it is not possible to solve directly by rearranging into the form X = something in terms of A and B.

This is a type of Sylvester equation. Check out https://en.wikipedia.org/wiki/Sylvester_equation?wprov=sfti1#

I could be wrong here, but I’m like 75% sure that the only solution is the n × m zero matrix if no further information about A and B are given

Do you understand how to view T as a matrix? Parts a and b are just calculating the matrix products TS and ST and c is a system of 3 linear equations in 3 variables, the general solution is finding the inverse of T (there is an arguably easier one available in this case though).

No inverse operators are needed for this question. Can you be specific with what you are stuck at?

Let "r := [x; y; z]^T ". If "MS; MT" are the matrices of "S; T" in canonical base:

A(r)  =  (T o S)(r)  =  T(S(r))  =  MT . (MS . r)  =  (MT . MS) . r
B(r)  =  (S o T)(r)  =  S(T(r))  =  MS . (MT . r)  =  (MS . MT) . r

For c), you need to solve "[8; 9; 5]^T = T(r) = MT . r" -- can you take it from here?

>I’ve followed all the steps I know

Okay, can you share with us what exactly you have done?

>but I'm stuck now, cant find a solution.

What exactly are you stuck on?

If you rewrite the equations you'll have as much work if not more. Matlab is very optimized to work with matrices and using it as such is probably the best way to go.

i assume you talk about z system of linear equations. Why are you reluctant to use something that should work? What would be the benefits of solving this 'manually' when a matrix form allows to handle zlmost any number of equations seamlessly?

Can your describe your problem better?

In the time it took to post to reddit, wait for & reply to responses, implement/test/debug proposed solutions, you could have typed up a 21x21 matrix and solved it directly. I just don’t get these kinds of questions.

7 times the first equation minus the third equation equals the second equation. So, the only equation left would be r+s-2t=5, which has many possible solutions.

To solve for N unknowns, you need N equations.

In the first case, you have 3 equations for 4 unknowns. In the second, 3 equations for 5 unknowns.

You either need more equations or you need to decide that some of your variables are not to be considered unknowns. For example, in the first case, you could assume u is a known constant and then solve for r,s,t in terms of u.

u/jacklope_hunter69 u/KilonumSpoof u/Alkalannar u/grebdlogr u/Party_Limit_3432 thank you for your input, now I get that there isn't a single possible solution

u=r-4

t=r/2+s/2-5/2

[2 3 -6 1 | 11]
[1 1 -2 0 | 5]
[-3 4 -7 7 | -8]

[1 1 -2 0 | 5]
[0 1 -4 1 | 1]
[0 7 -13 7 | 7]

[1 0 2 -1 | 4]
[0 1 -4 1 | 1]
[0 0 15 0 | 0]

[1 0 0 -1 | 4]
[0 1 0 1 | 1]
[0 0 1 0 | 0]

r = u + 4
s = -u + 1
t = 0
u can be anything.

##Off-topic Comments Section

---

All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9.

---

^(OP and Valued/Notable Contributors can close this post by using /lock command)

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

/lock

Answer: Let's call the first matrix M, let's also set v=[x y]^(T) and a=[16 5]^(T)

We know that Mv=a, so M^(-1)Mv=v=M^(-1)a

Transform columns into REF from left to right:

        -->          1  -3  2 |  -22            ->             1 -3  2 | -22
  I' =         II    0   1  9 |   18    III' = III/3 + 5*II    0  1  9 |  18
 II' =   I - 2*II    0 -15  6 | -129                           0  0 47 |  47
III' = III + 4*II

The last row yields "47z = 47", i.e. "z = 1". Insert successively into the rows above:

y  =  18 - 9z  =  9,      x  =  -22 + 3y - 2z  =  3

It's easier to understand when you leave upper row untouched, because you need to get identity matrix, so your left upper number must not be 0.

Or rearrange your rows in that way, so after first eleimination you get

-4 -3 -2 | -41

1 - 3 2 | -22

0 -13 24 | -93

Algorithm asks you to subtract first row multiplied for some constant from second and third rows such that first elements in them become zeroes:

https://preview.redd.it/xf8efa9coyqe1.png?width=1080&format=png&auto=webp&s=ed54d3173ad4425a91d0ccdfb4cab224e4029c19

So we multiply it by 2 and and add to last row, then (for integer calculations) multiply second row by 2 and subtract first row from it

Now you need to do the same for second row (for example, from here you may multiply second row by 5 and subtract it from first row or multiply it by 14 and subtract it from third row)

Okay, I think I get it. For your first step you added the third row to twice the first - but then you replaced the first with that sum. Because it was the first row, and not the third, that you doubled, the sum should be used to replace the third row, and not the first. Remember: Always replace a row with itself, plus a multiple of another row.

Meanwhile, DON'T scale the second row, as it already starts with a 1, which is what you want. Ultimately, your goal is that each row start with any number of zeroes (or none at all), followed by a 1. (It might help to swap the second row with the first, which is the preferred form.)

Any more questions, please ask.

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.

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

If you're asking for help with LibreOffice, please make sure your post includes lots of information that could be relevant, such as:

1. Full LibreOffice information from Help > About LibreOffice (it has a copy button).
2. Format of the document (.odt, .docx, .xlsx, ...).
3. A link to the document itself, or part of it, if you can share it.
4. Anything else that may be relevant.

(You can edit your post or put it in a comment.)

This information helps others to help you.

Thank you :-)

Important: If your post doesn't have enough info, it will eventually be removed (to stop this subreddit from filling with posts that can't be answered).

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

I found a way to solve it numerically. Thanks everyone

Matrix is math. Find articles about it.

Youtube have some vids too.

Perhaps class in linear algebra

Let say you want to implement A = I + J. Start with 1 dimention, you first task will be figure out:

For each index i, what will A[i] be? This is simple because A[i] = I[i] + J[i]. Then you just do a loop then you solved 1 dimention. Now you need do the same for 2 dimentions. And other operations.

For each one, your goal is simple, find out the equation of A[i][j] = ??? (A mathmatic expression using matrix I and J)