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Table of Contents
Step 4: the determinant of an $n\times n$ matrix
Definition
If $\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}A=\mat{a_{11}&a_{12}&\dots&a_{1n}\\\vdots&&&\vdots\\a_{n1}&a_{n2}&\dots&a_{nn}}$ is an $n\times n$ matrix, then \[\det A=a_{11}C_{11}+a_{12}C_{12}+\dots+a_{1n}C_{1n}.\] Here $C_{ij}$ are the cofactors of $A$.
This formula is called the Laplace expansion of $\det A$ along the first row, since $a_{11}, a_{12},\dots,a_{1n}$ make up the first row of $A$.
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 $\det(A)=a_{11}C_{11}+0+\dots+0=a_{11}C_{11}$. Since $C_{11}$ 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 $a_{22},a_{33},\dots,a_{nn}$, we have $C_{11}=a_{22}a_{33}\dots a_{nn}$, by our inductive assumption. 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 $\det(5I_n)=5^n$.
Theorem
Let $A$ be an $n\times n$ matrix.
- $A$ is invertible if and only if $\det(A)\ne0$.
- If $A'$ is the same as $A$, except with two rows swapped, then $\det(A')=-\det(A)$.
- If $c$ is a scalar and $A'$ is the same as $A$ except with one row multiplied by $c$, then $\det(A')=c\det(A)$.
- If $A'$ and $A''$ are the same as $A$ except in row $i$, and $row_i(A'')=row_i(A)+row_i(A')$, then $\det(A'')=\det(A)+\det(A')$.
- $\det(A^T)=\det(A)$. So we can swap “row” with “column” in these properties.
- If $B$ is another $n\times n$ matrix, then $\det(AB)=\det(A)\det(B)$.