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We have a set of equations as

a x + b y = c z + d w
e x + f y = g z + h w

We wish to have a matrix as below :

enter image description here

How can we reach the m (matrix)

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    $\begingroup$ Definig equations in Mathematica you must use == instaed of =! $\endgroup$ – Ulrich Neumann Nov 9 '18 at 10:23
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    $\begingroup$ First solve for {z,w} and then extract the appropriate coefficient array. In[27]:= solzw = {z, w} /. First[Solve[{a x + b y == c z + d w, e x + f y == g z + h w}, {z, w}]]; Normal[CoefficientArrays[solzw, {x, y}][[2]]] Out[28]= {{(d e)/(d g - c h) - (a h)/(d g - c h), (d f)/( d g - c h) - (b h)/(d g - c h)}, {-((c e)/(d g - c h)) + (a g)/( d g - c h), -((c f)/(d g - c h)) + (b g)/(d g - c h)}} $\endgroup$ – Daniel Lichtblau Nov 9 '18 at 15:46
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If you rewrite your euations as a list

eq = {-d w + a x + b y - c z, -h w + e x + f y - g z}


M = D[eq, {{x, y}}]
r = eq /. {x -> 0, y -> 0}

check result

M.{x,y}+r ==eq
(* True*) 
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This one helps you

 X = {{x},{y}} ; Z ={{z},{w}};

Consider the equations as

eq1 = a x + b y - c z - d w == 0.0;
eq2 = e x + f y - g z - h w == 0.0;

To find the coefficients of X and Z matrix

In[1]:= am = Normal[CoefficientArrays[{eq1, eq2}, {x, y}]][[2]]

Out[1]= {{a, b}, {e, f}}

In[2]:= bm = Normal[CoefficientArrays[{eq1, eq2}, {z, w}]][[2]]

Out[2]= {{-1. c, -1. d}, {-1. g, -1. h}}

Standard form othe equation will be like

am.X == bm.Z 

{{a x + b y}, {e x + f y}} == {{-1. d w - 1. c z}, {-1. h w - 1. g z}}

to find the equation for your requested form you have to calculate inverse of bm

In[3]:= bmInv = Inverse[bm]

Out[3]:= {{-((1. h)/(-1. d g + 1. c h)), (1. d)/(-1. d g + 1. c h)}, {1. g)/(-1. d g + 1. c h), -((1. c)/(-1. d g + 1. c h))}}

multiply bmInv with am to get matr

mat = MatrixForm[bmInv.am]

lastly you can get the standard form of equation as you expected using below line

mat.X == Z
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