For 2x2-systems, use the general formula for the matrix inverse -- 1 step, done.

If you just need one component of the solution vector, and your system is 2x2 or 3x3, use Cramer's Rule. For larger systems, don't bother, since the determinants quickly get more involved to compute than just using Gauss-Jordan.

Otherwise/in general, Gauss-Jordan until upper echelon form, then back-substitution.

It's the best for solving by hand, if you need the entire solution vector. If you only deal with systems of equations over "Z", you can slightly modify the approach to obtain the Smith Normal Form, but that's probably more advanced than what you're looking for.

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Try Matlab?

scipy.linalg.solve

For a general dense matrix, gaussian elimination with partial pivoting is the standard. Takes (2/3)n^(3) arithmetic operations, where n is the number of unknowns. https://en.wikipedia.org/wiki/Gaussian_elimination#Computational_efficiency

For some types of linear systems you can do much better.

In my experience, if you have access to Hessians, Newton-Raphson iterations work amazingly well for solving non-linear optimization problems. Of course, that assumes you have second-order differentiability.

It depends on your usage and problem. There Is no best, some are slower, others are fast to calculate and converge. For simple root finding methods, There's two methods commonly used: Newton and Bisection.

Newton Raphson is quite popular. It has a few issues to be wary of: overshoot, local minima, and when the line is nearly tangent to x axis. It may not work for every problem.

Bisection is another method, said to be slower in iterations (quadratic convergence) but robust. It avoids the problems of Newton such as overshooting and having the line nearly tangent to x axis problem, but it can be limited by range (if the solution isn't in the range you defined, it's not gonna solve it unlike newton). So it can't solve every problem either.

Secant.. I'm not so knowledgeable.

There's a few hybrid methods. Dekker-Brent uses secant+ bisection, and There's Newton+bisection hybrid. Bisection can also be parallelized as well with more intervals so it's another way to speed up calculations. These methods can be more robust.

Kinda sidetracking, There's also a whole family of Newton solvers known as householder methods, which include Newton's method and Halley's method (yep! Another root finding. Lower iterations than newton but more mathematically intense).

OSCAR Veliz's YouTube channel is a great resource on most of these traditional methods.

Other optimizers and algorithms can be used to solve root finding as well (but not the other way around). They can be considered overkill and difficult to calculate by hand. But they can sometimes converge to the optimal solution better than newton and bisection. Python's sci-kit has a lot of them available.

There's a "simple" solution but it's a bit annoying to use. A system of equation can be expressed in Matrix form (A*X=B where X is the vertical vector (1×3 here) of the variables x, y and z, A is the coefficients of x, y and z and B is everything not attached to a variable). According to Crammer, such a matrix equation has a solution if and only if the determinant of A is different of zero.

This gives us a polynomial dependent on k with roots -1 and 0. To know which one leads to impossible solutions and which one gives us a multiplicity of solutions is a bit tricky but the way I'd go about it would be to plug the values in the system and see what happens. If you plug 0 in, you'll see that equation 1 becomes the negation of equation 3 (-x + 3y +2z = 0 and x -3y -2z = 0) which means that it's redundant information and thus does not bring anything to the table. This gives us our multiplicity of solutions.

If you plug -1 in, looking at the same equations you'll notice their results are not coherent. You get -x + 3y +2z = -1 and x - 3y - 2z = 0 which is of course impossible. This gives us the answer to the first question.

The answer to the second one is literally everything else.

Another way of doing it is to do it intuitively and use the way the questions are written to your avantage. Thanks to 1 and 3 you know there's only 1 impossible value and 1 undetermined value while the rest gives you a unique solution. So try to find that.

Again, playing with equations 1 and 3 gives you the value of z(k) from which you can determine that -1 is an impossible value (leads to a division by 0). It's the simplest way to get a variable so it's not an unreasonable assumption that someone would go for that one first when solving. Additionally, intuitively, you'll see that 1 and 3 are identical aside from the presence of the addends in k. If they are set to 0, the equations are identical and 3 becomes redundant, leading to an infinity of solutions.

These two give you the answer to the second question which is everything that's not -1 and 0.

But the "proper" way to do it is to use Cramer and show that the determinant is different from 0 in all cases save -1 and 0.

Can you show your work? I would do the row reduction with the augmented matrix. At some point, you would divide by 2k+2, so there is a special case when k = -1. At another point, you would divide by k(k+1) so there is another special case when k =0.

I agree with u/EatThePinguin’s comment. Analytical solutions can be found for only a small class of ODEs, with the simplest being linear and systems of linear ODEs.

If you are dealing with more general differential equations, it’s more practical to find numerical or asymptotic solutions. Another approach is to look at stability and nature of solutions(i.e., periodic or more complicated orbits and bifurcation) which falls under the subject of dynamical systems.

If your diff eq are non linear and have no specific other properties, I'd expect most to have no known analytic solution, and definitely no general solution method.

Are you going to encounter a certain class of diff eq? What is the problem area?

using the DSolve and NDSolve commands in mathematica is a good method

Integrating factors and variation of parameters are completely useless. Matrix methods, as you call them, are just higher dimensional ODEs.

Perturbation theory and homotopy methods (the original 1999 paper by He has well over 2000 citations so far)

I'm not sure on the specifics, as this particular topic is outside of my knowledge, but in general, this appears to be Lagrange's method of undetermined coefficients, it's usually used in optimisations.

Eigen vector/values

This is Lagrange multipliers used on a positive semidefinite quadratic form with matrix notation. That's what you should Google. If you search for "capon beamformer Lagrange multipliers" or "MVDR Lagrange multipliers" I am confident you will find a fairly detailed, step by step solution somewhere on the web. I've seen it, but I don't have a reference handy. It's out there. Bear in mind that MVDR is just one example of such a problem. It's not the same as what you're looking at, but it's close enough not to matter a whole lot.

They are just using matrices for all their notation. If they weren't doing that, it'd be ordinary calc 3.

Pretty sure it is the regular lagrange multipiers, lamba is just a vector, to make Cu into a scalar, if you differentiate with respect to u's components and lambd's components you'll get the same system

Eq. 9.2 appears frequently in the numerical optimization and nonlinear programming literature. The specific equation is formulated/solved as a quadratic program and constructing eq. 9.3 is one approach. I believe, modern linear solvers can operate directly on the symmetric indefinite matrix to arrive at solutions to the variables (displacements in this case) and the Lagrange multipliers.

You say you’re trying to develop your own solver. Is this just a learning exercise? I would review the literature on symmetric indefinite matrix equation solvers. If you have access to back issues of SIAM journals, start there. Otherwise, if you’re just trying to get the job done, your time might be better spent looking for high quality computer solutions that are available.

Edited to include: look at routines MA21 and MA57 from the Harwell Subroutine Library. I.A Duff

Looks good to me

If you’re ever in doubt, sub back in. One wrong answer is enough to know you’re wrong (no amount of rights will prove it though)

Can you give an example of the sort of thing that isn’t clicking?

you mean like systems of linear equations?

Think of what an equals sign means.

If you do the same thing to what is on either side of that sign, then the actual equality must still hold true.

Perhaps you could give an example where it doesn't make sense, and someone can explain it to you.

Please show us a case where you were mystified. We can definitely explain what’s going on

First let's clear up a prerequisite. When you add one equation to another, you actually are doing the same thing to both sides. Because the two sides of an equation are equal, you are adding the same thing to both sides. Is that what you're confused about there?

Anyway, it sounds like you are trying to learn "steps" to math problems, but that's not how math or problem solving works. You just have tools and goals and then you do something.

For linear systems, the two main tools are substitution and combination. Good old substitution is totally fine! You just pick a variable, solve for it, plug it into the next one, and so on, and you'll be done!

The purpose of combination is that it's faster, not that it's necessary. But the order doesn't matter at all. There are, I think, 108 ways to solve a 3-variable problem, and at least 4 ways to write each step. So you have to just let go of looking for a "the way" to do a problem. Ultimately the point is to clear the equations column-by-column. But people jump around if they see a shortcut. Like if you see a 4 in one row and a -4 in another, you might as well get an easy 0 now.

If you don't know where to start with combinations, just go from the left column to the right, and from the top number to the bottom. Divide the top row by the first number to make it a 1, then use that to kill the rest of the column, repeat with the second number in the second column, and so on. That's how a computer does it. It's just that fractions are annoying to humans, so you can jump around a little.

And if you don't want to do combinations at all, just do substitution. But you absolutely cannot just try to emulate someone else's solution. The though process is "this is what I have, this is what I want, here is the list of theorems and tricks that might possibly move me in that direction, I'll try this one". If it helps, you repeat, if it doesn't, you go back and try a different one. Which is fine and normal.

Aus der Schule kennst du doch bestimmt noch Aufgaben/gleichungen wie : 4x=8. Die musste man lösen und man hat dann durch 4 geteilt und hat x=2 bekommen. Das war 1 Gleichung mit 1 variable.

Dann kamen Aufgaben mit 2 Gleichungen und 2 Variablen

Zum bsp 4x+2y=8 und 3x+2y=6. Jetzt ist es Aufgabe ein wertepaar (x/y) zu finden sodass beide Gleichungen erfüllt sind x=2, y=0, Wichtig hierbei ist,dass die Lösung für beide Gleichungen funktioniert. Zum bsp funktioniert das wertepaar x=0, y=4 nur für die erste genannte Gleichung. Jede Gleichungen für sich allein betrachtet hat unendlich viele Lösungen. Nur wenn man sagt es sollen beide gelten gibt es hier(gibt auch andere Fälle) genau eine Lösung die für beide Gleichungen gleichzeitig funktioniert.

Jetzt kann mam das beliebig hochschrauben. Man nimmt 3 Gleichungen und 3 Variablen. Wieder sollen alle diese 3 Gleichungen für ein wertepaar(eigtl.tripel) Gelten (x/y/z) .

Zum bsp könnten x y und z Einkaufspreise sein.

1 Person kauft 3x+4y+7z= 20 und zahlt 20 euro und wenn man den Einkauf noch von zwei weiteren Leuten kennt(sie kaufen nur die 3 genannten Produkte) kann man nun ausrechnen wie teuer die Produkte waren. Das ist jetzt zwar nicht der klassische Anwendungsfall aber man kann es sich so ganz gut vorstellen

Die Antworten lesen sich recht philosophisch, daher hier mal eine etwas stumpfere Antwort:

Ein Gleichungssystem (egal, ob linear oder nicht) ist eine Ansammlung an Gleichungen mit mehreren Variablen (du kannst es als mehrdimensionale Abbildung auffassen), die du alle gleichzeitig zu lösen versuchst. Hast du ein solches System also aufgelöst, erhältst du die Menge aller Kombinationen an Werten für die einzelnen Variablen, die dafür sorgen, dass alle diese Gleichungen gleichzeitig wahr sind.

Im Fall von linearen Systemen kannst du sogar etwas konkreter werden: Du suchst zu einer gegebenen Matrix einen Vektor, sodass nach Matrixmultiplikation ein entsprechender anderer Vektor herauskommt. Wozu braucht man das? Lineare Algebra ist der Grundstein für eine riesige Menge an Bereichen, von klassischer Mechanik und Elektrodynamik über Quantenmechanik und Optik zur allgemeinen Relativitätstheorie. Überall lassen sich Gleichungssysteme aufstellen, und überall lassen sich die fundamentalen Ideen, die man beim Lösen von LGS anwendet, weiterverwenden.

Du suchst den Schnittpunkt von zwei Geraden.

Eine lineare Gleichung kann als Gerade in einem Koordinatensystem dargestellt werden.

Zwei Gleichungen sind zwei Geraden - die treffen sich irgendwo, sind parallel oder gleich.

Praktisch können das Kosten sein, oder Wegstrecken oder Kräfte - irgendwo sind sie gleich und dieser Punkt ist oft interessant, weil sich dort die Praxis ändert: erst war das eine teurer, danach ist es das andere.

Das hängt davon ab von welchem Blickwinkel du das betrachten möchtest.
Nach dem ersten Semester Mathestudium würde man sagen, man rechnet mit Matrizen.
Wenn man Mathematik ganz funktional als Modell der Wirklichkeit betrachtet würde man das wohl abhängig machen von den Informationen die man da gerade mit einem Gleichungssystem modelliert. Dann würdest du wenn du ein Gleichungssystem löst, z.B. die Fahrzeit deines Autos auf dem Weg von A nach B berechnen.
Wenn man Mathematik fundamental betrachtet, dann nutzt man einfach verschiedene Eigenschaften der Identiätsrelation aus um neue Identiätsausagen aufzustellen. Transitvität, Symmetrie, Reflexivität usw.
Ich als Philosoph würde einfach die Antwort geben, dass du aus einer Menge von Informationen die du als wahr akzeptierst, also Prämissen, eine Information steng logisch erschließt. Du stellst also beim "auflösen" eines Gleichungssystems nur ein deduktiv gültiges Argument auf.

Wann immer du einen Sachzusammenhang mit mehreren Bedingungen (=Gleichungen) hast, findest du durch das Lösen des Gleichungssystems Werte für die Variablen, sodass alle Bedingungen erfüllt sind.