Linear Algebra Notes - Determinants

by Keith50, May 12, 2021, 12:34 PM

$\Large \textbf{\color{blue}{Linear Algebra - Determinants}}$
$\large \textbf{\color{magenta}{(a) 2-order and 3-order Determinants}}$
We know that if we are given a system of two linear equations $\begin{cases} a_{11}x_1+a_{12}x_2=b_1 \\ a_{21}x_1+a_{22}x_2=b_2 \end{cases}$ we can solve it by elimination method and get: $x_1=\frac{b_1a_{22}-a_{12}b_2}{a_{11}a_{22}-a_{12}a_{21}}, x_2=\frac{a_{11}b_2-a_{21}b_1}{a_{11}a_{22}-a_{12}a_{21}}$ notice that the expressions in the numerator and the denominator of $x_1,x_2$ consist of $4$ numbers with $2$ numbers multiplied together respectively and subtracted with the other. In particular, the numbers in the expression $a_{11}a_{22}-a_{12}a_{21}$ are the coefficients in our system of two linear equations, so let's put them in rows and columns of a matrix, $\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ we can then represent $a_{11}a_{22}-a_{12}a_{21}$ as a $2 \times 2$ determinant, $\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}.$ Therefore, we can rewrite $x_1,x_2$ as $x_1=\frac{\begin{vmatrix} b_{1} & a_{12} \\ b_{2} & a_{22} \end{vmatrix}}{\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}}, x_2=\frac{\begin{vmatrix} a_{11} & b_{1} \\ a_{21} & b_{2} \end{vmatrix}}{\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}}.$ Now, we then define the 3x3 determinant as $\begin{vmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}= a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32}.$ $\large \textbf{\color{magenta}{(b) Permutations and Inversions}}$
Let's consider the following example:
Given three numbers $1,2,3$ , how many different arrangements are there for these numbers?
It's easy, there are $3!=6$ of these, $123, 231, 312, 132, 213, 321.$ In general, these numbers like in the example above, $1,2,3$ are called elements. So, given $n$ different elements, there are $n!$ arrangements and these arrangements are also called permutations of the elements.
We then define the inversion number of a permutation. First, an inversion in a permutation is a pair of numbers such that the larger number appears to the left of the smaller one in the permutation. The inversion number of a permutation is the total number of inversions. One way to help calculate the inversion number is to look at each position in the permutation and count how many bigger numbers are to the left, and then add those numbers up.

Exercise wrote:

Calculate the inversion number of the permutation, $9236815$ .
Answer

Lastly, we say that a permutation is odd if the inversion number is odd and a permutation of even if the inversion number is even.
$\large \textbf{\color{magenta}{(c) The Meaning of n-order Determinants}}$
Let's go back to the $3\times 3$ determinant, $\begin{vmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}= a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32}.$ It's easy to see that there are three elements in each term of the above expression and these three elements are from different row and columns. Thus, each term of the expression (neglecting the positive and negative signs) can be rewritten as $a_{1p_1}a_{2p_2}a_{3p_3}$ . Here, the row numbers of each term are assigned $123$ and the column numbers of each term are assigned $p_1p_2p_3$ , there are the permutations of elements $1,2,3$ . There are $6$ permutations which is the same as the number of terms.
By a close inspection, we realise that terms with positive sign have the column numbers being the permutations, $123,231,312$ and terms with negative sign are the permutations $132,213,321$ . Also, by some calculations, we find out that the first three are even permutations and the latter are odd permutations. Therefore, the sign of each term can be represented as $(-1)^t$ , where $t$ is the inversion number of the column numbers of each term.
Hence, the $3-$ order determinant can be rewritten as $\begin{vmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}= \sum (-1)^t a_{1p_1}a_{2p_2}a_{3p_3},$ where $t$ is the inversion number of the permutation $p_1p_2p_3$ .
In general, given a $n-$ order matrix, $\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n} \\ & \cdots & \cdots &\\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{pmatrix}$ we can form products of $n$ elements from different rows and columns and with their respective signs $(-1)^t$ , $(-1)^ta_{1p_1}a_{2p_2}\cdots a_{np_n}$ . The numbers $p_1p_2\cdots p_n$ are permutations of $1,2, \cdots ,n$ and $t$ as the inversion number. Since there are $n!$ permutations, there are total $n!$ terms and their sum $\sum (-1)^t a_{1p_1}a_{2p_2}\cdots a_{np_n}$ is the $n-$ order determinant, denoted by $D=\begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{vmatrix}.$

Exercise wrote:

Prove the following $n-$ order determinant $\begin{vmatrix}\lambda_1 & &\\ & \lambda_2 & \\ & &\ddots & \\ & & & \lambda_n \end{vmatrix}=\lambda_1 \lambda_2 \cdots \lambda_n$ $\begin{vmatrix} & & & \lambda_1 \\ & & \lambda_2 \\ & \iddots \\ \lambda_n & \end{vmatrix}=(-1)^{\frac{n(n-1)}{2}}\lambda_1 \lambda_2 \cdots \lambda_n,$ where the unfilled elements are all $0$ .

Exercise wrote:

Prove the following determinant of a lower triangular matrix: $D=\begin{vmatrix} a_{11} \\ a_{21} & a_{22} \\ \vdots & \vdots & \ddots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{vmatrix}=a_{11}a_{22}\cdots a_{nn}.$

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linear algebra

determinant

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Keith50

Linear Algebra Notes - Determinants

by Keith50, May 12, 2021, 12:34 PM

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