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Table of Contents
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
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 $\det(A)=a_{11}C_{11}+0+\dots+0=a_{11}C_{11}$. 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 $\det(A)\ne0$.
- 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 $\text{row}_i(A'')=\text{row}_i(A)+\text{row}_i(A')$, 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 $\det(AB)=\det(A)\det(B)$.
Corollary
If an $n\times n$ matrix $A$ has two equal rows (or columns), then $\det(A)=0$, and $A$ is not invertible.
Proof
Suppose $A$ has two equal rows. Let $A'$ be $A$ with the two equal rows swapped. By property 2, we have $\det(A')=-\det(A)$. But since these two rows are equal, we have $A'=A$! So $\det(A)=-\det(A)$. Solving this for $\det(A)$ gives $\det(A)=0$, and hence $A$ is not invertible, by property 1.
If $A$ has two equal columns, then $A^T$ has two equal rows, so $\det(A^T)=0$, so $\det(A)=0$ by property 5. ■
Examples
- Swapping two rows changes the sign, so $\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\vm{0&0&2\\0&3&0\\4&0&0} = -\vm{4&0&0\\0&3&0\\0&0&2}=-4\cdot 3\cdot 2 = -24$.
- Multiplying a row or a column by a constant multiplies the determinant by that constant, so \begin{align*}\vm{ 2&4&6&10\\5&0&0&-10\\9&0&81&99\\1&2&3&4} &= 2\vm{ 1&2&3&5\\5&0&0&-10\\9&0&81&99\\1&2&3&4} \\&= 2\cdot 5\vm{ 1&2&3&5\\1&0&0&-2\\9&0&81&99\\1&2&3&4}\\&= 2\cdot 5\cdot 9 \vm{ 1&2&3&5\\1&0&0&-2\\1&0&9&11\\1&2&3&4}\\&=2\cdot 5\cdot 9\cdot 2 \vm{ 1&1&3&5\\1&0&0&-2\\1&0&9&11\\1&1&3&4}\\&=2\cdot 5\cdot 9\cdot 2\cdot 3 \vm{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4}.\end{align*}
- The matrices $\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}A''=\mat{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4}$, $A=\mat{ 0&0&0&1\\1&0&0&-2\\1&0&3&11\\1&1&1&4}$ and $A'=\mat{ 1&1&1&4\\1&0&0&-2\\1&0&3&11\\1&1&1&4}$ are the same except in row $1$, and $\text{row}_1(A'')=\text{row}_1(A)+\text{row}_1(A')$. Moreover, $A'$ has two equal rows, so has determinant zero. Hence by property 4 in the theorem, \begin{align*}\vm{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4} &=\det(A'')=\det(A)+\det(A')\\&=\vm{ 0&0&0&1\\1&0&0&-2\\1&0&3&11\\1&1&1&4}+\vm{ 1&1&1&4\\1&0&0&-2\\1&0&3&11\\1&1&1&4}\\&=-1\vm{1&0&0\\1&0&3\\1&1&1}+0\\&=-\vm{0&3\\1&1} = -(-3)=3.\end{align*}
- Hence \[\vm{ 2&4&6&10\\5&0&0&-10\\9&0&81&99\\1&2&3&4} = 2\cdot 5\cdot 9\cdot 2\cdot 3 \vm{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4} = 2\cdot 5\cdot 9\cdot 2\cdot 3 \cdot 3 = 1620.\]