1
$\begingroup$

I want to solve the following linear equations:

$$\left\{\begin{aligned} x_{1}+3 x_{2}+x_{3} &=0 \\ 2 x_{1}+6 x_{2}+3 x_{3}-2 x_{4} &=0 \\ -2 x_{1}-6 x_{2}-4 x_{4} &=0 \end{aligned}\right.$$

The form of the solution given in the textbook is as follows:

$$\boldsymbol{X}=\left(\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}\right)=\left(\begin{array}{rr} -3 k_{1}-2 k_{2} \\ k_{1} & 2 k_{2} \\ k_{2} \end{array}\right)=k_{1}\left(\begin{array}{r} -3 \\ 1 \\ 0 \\ 0 \end{array}\right)+k_{2}\left(\begin{array}{r} -2 \\ 0 \\ 2 \\ 1 \end{array}\right), \quad k_{1}, k_{2} \text { are arbitrary constants }$$

Solve[{x1 + 3 x2 + x3 == 0, 
  2 x1 + 6 x2 + 3 x3 - 2 x4 == 0, -2 x1 - 6 x2 - 4 x4 == 0}, {x1, x2, 
  x3, x4}]
LinearSolve[{{1, 3, 1, 0}, {2, 6, 3, -2}, {-2, -6, 0, -4}}, {0, 0, 0}]
NullSpace[{{1, 3, 1, 0}, {2, 6, 3, -2}, {-2, -6, 0, -4}}]

$$\left\{\begin{aligned} x_{1}+x_{2}-2 x_{3}-x_{4}=& 4 \\ 3 x_{1}-2 x_{2}-x_{3}+2 x_{4}=& 2 \\ 5 x_{2}+7 x_{3}+3 x_{4}=-2 \\ 2 x_{1}-3 x_{2}-5 x_{3}-x_{4}=& 4 \end{aligned}\right.$$

The form of the solution given in the textbook is as follows:

$$\boldsymbol{X}=\left(\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}\right)=\left(\begin{array}{r} 1 \\ 1 \\ -1 \\ 0 \end{array}\right)+k\left(\begin{array}{r} -\frac{2}{3} \\ \frac{1}{3} \\ -\frac{2}{3} \\ 1 \end{array}\right)=\left(\begin{array}{r} 1 \\ 1 \\ -1 \\ 0 \end{array}\right)+k_{1}\left(\begin{array}{r} -2 \\ 1 \\ -2 \\ 3 \end{array}\right), \quad k, k_{1} \text { are arbitrary constants }$$

Solve[{{1, 1, -2, -1}, {3, -2, -1, 2}, {0, 5, 7, 
     3}, {2, -3, -5, -1}}.{x1, x2, x3, x4} == {4, 2, -2, 4}, {x1, x2, 
  x3, x4}]
LinearSolve[{{1, 1, -2, -1}, {3, -2, -1, 2}, {0, 5, 7, 
   3}, {2, -3, -5, -1}}, {4, 2, -2, 4}]
NullSpace[{{1, 1, -2, -1}, {3, -2, -1, 2}, {0, 5, 7, 
   3}, {2, -3, -5, -1}}]

But the solution form of the above code output is not the style of the textbook. How can we make the LinearSolve function output consistent with the result form of the textbook (arbitrary constant can be replaced by c)?

In addition, I have another problem. To prove that two linear equations have the same solution, we need the row vectors of their augmented matrices to be equivalent to each other.

For example, to prove that the linear equations represented by the following two matrices have the same system of solutions, I need to remove the meaningless zero row vector at the end:

RowReduce[{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}]
RowReduce[{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {2, 4, 6}}]

I wonder if there's any more clever way to get rid of the meaningless zero row vector at the end of a matrix:

Select[RowReduce[{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {2, 4, 
    6}}], # 0 != # &]

enter image description here

By the way, it's inconvenient to customize the display size of the inserted images.

$\endgroup$
2
$\begingroup$

One might Solve over the integers (like LinearSolve does in this case) to avoid fractions:

solvec = {x1, x2, x3, x4} /.
    Solve[
     {{1,  1, -2, -1},
      {3, -2, -1,  2},
      {0,  5,  7,  3},
      {2, -3, -5, -1}
     }.{x1, x2, x3, x4} == {4, 2, -2, 4},
     {x1, x2, x3, x4}, Integers
     ] // Normal // First
(*  {1 + 2 C[1], 1 - C[1], -1 + 2 C[1], -3 C[1]}  *)

Output formatting with CoefficientArrays:

vectorToColumnMatrix = Transpose@List[#] &;
columnMatrices = Transpose /@ List /@ Transpose[#] &;

Flatten@ MapThread[Apply]@ {
     {MatrixForm @* vectorToColumnMatrix, 
      Map[MatrixForm] @* columnMatrices},
     List /@ CoefficientArrays[solvec, Variables[solvec]]
     } . Prepend[Variables[solvec], 1] // Normal

One could add GeneratedParameters -> k to Solve and

k /: Format[k[n_]] := Subscript[k, n];

to get

$\endgroup$
0
$\begingroup$

I wrote a simple custom function to implement this requirement:

 linearSolve[mat_?MatrixQ, b_] := 
 Module[{m, n, L, k, generalsolution, specialsolution, 
   specialsolutionColumn},
  {m, n} = Dimensions[mat];
  L = Length[NullSpace[mat]];
  generalsolution = 
   MapThread[
    Defer[#1*MatrixForm[#2]] &, {Array["k" <> ToString[#] &, L], 
     Map[List[#]\[Transpose] &, NullSpace[mat]]}]; 
  specialsolution = LinearSolve[mat, b];
  specialsolutionColumn = {LinearSolve[mat, b]}\[Transpose];
  
  If[specialsolution.specialsolution == 0 || b.b == 0, 
   If[Length[generalsolution] != 1, 
    StandardForm[
     Defer @@ 
      MakeExpression[StringRiffle[generalsolution, "+"], 
       StandardForm]], StandardForm[Defer @@ First[generalsolution]]], 
   StandardForm[
    Defer @@ 
     MakeExpression[
      StringRiffle[
       PrependTo[generalsolution, MatrixForm[specialsolutionColumn]], 
       "+"], StandardForm]]]]

    linearSolve[{{1, 1, -2, -1}, {3, -2, -1, 2}, {0, 5, 7, 
       3}, {2, -3, -5, -1}}, {4, 2, -2, 4}]

    linearSolve[{{1, 0, -1}, {1, -1, 0}, {-2, 1, 1}}, {0, 0, 0}]

    linearSolve[{{1, 1, 1}, {2, 2, 2}, {3, 3, 3}}, {1, 2, 3}]

    linearSolve[{{1, 1, 1}, {2, 2, 2}, {3, 3, 3}}, {0, 0, 0}]
    
    {{1, 0, -1}, {0, 0, 0}, {0, 0, 1}, {0, 0, 
       0}} /. {a__List, {0 ..} ..} -> {a}

I'd like to see other people do that in a more subtle way.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.