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I'm trying to find a topic to study on that would deal on expressing equations as matrices determinants. Let me explain.

Let's say I have the equation:

x - 2y + 2z = 0

I can write the above equation in matrix form as:

$$ det \begin{bmatrix} x & 0 & 2 \\ z & 1 & 0 \\ y & 1 & 1 \end{bmatrix} = 0 $$

Is there anyway to create this matrix not being by trial and error? Is there a name for this operation? Maybe I'm cracking my head on something that Wolfram could give me the answer if I type the correct command. Or maybe someone created a code that does this and I just don't know what to look for.

Especifically, I need to create 3x3 matrices and each variable shoud be in different row. I can also create a matrix and then manipulate it by diving a column by a variable, etc.

Just to put it in context, I use this topic to create matrices and later use them as parametric equations to plot a nomogram. If I have a equation with 3 variables and create a matrix where each variable is in separate rows, I can create a nomogram with 3 axes, where I can choose 2 variables, connect them with a straight line and the place it crosses the third axis is the answer for the equation.

Obviously not every equation can be turned into a matrix, but doing it by trial and error is extremely exhausting and I can't ever be sure if I ran out of possibilities. (Actually I'm trying to create a matrix of an especific equation and I thought about posting it here, but since I'm not even sure this topic exists, so I thought about creating this question first)

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    $\begingroup$ You might be interested in the companion matrix which has a simple form for the characteristic polynomial associated with it. $\endgroup$ – CyclotomicField May 19 at 18:59
  • $\begingroup$ There are many kinds of equations, and indeed you've taken a single "homogeneous" linear equation in three variables and found another homogeneous equation involving a determinant of a $3\times 3$ matrix. Narrowing what "input" equations you would like to treat in this fashion would help Readers respond cogently. $\endgroup$ – hardmath May 19 at 19:04
  • $\begingroup$ This problem can be solved with linear algebra, by obtaining a basis of the kernel (which is 2 dimensional) for the $1 \times 3$ matrix $(1,-2,2)$ and using the theorem that the determinant of a matrix is 0 if its columns are linearly dependent. You don't want the columns to be linearly dependent, since the plane is a 2 dimensional subspace of $\mathbb{R}^3$, so you can make any "determinant" equation you want with 2 basis vectors of your choice and a third vector of variables. $\endgroup$ – Theo Diamantakis May 19 at 19:36
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Generalization

Yes.

The determinant traduce the idea of dependance if it is zero, independance elsewhere.

For instance

$$ D=\left| \begin{array}x x&0&2\\ y&1&0\\ z&1&1 \end{array} \right|=0$$

Means that $$\vec X= \left( \begin{array} xx\\ y\\ z\\ \end{array} \right),\vec A= \left( \begin{array} 00\\ 1\\ 1\\ \end{array} \right),\vec B= \left( \begin{array} 22\\ 0\\ 1\\ \end{array} \right)$$

are dependent (family is not free)

So if you have $$ x+ay+bz=0 \ (1) $$

You can take two vectors $\vec{A}= \left( \begin{array} xa\\ -1\\ 0\\ \end{array} \right)$ and $\vec{B}= \left( \begin{array} xb\\ 0\\ -1\\ \end{array} \right)$ , for which the coordinates satisfies the equation$(1)$

Vector $A$ and $B$ are taken to belong to the plan, and to be independant (it is necessary). So because they belong to the plan defined by the equation. If X belong to the plan, the determinant of X,A,B must be null because vectors are linearly dependant (Because 3 vectors belonging to the same 2 dimensional plan (hyperplan of 3D space))

Develop the determinant, it will give you an equation of the same plan.

N-th dimension

You can generalize for $n$ dimensions plans :

Given an hyperplan $H$ of dimension $n$ generated by the equation :

$$x_1+a_2 x_2+a_3 x_3+.....+a_{n} x_{n}=0$$

You will get by taking the $i$ vectors

$$\vec{X_i}= \left( \begin{array} 11\\ 0\\ .\\ a_i \ \text{at the i-th row}\\ .\\ .\\ 0 \\ \end{array} \right) $$

You get :

$$ det(\vec X,\vec X_1,...,\vec X_n)=0$$

Which gives an equation of the same plan.

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  • $\begingroup$ Thank you for taking your time to answer my question, but I didn't quite follow what you mean. Your answer seems to be on track on what I want and maybe wasn't clear enough on explaining my question. Thing is, I do not understand how you got vectors A and B for equation 1. $\endgroup$ – Artur Avelar May 20 at 18:52
  • $\begingroup$ I extended the anwser $\endgroup$ – EDX May 20 at 22:49
  • $\begingroup$ I understood it now. Thank you. It has really broaden my view on this topic. $\endgroup$ – Artur Avelar May 21 at 14:30
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If anything, you have to make it a very precise request because you can always rewrite an equation $P(\underline{x})=0$ trivially as $$ \det\left( \begin{array}[cc] {}P(\underline{x}) & 0 \\ 0 & 1 \end{array} \right)=0 $$ (not to mention the fact that "anything" is the determinant of a $1\times 1$ matrix whose only entry is that"anything").

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  • $\begingroup$ I think he means multvariable for plane equations (n dimensional at least) $\endgroup$ – EDX May 19 at 19:16

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