\section*{Algebra 04: Determinants and Linear Maps}

\subsection*{Definition}

Let \( A \in \mathcal{M}_2(\mathbb{K}) \), then:

\[
A = \begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{pmatrix}
\]

The determinant of \( A \), denoted \( \det A \), is given by:

\[
\det A = \left| \begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array} \right|
= a_{11} a_{22} - a_{12} a_{21}
\]
Let 
\[
B = \begin{pmatrix} 2 & 1 \\ 4 & 0 \end{pmatrix} \quad \Rightarrow \quad \det(B) = 2 \cdot 0 - 4 \cdot 1 = -4
\]

Now consider a matrix \( A \in \mathcal{M}_3(\mathbb{K}) \),  
\[
A = \begin{pmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{pmatrix}
\]

Then the determinant is:

\[
\det A = \left| \begin{array}{ccc}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array} \right|
= a_{11} \left| \begin{array}{cc}
a_{22} & a_{23} \\
a_{32} & a_{33}
\end{array} \right|
- a_{12} \left| \begin{array}{cc}
a_{21} & a_{23} \\
a_{31} & a_{33}
\end{array} \right|
+ a_{13} \left| \begin{array}{cc}
a_{21} & a_{22} \\
a_{31} & a_{32}
\end{array} \right|
\]

\section*{Example}

\[
A = \begin{pmatrix}
1 & 3 & 2 \\
4 & 1 & 3 \\
2 & 0 & 0
\end{pmatrix}
\]

\[
\det A = 1 \cdot \left| \begin{array}{cc} 1 & 3 \\ 0 & 0 \end{array} \right|
- 3 \cdot \left| \begin{array}{cc} 4 & 3 \\ 2 & 0 \end{array} \right|
+ 2 \cdot \left| \begin{array}{cc} 4 & 1 \\ 2 & 0 \end{array} \right|
\]

\[
= -6 + 18 + 12 = 24
\]
\section*{Determinant Properties}

\begin{enumerate}
  \item If there is equality between two columns or two rows, then $\det A = 0$.
  
  Example:
  \[
  A = \begin{pmatrix} 1 & 0 & 2 \\ 2 & 0 & 2 \end{pmatrix} \Rightarrow \det A = 0
  \]

  \item If $A \in \mathcal{M}_n(\mathbb{K})$ has a row or column of zeros, then $\det A = 0$.

  \item For a triangular matrix:
  \[
  A = \begin{pmatrix}
  a_{11} & \cdots & * \\
  0 & \ddots & * \\
  0 & \cdots & a_{nn}
  \end{pmatrix}
  \Rightarrow \det A = \prod_{i=1}^{n} a_{ii}
  \]

  \item If $A$ is a diagonal matrix:
  \[
  A = \begin{pmatrix}
  a_{11} & 0 & \cdots & 0 \\
  0 & a_{22} & \cdots & 0 \\
  \vdots & \vdots & \ddots & \vdots \\
  0 & 0 & \cdots & a_{nn}
  \end{pmatrix}
  \Rightarrow \det A = \prod_{i=1}^{n} a_{ii}
  \]
\end{enumerate}

\section*{Properties}

Let $A, B \in \mathcal{M}_n(\mathbb{K})$ and $\lambda \in \mathbb{K}$:

\begin{enumerate}
  \item $\det(AB) = \det A \cdot \det B$
  \item $\det(A^T) = \det A$
  \item $A$ is invertible $\Leftrightarrow \det A \ne 0$ \\
  If $A^{-1}$ exists, then $\det(A^{-1}) = \frac{1}{\det A}$
  \item $\det(\lambda A) = \lambda^n \cdot \det A$
  \item If $A$ and $B$ are similar, i.e., $\exists P \in GL_n(\mathbb{K})$ such that $B = P^{-1}AP$,
  \[
  \det B = \det(P^{-1}AP) = \det(P^{-1}) \cdot \det A \cdot \det P = \det A
  \]
  \item If $f : E \rightarrow E$ is a linear map, and $A = \text{Mat}(f, \mathcal{B}_c)$, then
  \[
  \det f := \det A = \det \text{Mat}(f, \mathcal{B}_c)
  \]
  where $\mathcal{B}_c$ is a basis of $E$.
\end{enumerate}
\section*{Applications: Inverse Matrix Calculation}

Let $A \in \mathcal{M}_n(\mathbb{K})$ with $\det A \neq 0$.

Let $A = (a_{ij})$, and $A_{ij}$ be the minor of $a_{ij}$:  
$A_{ij}$ is the determinant of the matrix obtained by removing row $i$ and column $j$.

The cofactor is $c_{ij} = (-1)^{i+j} A_{ij}$.

Let $C = (c_{ij})$ be the cofactor matrix of $A$.

Then:
\[
A^{-1} = \frac{1}{\det A} \cdot C^T
\]

\subsection*{Example 1}
\[
A = \begin{pmatrix} 1 & 3 \\ 2 & 5 \end{pmatrix}, \quad \det A = -1
\]

\[
\begin{aligned}
&c_{11} = (-1)^{2} \cdot 5 = 5, \quad c_{12} = (-1)^{3} \cdot 2 = -2 \\
&c_{21} = (-1)^{3} \cdot 3 = -3, \quad c_{22} = (-1)^{4} \cdot 1 = 1
\end{aligned}
\]

\[
C = \begin{pmatrix} 5 & -2 \\ -3 & 1 \end{pmatrix}, \quad
A^{-1} = \begin{pmatrix} -5 & 3 \\ 2 & -1 \end{pmatrix}
\]

\subsection*{Example 2}
\[
A = \begin{pmatrix} 2 & 4 & 3 \\ 0 & 2 & 1 \\ 2 & 2 & -1 \end{pmatrix}, \quad \det A = -6
\]

Cofactors:

\[
\begin{aligned}
&c_{11} = -3,\quad c_{12} = 2,\quad c_{13} = -2 \\
&c_{21} = 10,\quad c_{22} = -8,\quad c_{23} = 4 \\
&c_{31} = 1,\quad c_{32} = -2,\quad c_{33} = 2
\end{aligned}
\]

\[
C = \begin{pmatrix} -3 & 2 & -2 \\ 10 & -8 & 4 \\ 1 & -2 & 2 \end{pmatrix}, \quad
A^{-1} = \frac{1}{-6} \cdot
\begin{pmatrix}
-3 & 10 & 1 \\
2 & -8 & -2 \\
-2 & 4 & 2
\end{pmatrix}
\]

\subsection*{Note}
If $D = \begin{pmatrix}
a_{11} & 0 & \cdots & 0 \\
0 & a_{22} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{nn}
\end{pmatrix}$ is diagonal, then:
\[
\det D = \prod_{i=1}^{n} a_{ii}
\]
\section*{Matrix Inversibility and Linear Systems}

Let $D = \text{diag}(a_{11}, a_{22}, \dots, a_{nn})$.  
If $D$ is invertible, then $\det D \neq 0$.

\[
D = \begin{pmatrix}
a_{11} & 0 & \cdots & 0 \\
0 & a_{22} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{nn}
\end{pmatrix}
\]

\subsection*{Matrix Powers}

Let $A \in \mathcal{M}_n(\mathbb{K})$.

\[
A^n = A \cdot A \cdots A \quad (n \text{ times}), \quad n \in \mathbb{N}^*
\]
By convention: $A^0 = I_n$ if $A$ is invertible

\[
(A^{-1})^n = (A^n)^{-1}, \quad \text{for } n \in \mathbb{N}^*
\]

\section*{Applications: Linear Systems}

\subsection*{Definition}

A linear system with $n$ unknowns and $p$ equations is a system of the form:

\[
(S) \left\{<br>\begin{aligned}
a^1_1 x_1 + a^1_2 x_2 + \cdots + a^1_n x_n &= b_1 \\
a^2_1 x_1 + a^2_2 x_2 + \cdots + a^2_n x_n &= b_2 \\
\vdots \\
a^p_1 x_1 + a^p_2 x_2 + \cdots + a^p_n x_n &= b_p \\
\end{aligned}
\right.
\]

With $n, p \in \mathbb{N}^*$ and $a^i_j \in \mathbb{K}, \forall i = 1..p, \forall j = 1..n$

Let:

\[
A = \begin{pmatrix}
a^1_1 & a^1_2 & \cdots & a^1_n \\
a^2_1 & a^2_2 & \cdots & a^2_n \\
\vdots & \vdots & \ddots & \vdots \\
a^p_1 & a^p_2 & \cdots & a^p_n
\end{pmatrix}, \quad
X = \begin{pmatrix}
x_1 \\ x_2 \\ \vdots \\ x_n
\end{pmatrix}, \quad
B = \begin{pmatrix}
b_1 \\ b_2 \\ \vdots \\ b_p
\end{pmatrix}
\]

Then:
\[
(S) \Longleftrightarrow A X = B
\]

\subsection*{Resolution of (S)}

\textbf{Case 1:} $n = p$

\[
A = \begin{pmatrix}
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{nn}
\end{pmatrix}, \quad
X = \begin{pmatrix}
x_1 \\ \vdots \\ x_n
\end{pmatrix}, \quad
B = \begin{pmatrix}
b_1 \\ \vdots \\ b_n
\end{pmatrix}
\]

In this case:

\textbf{(a)} If $\det A \neq 0$, then (S) is a \textbf{Cramer system}.
\section*{(S) admits a unique solution}
The solution of (S) is given by:
\[
x_i = \frac{\det(c_1, c_2, \ldots, c_{i-1}, B, c_{i+1}, \ldots, c_n)}{\det(A)}
\]

\subsection*{Example:}
\[
(S) \left\{<br>\begin{array}{l}
2x + 7y + 3z = 2 \\
3x + 9y + 4z = 2 \\
x + 5y + 3z = 5
\end{array}
\right.
\]

\[
X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad 
B = \begin{pmatrix} 2 \\ 2 \\ 5 \end{pmatrix}, \quad
A = \begin{pmatrix}
2 & 7 & 3 \\
3 & 9 & 4 \\
1 & 5 & 3
\end{pmatrix}
\]

\[
\det A = 2(27 - 20) - 7(9 - 4) + 3(15 - 9) = 14 - 35 + 18 = -3 \neq 0
\]

So (S) is a Cramer system, and it admits a unique solution.

\[
x = \frac{\left|<br>\begin{array}{ccc}
2 & 7 & 3 \\
2 & 9 & 4 \\
5 & 5 & 3
\end{array}
\right|}{-3} = \frac{7 \cdot 9 \cdot 5 - 105}{-3} = 0
\]

\[
y = \frac{\left|<br>\begin{array}{ccc}
2 & 2 & 3 \\
3 & 2 & 4 \\
1 & 5 & 3
\end{array}
\right|}{-3} = \frac{27 - 5 + 39}{-3} = -2
\]

\[
z = \frac{\left|<br>\begin{array}{ccc}
2 & 7 & 2 \\
3 & 9 & 2 \\
1 & 5 & 5
\end{array}
\right|}{-3} = \frac{70 - 9 + 6}{-3} = -5
\]

\[
\Rightarrow x = 0, \quad y = -2, \quad z = -5
\]

\subsection*{b) Case where $\det A = 0$}

\subsection*{Example:}

\[
(S) \left\{<br>\begin{array}{l}
2x + y - 2z = 4 \\
x + 2y - 3z = 6 \\
3x - 3y - 5z = 10
\end{array}
\right.
\]

\[
A = \begin{pmatrix}
2 & 1 & -2 \\
1 & 2 & -3 \\
3 & -3 & -5
\end{pmatrix}, \quad
B = \begin{pmatrix} 4 \\ 6 \\ 10 \end{pmatrix}
\]

\[
\det A = 2(2 \cdot -5 - (-3) \cdot -3) - 1(1 \cdot -5 - (-3) \cdot 3) + (-2)(1 \cdot -3 - 2 \cdot 3) = -2 - 4 + 6 = 0
\]