matrix_linear_system_2.m

%% Solving for Two Equations and Two Unknowns
% *back to* <https://fanwangecon.github.io *Fan*>*'s* <https://fanwangecon.github.io/Math4Econ/ 
% *Intro Math for Econ*>*,*  <https://fanwangecon.github.io/M4Econ/ *Matlab Examples*>*, 
% or* <https://fanwangecon.github.io/MEconTools/ *MEconTools*> *Repositories*
%% 
% _See also_: <https://fanwangecon.github.io/Math4Econ/matrix_system_of_equations/matrix_linear_equations.html 
% System of Linear Equations>
% 
% _See also_: <https://fanwangecon.github.io/Math4Econ/matrix_system_of_equations/matrix_linear_system_2.html 
% Solving for Two Equations and Two Unknowns>
% 
% _See also_: <https://fanwangecon.github.io/Math4Econ/matrix_system_of_equations/matrix_row_echelon_form.html 
% System of Linear Equations, Row Echelon Form>
%% Intersection of two Linear Equations
% We have two line:
% 
% $$\begin{cases} y = a + b \cdot x \\ y = c + d \cdot x \end{cases}$$
% 
% Where do these lines intersect? Visually, given some values for $a,b,c,d$:

% Symbol
syms x 
% Parameters
a = 1.1;
b = 2;
c = 2;
d = -1;
% Define Equations
y1 = a + b*x
y2 = c + d*x
% Solve for analytical solutions using symbolic toolbox
solve_analytical_x = double(solve(y1 - y2 == 0));
solve_analytical_y = double(subs(y1, solve_analytical_x));
% Plot Figure
figure();
hold;
fplot(y1)
fplot(y2)
% Labeling
ylabel('y')
xlabel('x')
grid on;
title({'Intersection of 2 lines'...
      ,['a=' num2str(a)...
        ',b=' num2str(b)...
        ',c=' num2str(c)...
        ',d=' num2str(d)]...
       ,['x intersect=',num2str(solve_analytical_x)]...
       ,['y intersect=',num2str(solve_analytical_y)]});
%% Linear Equation in Matrix Form
% Sometimes we can write down our problem as a set of linear equations. A linear 
% equation is an equation where the unknown variables are multiplied by a set 
% of known constants and then added up to a known constant:
%% 
% * for example: $-2\cdot x +  \cdot y = 1$, has two unknowns.
%% 
% Using matrix algebra, we can express the above equation in matrix form:
%% 
% * $\left[ {\begin{array}{cc}-2 & 1 \\\end{array} } \right] \cdot \left[ {\begin{array}{c}x 
% \\ y \end{array} } \right] = -2\cdot x + 1 \cdot y  = 1$
%% Two Linear Equation in Matrix Form
% We have two equations above, we can write both of them using the matrix form, 
% given:
%% 
% * $\begin{cases} y = a + b \cdot x \\ y = c + d \cdot x \end{cases}$
%% 
% We can re-write these as:
%% 
% * $\left[ {\begin{array}{cc} 1 & -b \\ 1 & -d \\ \end{array} } \right] \cdot 
% \left[ {\begin{array}{c} y \\ x \end{array} } \right] = \left[ {\begin{array}{cc} 
% 1\cdot y - b \cdot x \\ 1\cdot y - d \cdot x \\ \end{array} } \right]  = \left[ 
% {\begin{array}{c} a \\ c \\ \end{array} } \right]$
%% 
% We can define these following matrixes to simplify notations:
%% 
% * $W = \left[ {\begin{array}{cc} 1 & -b \\ 1 & -d \\ \end{array} } \right]$
% * $\mathbf{X} = \left[ {\begin{array}{c}x \\ y \end{array} } \right]$, note 
% the use of bold letter to represent a vector of unknowns, we could have called 
% small $x$ and $y$, $x_1$ and $x_2$.
% * $v = \left[ {\begin{array}{c} a \\ c \\ \end{array} } \right]$
%% 
% And the linear system of equations is:
%% 
% * $W\cdot\mathbf{X} = v$
%% Linsolve: Matlab Matrix Solution for 2 Equations and Two Unknowns 
% Once you have transformed a system of equations, you can use matlab's linsolve 
% function to solve for the unknowns. As long as the two lines are not parallel 
% to each other, you will be able to find solutions:

W = [1, -b;1, -d]
v = [a; c]
solution = linsolve(W,v)
yIntersection = solution(1,1)
xIntersection = solution(2,1)
%% 
% The solution here should match the number in title of the graph plotted earlier.
% 
% When you do not have matlab,  you can solve for the optimal choices using 
% a combination of elementary row operations. 
% 
% _Note_: If we used elementary row operations, and arrived at the reduced row 
% echelon form, the analytical solution would be (and this is what linsolve is 
% doing):

% Analytical Results using elementary row operations
yIntersectionEro = a + b*(c-a)/(b-d)
xIntersectionEro = (c-a)/(b-d)
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