Idea

$\underbrace{a}_ {\text{scalar}} ~ \underbrace{x}_ {\text{unknown}} = \underbrace{b}_ {\text{scalar}}$

x unknown that needs to be solved

$a \neq 0, x=b / a$

generalize these

$\begin{aligned} & a x+b y=m\end{aligned} ~, \begin{aligned} & c x+d y=n\end{aligned}$

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

are unknowns solutions?

$A x=B \Rightarrow {x}=$ ?

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

Goal: How to solve for $x$ ?

Notation: $\underline{x} \equiv x$

(in appropriate context, we use ' $x$ ' to denote the vector value)

Example:

$\begin{aligned} & a_{11} x+a_{12} y+a_{13} z=b_1 \end{aligned}$ $\longrightarrow$ (1)

$\begin{aligned} & a_{21} x+a_{22} y+a_{23} z=b_2 \end{aligned}$ $\longrightarrow$ (2)

$\begin{aligned} & a_{31} x+a_{32} y+a_{33} z=b_3 \end{aligned}$ $\longrightarrow$ (3)

$\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right] \left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right]$

$A \times x = B$

In general, $n \times n$($A$) $n \times1$($x$) $n \times1$($B$)

$A \underline{x}=B \rightarrow \text { System of Equations }$

Consistency: A system of equations is said to be consistent if a solution exists (one or more solutions)

Inconsistency: A system of equations is said to be inconsistent if a solution does not exist

$A \underline{x}=B$

consistency, inconsistency?

How to check?

Case 1: A is non. singular $(\operatorname{det}(A) \neq 0)$ $\Rightarrow A^{-1}$ exists

$A^{-1} \times [A \underline{x}=B$

$\Rightarrow \underbrace{{A}^{-1}{A}} \underline{x}=A^{-1} B$

$\Rightarrow \underline{x}=A^{-1} B$

Case 2: $A$ is singular $(det(A)=0)$

$(adj A) A= det(A) \cdot I=0$

$\rightarrow \times[A \underline{x}=B$

$\Rightarrow(\underbrace{adj A) A}_0 \underline{x}=(adj A) B$

$\Rightarrow 0 = (adj A) B$

(2a) $y(adj A) B=0$

then we cannot say anything about consistency or inconsistency$\rightarrow(2 b) 4(\operatorname{adj} A) B \neq 0$

then we say is 'inconsistent'

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

$\ A \times x = B$

$\operatorname{det}(A)=-1-4=-5 \neq 0$

$\Rightarrow$ This has a solution $\Rightarrow$ consistent

what is oration? $\underline{x}=A^{-1} B$

$\ A^{-1}=\frac{-1}{5}\left[\begin{array}{cc}-1 & -1 \\ -4 & 1\end{array}\right]$

$\Rightarrow \ A^{-1} B=-\frac{1}{5}\left[\begin{array}{cc}-1 & -1 \\ -4 & 1\end{array}\right]\left[\begin{array}{c}10 \\ 0\end{array}\right]$

$\Rightarrow \ \underline{x}=-\frac{1}{5}\left[\begin{array}{l}-10 \\ -40\end{array}\right]=\left[\begin{array}{l}2 \\ 8\end{array}\right]$

Check: $\underline{x}=\left[\begin{array}{l}2 \\ 8\end{array}\right]$ satisfies the equation?

$x+y=10$

$4x-y=0$

$\therefore$ This solution satisfies be original equations.

Geometry of this example

$x+y=10$

$4x-y=0$

The point of intersection satisfies the equations of both the lines.

An inconsistent system?

$\ x + y = 10 ~ \text{and}~ \ x + y = 20$

$\ \text {}\left[\begin{array}{l} 1 & 1 \\ 1 & 1\end{array}\right]\text {}\left[\begin{array}{l} x \\y \end{array}\right]=\text {}\left[\begin{array}{l}10 \\ 20 \end{array}\right]$

$\ A \times x = B$

$\ \operatorname{det}(A)=1-1=0$

$\ \operatorname{adj}(A)=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right] \times\left[\begin{array}{ll}A & \underline{x}\end{array}\right]=B$

check adj $\ (A) \cdot B=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}10 \\ 20\end{array}\right]=\left[\begin{array}{c}-10 \\ 10\end{array}\right]$

$\Rightarrow \ 0=\left[\begin{array}{c}-10 \\ 10\end{array}\right]$?? $\Rightarrow \text { inconsistent }$

Infinitily many solutions?

$x+y =10$

$2x+2y =10 \times 2=20$

$\text {}\left[\begin{array}{l} 1 & 1 \\ 2 & 2\end{array}\right] \text {}\left[\begin{array}{l} x \\y \end{array}\right]=\text {}\left[\begin{array}{l}10 \\ 20 \end{array}\right]$

$A \times x = B$

$det(A)=0$

$\begin{aligned}\operatorname{adj}(A)=\left[\begin{array}{cc}2 & -1 \\ -2 & 1\end{array}\right]\end{aligned}$

$\begin{aligned}\operatorname{adj}(A). B =\left[\begin{array}{cc}0 \\ 0\end{array}\right]\end{aligned}$

as per the procedure, comment concluds about consistant/inconsistent from here

Previous example

3 variations

summary,

1 $\operatorname{det}(A) \neq 0 \Rightarrow \underline{x}=A^{-1} B$ (one point of intersection)

2 (b) $\operatorname{det}(A)=0 \rightarrow$ (inconsistent) (parallel lines)

2 (a) $\operatorname{det}(A)=0 \rightarrow$ no conclusion ( some line $\Rightarrow$ infinitily many solutions)

Briefly,

$\alpha x+2 y=\lambda$

$3 x-2 y=\mu$

$\alpha, \lambda, h \in \mathbb{R}$

Is the following true?

if $\alpha \neq-3$ then the system has a only one unique solution for all $\lambda$ & $\mu$.

$\left[\begin{array}{cc}\alpha & 2 \\ 3 & -2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}\lambda \\ \mu\end{array}\right]$

$A \times x = B$

$\operatorname{det}(A)=-2 \alpha-6$

$\operatorname{det}(A) \neq 0$ if $\alpha \neq-3$ $(\alpha \neq-3)$

if $\operatorname{det}(A) \pm 0 \Rightarrow$ consistent system.

$\underline{x}=A^{-1} B=\frac{1}{-2 \alpha-6} \left[\begin{array}{ll}-2 & -2 \\ -3 & \alpha\end{array}\right]\left[\begin{array}{l}\lambda \\ \mu\end{array}\right]$

$=\frac{-1}{2 \alpha+6}\left[\begin{array}{l}-2 \lambda-2 \mu \\ -3 \lambda+\alpha \mu\end{array}\right]$

one solution for given $\lambda \mu \alpha(\neq-3)$. Statement is true