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I have this matrix equation $$ \left( \begin{array}{ccc} a & b & 0 \\ c & d & e \\ 0 & f & g \\ \end{array} \right)\left( \begin{array}{c} x \\ y \\ z \\ \end{array} \right)=0 $$ where all the parameters $\{a,b,c,d,e,f,g\}$ are nonzero; I am looking for a nontrivial solution for $\{x,y,z\}$ i.e. when the determinant of the coefficient matrix is zero. How can I ask this from Mathematica? Normally, without any assumptions, it just gives the trivial solution zero.

system = { a x + b y == 0  ,  c x + d y + e z == 0 , f y + g z == 0};
Solve[system, {x, y, z}]
(* {{x -> 0, y -> 0, z -> 0}} *)

P.S. NullSpace also gives the empty answer.

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    $\begingroup$ Just knowing det=0 is not enough to get the general solution, because this depends on the rank of the coefficient matrix. $\endgroup$
    – A. Kato
    Commented Jul 14 at 1:49

2 Answers 2

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The answer will depend how the determinant vanishes:

mat = {{a, b, 0}, {c, d, e}, {0, f, g}};
Det[mat]

(* -a e f - b c g + a d g *) 

cases = Solve[% == 0]

enter image description here

Grid[{#, NullSpace[mat /. #]} & /@ cases, Dividers -> All]

enter image description here

Remark: All the solutions above corresponds to generic cases, i.e. the matrix is assumed to be of rank two.

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  • $\begingroup$ Thank you very much. So, since all the parameters are non-zero, only the first case in your list can be a nontrivial solution here. $\endgroup$
    – charmin
    Commented Jul 14 at 2:18
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    $\begingroup$ Your welcome. You are right. I overlooked the non-zero condition. $\endgroup$
    – A. Kato
    Commented Jul 14 at 2:22
  • $\begingroup$ May I ask you to please explain where the matrix rank plays a role in your code? $\endgroup$
    – charmin
    Commented Jul 14 at 12:51
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    $\begingroup$ @charmin, I'm sorry if I misled you. In this code, the rank does not play any role. Also, with your positivity assumption, mat is guaranteed to have rank two. I just pointed out: Solve and NullSpace treat all parameters, which are not explicitly specialized, as generic values. The generic 3x3 matrix with det=0 has rank two, so the solution space is always one dimensional. But, if you need to care about more special cases, say a=b=f=g=0, then rank of mat becomes one, and the solution space is now two dimensional spanned by {-e/c,0,1} and {-d/c,1,0}. $\endgroup$
    – A. Kato
    Commented Jul 15 at 1:29
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Try Reduce

solu = Reduce[{{a, b, 0}, {c, d, e}, {0, f, g}} . {x, y, z} - {0, 0,0} == 0 , {x, y, z}, Backsubstitution -> True];
Simplify[solu, Map[# != 0 &, {a, b, c, d, e, f, g }]]

(*(e f != d g && y == (c g x)/(e f - d g) && z == (c f x)/(-e f + d g)&&a == (b c g)/(-e f + d g)) 
|| (a e f + b c g != a d g && x == 0 &&y == 0 && z == 0) *)
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  • $\begingroup$ Thank you very much! simple and elegant! $\endgroup$
    – charmin
    Commented Jul 14 at 13:10

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