Ax+by=c; dx+ey=f

[DivineNathicana]

For the equations

 

ax+by=c

and

dx+ey=f,

 

explain why the coordinate pair (x,y) can be derived by completing

 

((ce-bf)/ae-bd), (af-cd)/(ae-bd)).

 

I don't need to prove that it works - I need to explain why it does. Help! = /

 

- Alisa

[maddog]

First, if you could explain why you could prove. These are nearly the same.

 

however, one thing I see if you think of this system of equations as a matrix

 

( a b )

(d e)

 

the Determinant of this matrix becomes your norm |M| = ae - bd.

I thought I had the rest. I notice that the numerators are the

Determinant of the solution for X and Y

 

|X| = bf - ce where X = {b c | e f}

|Y| = af - cd where Y = {a c| d f}

 

(- (ce - bf), af - cd) = W

 

The normal vector W/ |M| is your answer.

 

Maddog

[Qfwfq]

your pair of equations may be regarded as a matrix applied to a column (x, y) giving another column (c, d) where, by applying a matrix to a column, we mean a row-by-column product:

 

a b | x = c

d e | y = f

 

This translates exactly into your two equations: the first row (a, :hyper: by the column (x, y) equals c. Doing likewise with the second row (d, e) equals f.

 

Of course, you have c and d and must find x and y, quite the opposite of the above matrix equation. Therefore you want to invert the matrix, which can be done if and only if it's determinant D = ae - bd isn't zero. Inverting the matrix gives:

 

e/D -d/D

 

-b/D a/D

 

and applying it to the column (c, f) to give the column (x, y), instead of vice-versa, gives the recipe.

 

How did I invert that matrix? Wow! That's a military secret! :cup:

[Bo]

altough the determinantmethod is complketely correct, it is perhaps more insightfull to jst solve this algabraicly:

(for y):

dax/a+ey=f

substitute x:

dc/a-dby/a+ey=f

solve for y:

(ea-db)/a *y = f-dc/a

y=(fa-dc)/(ea-db)

 

Bo