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I define three matrices:

A1={{-2, -5}, {1, 4}}
A2={{2, 3}, {-1, -2}}
A3={{3, -9}, {2, 3}}

Then I plot their solution curves using parametric plot:

Grid@Table[{ParametricPlot[A . {x, y}, {x, -3, 3}, {y, -3, 3},
PlotLabel -> Row[{"A = ", A}]]}, {A, {A1, A2, A3}}]

But the result, although intriguing is bizarre, it does not look like solution curves at all. Any idea what happened, and how to get solution curves?

Thanks

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    $\begingroup$ What are "solution curves" ? Perhaps ContourPlot[A1 . {x, y}, {x, -3, 3}, {y, -3, 3}] is what you are looking for? $\endgroup$ Commented Oct 13, 2022 at 9:50
  • $\begingroup$ You are right, this is what I was looking for. Thanks! $\endgroup$ Commented Oct 13, 2022 at 9:54
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    $\begingroup$ You are welcome! $\endgroup$ Commented Oct 13, 2022 at 10:08
  • $\begingroup$ You are making a parametric plot with 2 variables. This specifies a area, not a curve. What you see is actually how a rectangle of 6x6 gets deformed by the linear transformation given by A. $\endgroup$ Commented Oct 13, 2022 at 10:48

1 Answer 1

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Using Mesh -> {{0}}, MeshFunctions -> {#3 &, #4 &} to draw the line of A . {x, y}==0

Table[ParametricPlot[A . {x, y}, {x, -3, 3}, {y, -3, 3}, 
  Mesh -> {{0}}, MeshFunctions -> {#3 &, #4 &}, 
  MeshStyle -> Directive@{Thick, Red}, PlotStyle -> Yellow, 
  PlotLabel -> Row[{"A = ", A}]], {A, {A1, A2, A3}}]
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