=== Example ===

\begin{align*}\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\det\mat{1&2&3\\7&8&9\\11&12&13} &= 1\cdot C_{11} + 2 C_{12} + 3 C_{13}\\
&= 1 \cdot (+M_{11}) + 2 \cdot (-M_{12}) + 3 \cdot(+M_{13})\\
&= M_{11}-2M_{12}+3M_{13}\\
&= \det\mat{8&9\\12&13} -2\det\mat{7&9\\11&13} + 3\det\mat{7&8\\11&12}\\
&= (8\cdot 13-9\cdot 12) -2(7\cdot 13-9\cdot 11)+3(7\cdot 12-8\cdot 11)\\
&=-4 -2(-8)+3(-4)\\
&=-4+16-12\\
&=0.\end{align*}

From this, we can conclude that $\mat{1&2&3\\7&8&9\\11&12&13}$ is //not// invertible.

=== Notation ===

To save having to write $\det$ all the time, we sometimes write the entries of a matrix inside vertical bars $|\ |$ to mean the determinant of that matrix. Using this notation (and doing a few steps in our heads), we can rewrite the previous example as:

\begin{align*}\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\vm{1&2&3\\7&8&9\\11&12&13} &= 1\vm{8&9\\12&13} -2\vm{7&9\\11&13} + 3\vm{7&8\\11&12}\\
&=-4 -2(-8)+3(-4)\\
&=0.\end{align*}

==== Step 4: the determinant of an $n\times n$ matrix ====


===Definition===

{{page>determinant of an nxn matrix}}

===Example===

\begin{align*}
\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}
\vm{\color{red}1&\color{red}0&\color{red}2&\color{red}3\\0&2&1&-1\\2&0&0&1\\3&0&4&2} &= \color{red}1\vm{\color{blue}2&\color{blue}1&\color{blue}-1\\0&0&1\\0&4&2}-\color{red}0\vm{0&1&-1\\2&0&1\\3&4&2}+\color{red}2\vm{\color{orange}0&\color{orange}2&\color{orange}{-1}\\2&0&1\\3&0&2}-\color{red}3\vm{\color{purple}0&\color{purple}2&\color{purple}1\\2&0&0\\3&0&4}\\
&= 1\left(\color{blue}2\vm{0&1\\4&2}-\color{blue}1\vm{0&1\\0&2}\color{blue}{-1}\vm{0&0\\0&4}\right)-0+2\left(\color{orange}0-\color{orange}{2}\vm{2&1\\3&2}\color{orange}{-1}\vm{2&0\\3&0}\right)-3\left(\color{purple}0-\color{purple}2\vm{2&0\\3&4}+\color{purple}1\vm{2&0\\3&0}\right)\\
&=1(2(-4)-0-0)+2(-2(1)-0)-3(-2(8)+0)\\
&=-8-4+48\\
&=36.
\end{align*}

==== Theorem: Laplace expansion along any row or column gives the determinant ====

  - For any fixed $i$: $\det(A)=a_{i1}C_{i1}+a_{i2}C_{i2}+\dots+a_{in}C_{in}$ (Laplace expansion along row $i$)
  - For any fixed $j$: $\det(A)=a_{1j}C_{1j}+a_{2j}C_{2j}+\dots+a_{nj}C_{nj}$ (Laplace expansion along column $j$)

=== Example ===

We can make life easier by choosing expansion rows or columns with lots of zeros, if possible. Let's redo the previous example with this in mind:

\begin{align*}
\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}
\vm{1&\color{red}0&2&3\\0&\color{red}2&1&-1\\2&\color{red}0&0&1\\3&\color{red}0&4&2} &= 
-\color{red}0+\color{red}2\vm{1&2&3\\\color{purple}2&\color{purple}0&\color{purple}1\\3&4&2}-\color{red}0+\color{red}0\\
&=2\left(-\color{purple}2\vm{2&3\\4&2}+\color{purple}0-\color{purple}1\vm{1&2\\3&4}\right)\\
&=2(-2(-8)-(-2))\\
&=36.
\end{align*}

==== Definition: upper triangular matrices ====

An $n\times n$ matrix $A$ is **upper triangular** if all the entries below the main diagonal are zero.


==== Definition: diagonal matrices ====

An $n\times n$ matrix $A$ is **diagonal** if the only non-zero entries are on its main diagonal.

==== Corollary: the determinant of upper triangular matrices and diagonal matrices ====

  - The determinant of an upper triangular $n\times n$ matrix is the product of its diagonal entries: $\det(A)=a_{11}a_{22}\dots a_{nn}$.
  - The determinant of an $n\times n$ diagonal matrix is the product of its diagonal entries: $\det(A)=a_{11}a_{22}\dots a_{nn}$.

=== Proof ===

  - This is true for $n=1$, trivially. For $n>1$, assume inductively that it is true for $(n-1)\times (n-1)$ matrices and use the Laplace expansion of an upper triangular $n\times n$ matrix $A$ along the first column of $A$ to see that $\det(A)=a_{11}C_{11}+0+\dots+0=a_{11}C_{11}$. Now $C_{11}$ is the determinant of the $(n-1)\times (n-1)$ matrix formed by removing the first row and and column of $A$, and this matrix is upper triangular with diagonal entries $a_{22},a_{33},\dots,a_{nn}$. By our inductive assumption, we have $C_{11}=a_{22}a_{33}\dots a_{nn}$. So $\det(A)=a_{11}C_{11}=a_{11}a_{22}a_{33}\dots a_{nn}$ as desired.
  - Any diagonal matrix is upper triangular, so this is a special case of statement 1. ■

==== Examples ====

  - For any $n$, we have $\det(I_n)=1\cdot 1\cdots 1 = 1$.
  - For any $n$, we have $\det(5I_n)=5^n$.
  - $\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\vm{1&9&43&23434&4&132\\0&3&43&2&-1423&-12\\0&0&7&19&23&132\\0&0&0&2&0&0\\0&0&0&0&-1&-903\\0&0&0&0&0&6}=1\cdot3\cdot7\cdot2\cdot(-1)\cdot6 = 252$.


==== Theorem: important properties of the determinant ====

Let $A$ be an $n\times n$ matrix. 

  - $A$ is invertible if and only if $\det(A)\ne0$.
  - $\det(A^T)=\det(A)$
  - If $B$ is another $n\times n$ matrix, then $\det(AB)=\det(A)\det(B)$

==== Corollary on invertibility ====

  - $A^T$ is invertible if and only if $A$ is invertible
  - $AB$ is invertible if and only if **both** $A$ and $B$ are invertible

=== Proof ===

  - We have $\det(A^T)=\det(A)$. So $A^T$ is invertible $\iff$ $\det(A^T)\ne0$ $\iff$ $\det(A)\ne 0$ $\iff$ $A$ is invertible.
  - We have $\det(AB)=\det(A)\det(B)$. So $AB$ is invertible $\iff$ $\det(AB)\ne 0$ $\iff$ $\det(A)\det(B)\ne0$ $\iff$ $\det(A)\ne0$ and $\det(B)\ne 0$ $ \iff$ $A$ is invertible and $B$ is invertible. ■