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Find elements of a matrix such that its determinant is zero

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  • #26
Dick
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Okay, so the equation becomes:
##Ax_1^2+Cx_2^2+Dx_1+Ex_2+F=0##

I found this online
##B^2 - 4AC > 0##, hyperbola
##B^2 - 4AC = 0##, parabola
##B^2 - 4AC < 0##, ellipse or circle (circle only if B = 0 and A = C)

B=0, so we have to find A&C, to determine the form of the conic section.

So the other constants are combinations of a's, b's, and c's. Would I need to look have to look at the expanded form of the determinant to match the corresponding coefficients or is thinking of which points on a conic a method that will lead me to the same solution, but in a much easier way?
You could look at your expansion to find a relation between A and C. What is it? But you could also determine that just by looking at the matrix you are taking the determinant of and figuring what factors multiply ##x_1^2## and ##x_2^2##. Think about expansion by minors. Once you've figured out what kind of conic you have - you still need to know some points it passes through to determine the conic. That's a separate issue - but again there's an obvious answer by staring at the unexpanded determinant.
 
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You could look at your expansion to find a relation between A and C. What is it? But you could also determine that just by looking at the matrix you are taking the determinant of and figuring what factors multiply ##x_1^2## and ##x_2^2##. Think about expansion by minors. Once you've figured out what kind of conic you have - you still need to know some points it passes through to determine the conic. That's a separate issue - but again there's an obvious answer by staring at the unexpanded determinant.
Oh got it, from looking at the unexpanded determinant (and later verified by looking at the expanded determinant), I see that the following coefficients for ##x_1^2## and ##x_2^2##

For ##x_1^2##, there is ##-a_1b_1, a_1c_2, a_2b_1, -a2c1, -b_1c_2, b_2c_1## looks like just (# of points factorial, i.e., 3!).

For ##x_2^2##, there is ##-a_1b_2, a_1c_2, a_2b_1, -a2c1, -b_1c_2, b_2c_1## looks like just (# of points factorial, i.e., 3!).

So it is the same for both ##x_1^2## and ##x_2^2##.

So ##A=C=-a_1b_1+a_1c_2+a_2b_1+a2c1+b_1c_2+b_2c_1##.

So this is a circle.



So far so good?



Still kinda lost about how to find some identifying points on this conic.
Well actually, it's probably got something to do with (from a purely linear algebra perspective without considering geometry) linearly independency and if the set of x points equals the set of a, b, or c points, we'd get a 0 determinant.

So I'm guessing we can get 3 set of points right away that ##[x_1, x_2]=[a_1, a_2] or [b_1, b_2] or [c_1, c_2]##
 
  • #28
Dick
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Oh got it, from looking at the unexpanded determinant (and later verified by looking at the expanded determinant), I see that the following coefficients for ##x_1^2## and ##x_2^2##

For ##x_1^2##, there is ##-a_1b_1, a_1c_2, a_2b_1, -a2c1, -b_1c_2, b_2c_1## looks like just (# of points factorial, i.e., 3!).

For ##x_2^2##, there is ##-a_1b_2, a_1c_2, a_2b_1, -a2c1, -b_1c_2, b_2c_1## looks like just (# of points factorial, i.e., 3!).

So it is the same for both ##x_1^2## and ##x_2^2##.

So ##A=C=-a_1b_1+a_1c_2+a_2b_1+a2c1+b_1c_2+b_2c_1##.

So this is a circle.



So far so good?



Still kinda lost about how to find some identifying points on this conic.
So far so good. Yes, it's a circle. Now think about linear dependence. Suppose ##[x_1,x_2]=[a_1,a_2]##?
 
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So far so good. Yes, it's a circle. Now think about linear dependence. Suppose ##[x_1,x_2]=[a_1,a_2]##?
Yes got it! Just edited my previous post right as you posted that.
 
  • #30
Dick
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Yes got it! Just edited my previous post right as you posted that.
Right, it's the circumcircle of the triangle formed by a,b and c. Can you figure out how to do this without using the tedious expansion of the determinant?
 
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Right, it's the circumcircle of the triangle formed by a,b and c. Can you figure out how to do this without using the tedious expansion of the determinant?
So it is an equilateral triangle?
That means the points lie at 120degrees apart from each other.
 
  • #33
Dick
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So it is an equilateral triangle?
That means the points lie at 120degrees apart from each other.
No, the triangle is a given. ##a##, ##b## and ##c## are ANY points.

Had to look up what a circumcircle is and the first link is pretty much my homework problem http://mathworld.wolfram.com/Circumcircle.html
Right. But you worked it through independently.
 
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No, the triangle is a given. ##a##, ##b## and ##c## are ANY points.



Right. But you worked it through independently.
This is going to be a dumb question.

After I find the coefficients, A, C, D, E, and F, I will have some equation that describes a circle. Does this equation by itself satisfy the requirements of the problem, where I am asked for ##[x_1; x_2]##? The equation for the circle will be an implicit equation.
 
  • #35
Dick
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This is going to be a dumb question.

After I find the coefficients, A, C, D, E, and F, I will have some equation that describes a circle. Does this equation by itself satisfy the requirements of the problem, where I am asked for ##[x_1; x_2]##? The equation for the circle will be an implicit equation.
The problem says 'Describe the set'. If I were grading the problem, the description of the set as the circle that passes through the points ##a##, ##b## and ##c##, along with the reasons you think so would be enough. I don't think the actual equation of the circle is needed.
 
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The problem says 'Describe the set'. If I were grading the problem, the description of the set as the circle that passes through the points ##a##, ##b## and ##c##, along with the reasons you think so would be enough. I don't think the actual equation of the circle is needed.
Oh I see. Originally I thought the following would also satisfy the problem:
##[x_1 x_2]^T=u[a_1 a_2]+v[b_1 b_2]+w[c_1 c_2]## for some arbitrary constants u,v,w, such that not all of them are zero.
 
  • #37
Dick
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Oh I see. Originally I thought the following would also satisfy the problem:
##[x_1 x_2]^T=u[a_1 a_2]+v[b_1 b_2]+w[c_1 c_2]## for some arbitrary constants u,v,w, such that not all of them are zero.
I don't think that's the best way to describe the solution.
 
  • #38
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I don't think that's the best way to describe the solution.
Yes, I thought it was too simple to describe it like that as I did not go about finding constants u,v,w.

For the equation of the circle, it seems I still am not finding the constants, but seems a bit more informative than what I had formerly planned on doing.
 
  • #39
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No, the triangle is a given. ##a##, ##b## and ##c## are ANY points.



Right. But you worked it through independently.
I just thought of something. Couldn't the coefficients in front of the ##x_1^2## and ##x_2^2## be zero? If so, then the determiannt expansion when set to zero would yield an equation for a line that passes through those 3 distinct points.
 
  • #40
Dick
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I just thought of something. Couldn't the coefficients in front of the ##x_1^2## and ##x_2^2## be zero? If so, then the determiannt expansion when set to zero would yield an equation for a line that passes through those 3 distinct points.
Yes, you might also give some thought to 'degenerate cases'. If the three points ##a##, ##b## and ##c## are collinear then the 'circumcircle' becomes a line. And yes, the coefficient of ##x_1^2## and ##x_2^2## will be zero.
 

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