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11621371 No.11621371 [Reply] [Original]

I can't figure this shit out and i can't find it on google.

like if I have the last polynomial in pic related would the equivalent matrix be
[2]
[-5]
[1]
[3]

thanks buster browns :^)

>> No.11621396

https://www.mathsisfun.com/algebra/systems-linear-equations-matrices.html

you couldn't figure this out on google?

>> No.11621402

>>11621371
you can't as far as i know. matrices work for multivariable linear equations. someone correct me if im wrong

basically:

2a + 3b + c = 0
a -5b + 0 = 2

would be a matrix:

2 | 3 | 1 | 0
1 | -5 | 0 | 2

and you do operations to get it to a triangular form

>> No.11621408

>>11621396
these are linear, he's asking about polynomials

>> No.11621419

>>11621371
Consider a vector space with basis e_i = {x^i}_i
Then the coordinates are just the coefficients of the polynomial

>> No.11621422
File: 3.07 MB, 776x5164, mathz.png [View same] [iqdb] [saucenao] [google]
11621422

>>11621371
Putting a system of linear equations into matrix form is super useful when you want to solve that system.
You simply take the coefficients, and insert them into a matrix. Note that the cofficients must be attatched to the same "kind" of variable (my math education was not in english, sorry). Your matrix would look like so:

0 0 -2 7
0 1 3 0
0 4-1 6
2 -5 1 3

Where the first column represents all x^3, second row all x^2 and so on.

this is pretty elementary stuff, and will be in the first chapter of EVERY book on linear algebra.

The "matrix" that you made is wrong. Primarily because it's not a matrix, but a vector, and it's a row vector, not a column vector.

>> No.11621425

>>11621422
>Where the first column represents all x^3, second row all x^2 and so on.

*Column, not row, sorry.

>> No.11621430
File: 56 KB, 960x720, bbbbbb.jpg [View same] [iqdb] [saucenao] [google]
11621430

>>11621396
yes this is surpsingly difficult to find out for some reason. It's not in my book. I swear i saw my teacher do this in a few seconds but he never explained it. I think I found a pic of it though finally?

if this is how it works I guess i was doing it backwards

>> No.11621727

>>11621371
>>11621402
>>11621408
It's not to difficult to verify that monomials of a single variable are linearly independent, so when we consider polynomial equations of a single variable, we can take a basis [math] \left\{ x^j \mid j \leq k \right\} [/math], where k is the highest degree in the system of polynomials. You pad out all the lower degree polynomials with zeros. The resulting matrix is
[math]
\begin{bmatrix} 0 & 0 & -2 & 7 \\ 0 & 1 & 3 & 0 \\ 0 & 4 & -1 & 6 \\ 2 & -5 & 1 & 3 \end{bmatrix}
[/math]

Do you see how the coefficients of the polynomials coincide with the entries?

>> No.11621755

>>11621727
yes, i realized that x^2 and x are effectively different variables.

>> No.11621904

>>11621727
I didn't know you could do matrices in tex here

>> No.11621935
File: 31 KB, 1000x1000, go vegan.jpg [View same] [iqdb] [saucenao] [google]
11621935

>>11621402
why not just do like pic related?

>> No.11621941

>>11621935
2 in r2c2 is obviously supposed to be a 1, I am sleep deprived and retarded

>> No.11622146

[math]
\begin{pmatrix}x^2 & x & 1\end{pmatrix}
\begin{pmatrix}1 \\ 3 \\ 0\end{pmatrix}=x^2+3x
[/math]

>> No.11622270

[math] \displaystyle

\left [
\left .
\begin{matrix}
1 & 3 & 1 \\
1 & 1 & -1 \\
3 & 11 & 5
\end{matrix}
\right |
\begin{matrix}
1 \\
2 \\
3
\end{matrix}
\right ]
\begin{matrix}
\cdot 7 \\
\cdot 8 \\
\cdot 9
\end{matrix}

[/math]