Find elements of a matrix such that its determinant is zero

[Clandry]

Homework Statement

Please see the attached file if my inline insertion does not work.

Homework Equations

$det(A)=det(A^T)$[/B]

The Attempt at a Solution

Since a matrix has a determinant of zero only when it's columns are linearly dependent, we look for a set of points [x1 x2] such that the first column of the matrix is linearly dependent. This is about as far as I got and I was wondering if this is the right approach or if there's a much more simpler way to approach this problem.[/B]

 

[BvU]

Well, linear dependency means what in this case ? That you can write column 1 as a linear combination of columns 2,3,4. What do you get ?

 

[Clandry]

Yes, it means that. [x1 x2] can be any scalar multiple of [a1 a2], [b1 b2], and/or [c1 c2].

Could I say that [x1 x2] is spanned by the vectors [a1 a2], [b1 b2], and[c1 c2]?

 

[BvU]

[x1 x2] can be any scalar multiple of [a1 a2], [b1 b2], and/or [c1 c2].

That's 2 equations with 3 unknowns. You can pick the first row or the last row for the third unknown.

 

[Clandry]

That's 2 equations with 3 unknowns. You can pick the first row or the last row for the third unknown.

I'm sorry I do not understand. What are you referring to as 2 equations and 3 unknowns?

 

[BvU]

[x1 x2] can be any scalar multiple of [a1 a2], [b1 b2], and/or [c1 c2]

any ?

Column one has four elements. you write down something for two of them only!

When you work out what the three coefficients are

$$\begin{pmatrix}x_1\\x_2\end{pmatrix} = u \begin{pmatrix}a_1\\a_2\end{pmatrix} + v \begin{pmatrix}b_1\\b_2\end{pmatrix} + w \begin{pmatrix}c_1\\c_2\end{pmatrix}$$

2 eqns, 3 unknowns. Not good enough...

 

[Clandry]

any ?

Column one has four elements. you write down something for two of them only!

When you work out what the three coefficients are

$$\begin{pmatrix}x_1\\x_2\end{pmatrix} = u \begin{pmatrix}a_1\\a_2\end{pmatrix} + v \begin{pmatrix}b_1\\b_2\end{pmatrix} + w \begin{pmatrix}c_1\\c_2\end{pmatrix}$$

2 eqns, 3 unknowns. Not good enough...

Oh! I see what you are saying.
We need one additional equation and that can be obtained by formulating an equation using the 1st row (i.e., 1=u+v+w) or the last column (i.e., $x_1^2+x_2^2 = u(a_1^2+a_2^2)+v(b_1^2+b_2^2)+w(c_1^2+c_2^2)$)

 

[BvU]

Yep. There are two more equations, so one of them to solve, the other to find a relationship between x₁ and x₂.

 

[Dick]

Oh! I see what you are saying.
We need one additional equation and that can be obtained by formulating an equation using the 1st row (i.e., 1=u+v+w) or the last column (i.e., $x_1^2+x_2^2 = u(a_1^2+a_2^2)+v(b_1^2+b_2^2)+w(c_1^2+c_2^2)$)

You might also try the approach of trying to guess what the answer is. If you expand that determinant you are going to get a quadratic form in $x_1$ and $x_2$. The solution to that is a conic section. Can you deduce enough facts about that conic section to guess what it is without cranking through a lot of algebra? Knowing that might guide you with the algebra. If you deduce enough you might be able to show it without the algebra.

 

[Dick]

If you people have given up on this you should come back. There is actually a very neat way to solve this one.

 

[Mark44]

Since a matrix has a determinant of zero only when it's columns are linearly dependent, we look for a set of points [x1 x2] such that the first column of the matrix is linearly dependent.

It makes no sense to talk about a single column being dependent (excepting the possibility that all of the elements of the column are zero).

 

[Clandry]

Hello all, sorry for the late reply. I was caught up in other courses and neglected this one. I am doing this now, first using the 3 equation, 3 unknowns approach.

I haven't attempted to expand, as I was intimated by a 4x4 matrix of all non-zeros.

 

[Clandry]

If you people have given up on this you should come back. There is actually a very neat way to solve this one.

Is the neat way by expanding out the determinant?

 

[Clandry]

I took the determinant of this matrix using matlab.
Here is what I got:

a1^2*b1*c2 - a1^2*b1*x2 - a1^2*b2*c1 + a1^2*b2*x1 + a1^2*c1*x2 - a1^2*c2*x1 - a1*b1^2*c2 + a1*b1^2*x2 - a1*b2^2*c2 + a1*b2^2*x2 + a1*b2*c1^2 + a1*b2*c2^2 - a1*b2*x1^2 - a1*b2*x2^2 - a1*c1^2*x2 - a1*c2^2*x2 + a1*c2*x1^2 + a1*c2*x2^2 + a2^2*b1*c2 - a2^2*b1*x2 - a2^2*b2*c1 + a2^2*b2*x1 + a2^2*c1*x2 - a2^2*c2*x1 + a2*b1^2*c1 - a2*b1^2*x1 - a2*b1*c1^2 - a2*b1*c2^2 + a2*b1*x1^2 + a2*b1*x2^2 + a2*b2^2*c1 - a2*b2^2*x1 + a2*c1^2*x1 - a2*c1*x1^2 - a2*c1*x2^2 + a2*c2^2*x1 - b1^2*c1*x2 + b1^2*c2*x1 + b1*c1^2*x2 + b1*c2^2*x2 - b1*c2*x1^2 - b1*c2*x2^2 - b2^2*c1*x2 + b2^2*c2*x1 - b2*c1^2*x1 + b2*c1*x1^2 + b2*c1*x2^2 - b2*c2^2*x1OUCH! There must be an easier way to do this.

 

[Dick]

I took the determinant of this matrix using matlab.
Here is what I got:

a1^2*b1*c2 - a1^2*b1*x2 - a1^2*b2*c1 + a1^2*b2*x1 + a1^2*c1*x2 - a1^2*c2*x1 - a1*b1^2*c2 + a1*b1^2*x2 - a1*b2^2*c2 + a1*b2^2*x2 + a1*b2*c1^2 + a1*b2*c2^2 - a1*b2*x1^2 - a1*b2*x2^2 - a1*c1^2*x2 - a1*c2^2*x2 + a1*c2*x1^2 + a1*c2*x2^2 + a2^2*b1*c2 - a2^2*b1*x2 - a2^2*b2*c1 + a2^2*b2*x1 + a2^2*c1*x2 - a2^2*c2*x1 + a2*b1^2*c1 - a2*b1^2*x1 - a2*b1*c1^2 - a2*b1*c2^2 + a2*b1*x1^2 + a2*b1*x2^2 + a2*b2^2*c1 - a2*b2^2*x1 + a2*c1^2*x1 - a2*c1*x1^2 - a2*c1*x2^2 + a2*c2^2*x1 - b1^2*c1*x2 + b1^2*c2*x1 + b1*c1^2*x2 + b1*c2^2*x2 - b1*c2*x1^2 - b1*c2*x2^2 - b2^2*c1*x2 + b2^2*c2*x1 - b2*c1^2*x1 + b2*c1*x1^2 + b2*c1*x2^2 - b2*c2^2*x1OUCH! There must be an easier way to do this.

Yes, there is a much easier way. Your expansion is quadratic in $x_1$ and $x_2$. Where it's zero is a conic section as I said before. I.e. it's an ellipse, hyperbola, etc. That narrows it down a bit. Now go back to thinking about the determinant. Can you name three points on the conic section by thinking about the determinant? Can you narrow down the form of the conic by thinking more about the coefficients of the quadratic terms?

 

[Clandry]

Yes, there is a much easier way. Your expansion is quadratic in $x_1$ and $x_2$. Where it's zero is a conic section as I said before. I.e. it's an ellipse, hyperbola, etc. That narrows it down a bit. Now go back to thinking about the determinant. Can you name three points on the conic section by thinking about the determinant? Can you narrow down the form of the conic by thinking more about the coefficients of the quadratic terms?

I am struggling to view this in terms of geometry. I'm not a very visual person when it comes to math.

Can you explain what you mean by "your expansion is quadratic in $x_1$ and $x_1$"?

 

[Dick]

I am struggling to view this in terms of geometry. I'm not a very visual person when it comes to math.

Can you explain what you mean by "your expansion is quadratic in $x_1$ and $x_1$"?

I mean that it is like $x^2+y^2=1$ is a circle. And $x^2-y^2=1$ is a hyperbola. Like conic sections. You must have done these at some point. Try to visualize the graph in $x_1$ and $x_2$ like it's a graph in $x$ and $y$. All the terms have degree 2 or less. It must be a conic section, yes?

 

[Clandry]

I mean that it is like $x^2+y^2=1$ is a circle. And $x^2-y^2=1$ is a hyperbola. Like conic sections. You must have done these at some point. Try to visualize the graph in $x_1$ and $x_2$ like it's a graph in $x$ and $y$. All the terms have degree 2 or less. It must be a conic section, yes?

Yes, quite awhile ago. I am familiar with conics as I do a numerical work with hyperbolic PDEs, but I am having a hard time tying geometrical concepts with this matrix.

So the 4 terms in the last row all represent a conic section?

 

[Dick]

Yes, quite awhile ago. I am familiar with conics as I do a numerical work with hyperbolic PDEs, but I am having a hard time tying geometrical concepts with this matrix.

So the 4 terms in the last row all represent a conic section?

ALL of terms put together represent a conic section. It has the form $Ax_1^2+Bx_1x_2+Cx_2^2+Dx_1+Ex_2+F=0$, yes? Where the constants are combinations of a's b's and c's? What is B? That's a start.

 

[Clandry]

ALL of terms put together represent a conic section. It has the form $Ax_1^2+Bx_1x_2+Cx_2^2+Dx_1+Ex_2+F=0$, yes? Where the constants are combinations of a's b's and c's? What is B? That's a start.

Oh I see.
I'm guessing B=0? But I am unsure why.

If that's right, then we would only have linear terms.

 

[Dick]

Oh I see.
I'm guessing B=0? But I am unsure why.

If that's right, then we would only have linear terms.

Look through your expansion. Do you see any $x_1 x_2$ terms? Think about the determinant if you want an even better reason.

 

[Clandry]

Look through your expansion. Do you see any $x_1 x_2$ terms? Think about the determinant if you want an even better reason.

No, I don't see any $x_1x_2$ terms in the determinant expansion I had previously posted. For this 4x4 matrix, we can't possibly have x_1*x_2 since we would never multiply those 2 elements in the calculation of the determinant, or more generally, any two elements from the same column in the calculation of a determinant. Is this the reason?

 

[Dick]

No, I don't see any $x_1x_2$ terms in the determinant expansion I had previously posted. For this 4x4 matrix, we can't possibly have x_1*x_2 since we would never multiply those 2 elements in the calculation of the determinant, or more generally, any two elements from the same column in the calculation of a determinant. Is this the reason?

Exactly, use properties of the determinant to figure out stuff about the conic. Any points you can think of that are on the conic?

 

[Clandry]

Exactly, use properties of the determinant to figure out stuff about the conic. Any points you can think of that are on the conic?

Thanks for all the help today! I am super sleepy and can't think anymore, so I will look at this again tomorrow.

 

[Clandry]

Exactly, use properties of the determinant to figure out stuff about the conic. Any points you can think of that are on the conic?

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?

 

[Dick]

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.

 

[Clandry]

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 $[x_1, x_2]=[a_1, a_2] or [b_1, b_2] or [c_1, c_2]$

 

[Dick]

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 $[x_1,x_2]=[a_1,a_2]$?

 

[Clandry]

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.

 

[Dick]

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?

 

[Clandry]

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.

 

[Clandry]

Had to look up what a circumcircle is and the first link is pretty much my homework problem http://mathworld.wolfram.com/Circumcircle.html

 

[Dick]

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.

 

[Clandry]

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.

 

[Dick]

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.