a ‘useful’ number associated with a square matrix

Example: System of equations-

$\begin{gathered} x+y=10 \\ 4 x=y \end{gathered}$

$\Rightarrow x+y=10$

$\Rightarrow 4 x-y=0$

$\left[\begin{array}{ll} 1 & 1 \\ 4 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]$= $\left[\begin{array}{c} 10 \\ 0 \end{array}\right]$

$\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]$= $\left[\begin{array}{c} M \\ N \end{array}\right]$

$\Rightarrow 2 \times 2$ square matrix

$\begin{gathered} d \times[a x+b y=M \\ -b \times[c x+d y=N \\ \Rightarrow(a d-b c) x=d M-b N \\ \text { if } a d-b c \neq 0 \\ \Rightarrow x=\frac{d M-b N}{a d-b c} \end{gathered}$

Similarly , for $y$,

$\begin{aligned} a x+b y & =M \\ c x+d y & =N \end{aligned}$

$\Rightarrow y=\frac{c M-a N}{b c-a d}$

This is for $a d-b c \neq 0$

$\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} M \\ N \end{array}\right]$

solutions found when $a d-b c \neq 0$

Thus, this is the determinant of the $2\times2$ matrix $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

Determinant provides a way to check for the existence of solution in a linear system of equations

Example 2: deteteminant as an area $\begin{aligned} & A=\left[\begin{array}{l} a \\ c \end{array}\begin{array}{l} b \\ d \end{array}\right] \\ & \text { area }=a d-b c \ & \end{aligned}$

$\begin{aligned} & \Delta=(a+b)(c+d)-a c-b d-2 b c \\ &=a c+a d+b c+b d-a c-b d-2 b c \\ &=a d-b c \\ & {\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \rightarrow \Delta = a d-b c } \end{aligned}$

Determinant gives the ‘area’ of parallelogram formed from the columns of a matrix

Go back to initial example

$\left[\begin{array}{ll} 1 & 1 \\ 4 & -1 \end{array}\right]_{2\times2}\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 10 \\ 0 \end{array}\right]$

determinant is

$1 \times(-1)-1 \times 4$

$=-1-4=-5 \neq 0$

$\therefore$ Solution exists.

Also, area is $|-5|$(absolute value)$=5$

Notation: $\operatorname{det}(A)$ or $|A|$

$A$ is a square matrix

$\longrightarrow$ Define a determinant

$\longrightarrow$ Properties

$\longrightarrow$ Applications

Definition

$A_{n \times n}=\left[a_{i j}\right]_{n\times n}$

$i^{th}$row , $j^{th}$ column

$\begin{gathered} =\left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end{array}\right] \end{gathered}$

$\quad \quad \quad j^{th} column$

$= i^{th} row\left[\begin{array}{ccc}& \vdots & \\ \cdots & a_{ij} & \cdots \\ & \vdots & \end{array}\right]$

Goal: $\operatorname{det}(A)$ in terms of $a_{i j}$

Cofactor , $A_{ij}$ is defined as $A_{i j}=(-1)^{i+j} M_{i j}$

In example before, $\begin{aligned} A_{12} & =(-1)^{1+2} \cdot M_{12} \\ & =(-1)^3 M_{12} \\ & =-M_{12} \\ & =-4 . \end{aligned}$

Example: $1 \times 1$ matrix which is a scalar. Its determinant was taken as the scalar itself

Example: $2 \times 2$ matrix $\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=a \cdot d+b\cdot (-c)$

cofactor of ' $a$ ': $d(-1)^{1+1}=d$

cofactor of ' $b$ ': $\quad c(-1)^{1+2}=-c$

Let's expand a column, $\begin{aligned} {\left[\begin{array}{ll} a & b \downarrow \ c & d \end{array}\right] } & =b(-c)+d(a) \ & =a d-b c ! ! \end{aligned}$

which is same as before

$\longrightarrow$ formal definition of determinant

$\longrightarrow$ a number to check existance of solutions

$\longrightarrow$ as area of parallelogram

Example: $3 \times 3$ matrix

${\left[\begin{array}{ccc} 1 & 0 & 2 \\ 3 & -1 & 2 \\ 5 & 2 & 0 \end{array}\right] }$ $\operatorname{det} ?$

$\operatorname{det}$ $= 1 \times(-1)^{1+1}\left|\begin{array}{cc} -1 & 2 \\ 2 & 0 \end{array}\right|$ $+0 \times 0 +2 \times(-1)^{1+3}\left|\begin{array}{cc} 3 & -1 \\ 5 & 2\end{array}\right|$

$= 1(-1 \times 0-2 \times 2)+2(3 \times 2-(-1) \times 5) \\ = -4+2(6+5)\\=-4+22=18$