$ nonsingular matrix with $n \in \mathbb{N}$ and let $\mathbf{b} \in \mathbb{R}^n$. Let $A_i$ be the matrix obtained by replacing the $i$th column of $A$ by $\mathbf{b}$. If $\mathbf{x}$ is the unique solution of $A\mathbf{x} = \mathbf{b}$ then
+>
+> $$
+>   x_i = \frac{\det(A_i)}{\det(A)}
+> $$
+>
+> for $i \in \{1, \dots, n\}$.
+
+??? note "*Proof*:"
+
+    Let $A$ be an $n \times n$ nonsingular matrix with $n \in \mathbb{N}$ and let $\mathbf{b} \in \mathbb{R}^n$. If $\mathbf{x}$ is the unique solution of $A\mathbf{x} = \mathbf{b}$ then we have
+
+    $$
+        \mathbf{x} = A^{-1} \mathbf{b} = \frac{1}{\det(A)} \text{ adj}(A) \mathbf{b}
+    $$
+
+    it follows that
+
+    $$
+    \begin{align*}
+        x_i &= \frac{b_1 A_1i + \dots + b_n A_{ni}}{\det(A)} \\
+            &= \frac{\det(A_i)}{\det(A)}
+    \end{align*}
+    $$
+
+    for $i \in \{1, \dots, n\}$.
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