Question

Linear Algebra:
Find Coefficients a,b,c,d such that the equation:
ax^2 + ay^2 + bx + cy + d = 0
runs through the points:
(-2,7)
(-4,5)
(4,-3)

Answer 1

What I did so far:
Create a matrix representing [a b c d] for each point:
\[\left[\begin{matrix}53 & -2 & 7 & 1 \\ 41 & -4 & 5 & 1 \\ 25 & 4 & -3 & 1 \end{matrix}\right]\]
Then I did some matrix operations until I arrive at this point:
\[\left[\begin{matrix}1 & 0 & * & * \\ 0 & 1 & * & * \\ 0 & 0 & 1 & *\end{matrix}\right]\]
Where * represents a real number.
But at this point, I think I noticed a problem. The 4th column (which represents the coefficient of 'd', will always be a real-number instead of 1 or 0.

Do you guys see the same problem?

Answer 2

I think I did my matrix wrong from the very beginning

Answer 3

have you checked it on symbolab for matrix ?

Answer 4

I got the augmented matrix as

\[\left[\begin{matrix}4 & 49&5&1\\ 16 & 25&1&1\\16&9&16&1\end{matrix}\right]\]

Answer 5

@Narad I think there's an issue with your matrix.
The first row:
[4 49 5 1]
seems to correspond to the point (-2,7) right?
ax^2 = -2^2 = 4
ay&2 = 7^2 = 49
but for the rest:
bx = -2
cy = 7
d = 1
So I'm not seeing how you made your first row.
I know the matrix is consisted of coefficients for a,b,c,d:
(x^2 + y^2)a + (x)b + (y)c + (1)d = 0
\[\left[\begin{matrix}x^2+y^2 & x & y & 1 & | & 0\end{matrix}\right]\]

Answer 6

That's how I got the matrix:
\[\left[\begin{matrix}(-2)^2+(7)^2 & -2 & 7 & 1 & | & 0\\ (-4)^2+(5)^2 & -4 & 5 & 1 & | & 0 \\ (4)^2+(-3)^2 & 4 & -3 & 1 & | & 0\end{matrix}\right]\]

Row 1 is for (-2,7), x = -2, y = 7
Row 2 is for (-4,5), x = -4, y = 5
Row 3 is for (4,-3), x = 4, y = -3

which becomes

\[\left[\begin{matrix}53 & -2 & 7 & 1 & | & 0 \\ 41 & -4 & 5 & 1 & | & 0\\ 25 & 4 & -3 & 1 & | & 0 \end{matrix}\right]\]

Answer 7

@xXMarcelieXx
So I took your advice and it turns out I made a mistake lol:

https://www.symbolab.com/solver/matrix-calculator/gauss%20jordan%20%5Cbegin%7Bpmatrix%7D53%26-2%267%261%260%5C%5C%2041%26-4%265%261%260%5C%5C%2025%264%26-3%261%260%5Cend%7Bpmatrix%7D

Even so I'm not sure how to find the values of a,b,c,d based on this matrix:


\[\left[\begin{matrix}1 & 0 & 0 & \frac{1}{29} & | & 0 \\ 0 & 1 & 0 & \frac{-2}{29} & | & 0\\ 0 & 0 & 1 & \frac{-4}{29} & | & 0 \end{matrix}\right]\]

Which is solving this system of equation:
\[a + \frac{1}{29}d = 0\]
\[b + \frac{-2}{29}d = 0\]
\[c + \frac{-4}{29}d = 0\]

I could set d = r, and make it a parametric equation.

Answer 8

Second row of the matrix corresponding to the point (-4,5)

a*(-4)^2 + b*(5)^2 +c(-4+5)+d=0

16a+25b+1c+1d=0

The coefficients are 16, 25, 1,1

And you can determine the coefficients of the the third point (4,-3)

Answer 9

But the equation of the circle I'm given is
ax^2 + ay^2 + bx + cy + d = 0
not
ax^2 + by^2 + c(x+y) + d

Answer 10

I am SO sorry I wrote my question wrong, I corrected it.

Answer 11

WOW PARAMATRIC EQUATIONS WORK LOLWUT:

https://www.desmos.com/calculator/rmerfpca9w

r can be anything except 0.

Answer 12

oh okay i see lol
i remember doing this like 4 semesters ago but its pretty easy but i forgot lol

Answer 13

@xXMarcelieXx oh wow so you've taken linear algebra before?

Mind giving me some advice on this course, like which topics are the hardest

Answer 14

Also I'm taking Mathematical Analysis which is similar to the questions you ask. (Proving mathematical statements"

Answer 15

yeah, i took a class that was combined with diff equations and linear algebra.
I dont remmeber much but i think the last section with longer equations were harder. My advice is do flashcards for certain problems and vocab stuff.

Math proofs? @mhchen

Answer 16

@xXMarcelieXx yes math proofs.

Real numbers, natural numbers, set theory, and I just started cardinality

Answer 17

oh, you are way ahead lol
im barely on direct proofs, conditonals and biconditonals.
Perhaps, do you have any notes ? what math textbook do you use?

Answer 18

I'm in like a level higher than you so the textbook I use is way too hard for you.
You can just look at the first page to see if you understand: http://cms.dm.uba.ar/academico/materias/verano2012/taller_de_calculo_avanzado/Libros/Abbott%20-%20Understanding%20Analysis.pdf

Yeah I know you're starting with truth tables, and learning about notations.

Answer 19

oh thats way too advanced for me lmao

Answer 20

Better for you cause then it's easier for me to help you out.