it's kind of like adding & which equals an equation, or adding 2& to another thing. If you add it to the other side, it should equal. Think of matrices as kind of like synthetic division

Matrix methods are extraordinarily appropriate when dealing with large systems. The example you gave may have only two variables, but many applications may have up to hundreds of variables. In these cases, using the tools and techniques of linear algebra and the matrix representation for these systems can be invaluable. In fact, methods like elimination are contained within the standard matrix tools for such a situation. Matrices provide a way to keep track of elimination on a grand scale, among other, more complex uses. In particular, using a matrix representation allows you to use the machinery of linear algebra to learn more about the system and manipulate it in more complex ways.

One way to think of it is that you're just taking away all useless information; you don't really care too much about the names of variables you're interested, for example. Once you eliminate all superfluous information, you can more easily dig out useful information, which you can learn more about in a linear algebra or numerical methods class.

For example, let's say you can't solve Ax = b (for example, there are no solutions), but you would be happy getting relatively close to it. You can use least squares to do this, and this is much more easily expressed in terms of matrices than as an explicit linear system.

It's slow because you're just learning it. It's extremely simple to use.

Note that finding matrix inverse is overkill for the purpose of solving a system of equation. Finding matrix inverse is equivalent to solving all system of equations with the same coefficients but different inhomogenous part, not surprisingly it's slower. You only need to do Gaussian elimination for solving a single system.

In fact, try programming a linear system solver. What you have is likely to be the same thing as a matrix (you probably will even use a 2 dimensional array). Substitution, elimination and matrix are literally the same method as far as computation are concerned (you compute the same formula for the same number), using matrices just mean your number are organized nicely into a table. Which is why I asked you to try to program it, because then you will realized that what you're doing isn't different. They just look different because the numbers are written in different places.

Invert A, then multiply it with A B = 5 I from the left, such that A^-1 A B = A^-1 5 I. We know that A^-1 A = I by definition, so afterwards you can read off x and y elementwise in the equation B = 5 A^-1 . Note that scalars may commute, but matrices generally do not.

[edit]^formatting

So you've found a system of two equations in two unknowns: x+3y=0 and 2x+y=5. You should have learned in algebra how to solve a system of equations like this -- do you remember how to do that?

Your numbers doesn't make sense for me. I sligtly modified it to work.
a) calculate the inverse matrix of the coefficents
b) matrix multiplication with the constant vector gives you the solution to your linear equations

modified coefficent matrix --> inverse matrix with =MINVERSE($A$2:$C$4)

-1	0.5	0.5		-2	-2	0
0.5	-0.5	-0.5		-1	-3	1
0.5	0	-1		-1	-1	-1

The inverse matrix multiplied with the constant vector gives you the solution =MMULT($F$2:$H$4;D$2:D$4)

-2	-2	0		1		-4
-1	-3	1		1		-3
-1	-1	-1		1		-3

Here's the solution which I found using Deepseek (unfortunately all formatting is gone):

-----------------------

In LibreOffice Calc, you can solve a system of 4 equations with 3 variables using the Solver add-in. Here’s how to set it up step by step:

Step 1: Define Variables and Equations

Assume your system looks like this (modify as needed):

a1x+b1y+c1z=0

a2x+b2y+c2z=0

a3x+b3y+c3z=0

a4x+b4y+c4z=1

Assign cells for variables:

Let x be in A1, y in B1, z in C1 (initially empty or guessed values).

Enter the equations in cells: (Replace a1, b1, c1, ... with actual coefficients from your equations.)

Equation 1: =a1*A1 + b1*B1 + c1*C1 → Store in D1

Equation 2: =a2*A1 + b2*B1 + c2*C1 → Store in D2

Equation 3: =a3*A1 + b3*B1 + c3*C1 → Store in D3

Equation 4: =a4*A1 + b4*B1 + c4*C1 → Store in D4

Step 2: Enable and Configure Solver

Go to Tools → Solver (ensure the Solver add-in is enabled in Tools → Add-ons).

Set up Solver parameters:

Target Cell: D4 (the equation that must equal 1).

Equal To: Value of → 1.

By Changing Cells: $A$1:$C$1 (variables x, y, z).

Add Constraints:

Click Add and set:

D1 = 0

D2 = 0

D3 = 0

(This ensures the first three equations hold.)

Choose Solving Method:

For linear systems, select "Evolutionary" or "DEPS" (nonlinear solver, but works for linear cases).

Click Solve → The Solver will adjust A1, B1, C1 to satisfy all equations.

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1. Full LibreOffice information from Help > About LibreOffice (it has a copy button).
2. Format of the document (.odt, .docx, .xlsx, ...).
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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.

If the matrices are upper or lower triangular (with a few permutations if needed) you can solve a system pretty quickly, and as far as I know it's the only technique to solve matrix system at hands fairly quickly. If you have to turn a system into an upper/lower triangular matrix, you'd have to do Gaussian elimination anyway.

If you have access to the inverse (or it's easy to compute with information given) you could solve it quickly I guess but in terms of computing it, inverse are in general a bit longer to do than Gaussian elimination.

I'm also curious why would anyone ask their student to solve systems in 2 minutes without a calculator, there's litteraly nothing to learn out of it, unless your teacher gave you a trick of some sorts with some specific matrix, but even then that's dumb.

I’ve been studying on Lem.ma. I like the idea presented there that objects should be treated on their own terms.

I think a matrix is a set vectors in Rn.

You can use a set of vectors in Rn to represent a set of linear equations.

edit/update:

Sorry I thougth about this a bit more.

Linear algebra, the whole subject is solving system of equations.

Matrix itself is jsut a container that help this.

Matrix can represent anything.

Como x = -w, você consegue manipular as equações de baixo pra chegar nesses 3 casos.

Pra ser impossível, deixe os lados esquerdos das equações de baixo múltiplos (uma vai ser a outra vezes -1) mas com resultados diferentes.

Pra ser indeterminado acredito que é só deixar as equações de baixo múltiplas dos dois lados.

Pra ser determinado até onde me lembro basta que nenhuma equação seja múltipla ou igual a outra.

Boa sorte!

Veja as respostas em r/LinearAlgebra.

começa isolando o x ou w na primeira equação, aí vai manipulando…

{{1,0,0,1}; {a,2,1,2}; {1,-2,-1,0}}*{x; y; z; w}= {0; b; c}

Faz o det

Edit: aqui tem tutorial se tiver duvidas: https://www.todamateria.com.br/sistemas-lineares/

Eu tenho pena de você

This is probably not the right answer for your particular needs, but from the perspective of pure mathematics, I think the way in which matrices are introduced is often really unmotivated, especially in linear algebra courses designed mainly for students from other disciplines (also it is not really true that their main applications are in programming - matrices, and more generally linear transformations, which they represent, are ubiquitous across all of mathematics and any discipline using it). 

The reason that matrix multiplication is the way it is is so that multiplication of matrices corresponds to composition of linear transformations (under a choice of bases). If you know what a linear transformation is, and what a basis is, then you can sit down and write out what sort of operation you need in terms of vector components with respect to given bases to completely describe the algebra of linear transformations, and you will more or less automatically rederive the rule for matrix multiplication. This perspective is laid out especially nicely in Sheldon Axler's Linear Algebra Done Right.

each column corresponds to an input, each row to an output. So the product of two matrices requires "wiring" the outputs of one to the inputs of another, which is why you multiply rows and columns like that. This also explains the dimension thing, if one matrix has n inputs and m outputs, and another has m inputs and k outputs, the m outputs can be "wired" to the m inputs so now you have a matrix with n inputs and k outputs.

the "multiply each combination of components" thing you thought of is actually a thing called the tensor product, which is a generalization of objects like matrices and vectors to more complex types of function.

note: you can also say rows are inputs and columns are outputs, depends on if the smaller objects you're working with are column or row vectors.

The way I look at it is: matrices are things that act on vectors. For example, one matrix might scale up all vectors by a factor of 2, while another might rotate vectors by a certain angle around a certain axis.

You've probably heard how 1 x n matrices are sometimes called "column vectors." If A is some random matrix and e_1 is the column vector (1,0,0...,0) (which looks more like a row vector, but that's only because I can't draw a column vector easily on Reddit, so please pretend it's a column), then what is the product Ae_1? If you look closely at the definition of matrix multiplication, you will discover that Ae_1 is simply the first column of A. And if e_2 is the column vector (0,1,0,...,0), then Ae_2 is the second column of A, and so on. So the matrix A can be interpreted as a list of column vectors, specifically the column vectors that A maps e_1, e_2, ... e_n to. Since any vector can be written as a linear combination of e_1, e_2, ...e_n, knowing where A sends those vectors tells you where A sends any vector. Explicitly, if v=(v_1,v_2,... v_n), then Av=v_1Ae_1+v_2Ae_2+...+v_nAe_n. So matrices are just compact representations of linear functions that take in a vector and output another vector. The height of a column is the length of the output vector, and the number of columns is the length of the input vector. Thus, a (b x a) matrix represents a function that takes as input a vector of length a and outputs a vector of length b.

Now, what happens if we have two matrices A and B and a vector v, and we want to figure out what happens when we start with V, apply A, and then apply B. Well, we can first read off the first column of A, a.k.a. Ae_1, and then compute what B does to that vector. That should be the first column of BA, since BAv should be the matrix that describes our process of applying A and then applying B. (In other words, we want (BA)v=B(Av), making multiplication associative). Work it out carefully and you will see that the matrix multiplication you've been taught is exactly the one that does this job. In particular, the height of A and the width of B have to be the same, because the output of the first function we're applying has to be a vector of the same length as the input of the second function.

This is all very abstract stuff, and I certainly haven't told you why we should care about matrices and vectors in the first place. So let me leave you with a concrete example. You probably know that 2-dimensional vectors of real numbers, e.g. (1,2.5), can be drawn as arrows in the 2D plane starting at the origin. One thing we can do with the 2D plane is rotate it around the origin clockwise by an angle t. From your knowledge of the unit circle, you may know that, if we start at the point (1,0) on the positive x-axis and rotate by t, we get to the point (cos(t) ,sin(t)). And, with a little more trig identity work and the knowledge that (0,1) is what you get by rotating (1,0) by pi/2, you can show that rotating by t sends (0,1) to (-sin(t),cos(t)). So, the matrix that represents "rotate clockwise by t radians" is just R_t:=[[cos(t),-sin(t)],[sin(t),cos(t)]].

Now, what happens if I rotate clockwise by t radians, and then rotate clockwise by another s radians? Well, I've rotated by (s+t) radians in all, right? Based on this, we should expect that R_sR_t=R_(s+t). And by using matrix multiplication and angle addition trig identities, one can check that this is the case.

I recommend going to YouTube, looking up the channel 3blue1brown, and finding in his playlists tab “The Essence of Linear Algebra”. I feel like I learned more about matrices in that video series than I learned in a university-level course.

A lot of great explanations, but I want to give some more insight into use cases.

Linear algebra is used to work with 3 (or more) dimensional space. We have found a lot of ways to model things this way, for example each column could be a datapoint and suddenly you're halfway to machine learning. In calculus of multiple variables linear algebra is used as the ground you stand on, putting variables into matrices. This leads into a multitude of applications in physics (like all of physics). Linear algebra also pops up in proofs in unrelated areas every now and again such as graph theory or combinatorics.

A lot of our computers are optimized for fast work on matrices, such as GPU's, it has pretty much become our standard way of dealing with math on large amounts of data, which is something some programmers will need to do so if you are interested in simulations, computer graphics, machine learning, high performance computing, etc then linear algebra is the most important math course you'll take.