Linear Algebra Question: Determinant

1. Jul 10, 2008

FrogginTeach

The question on my homework says:

What can you say about the graph of the equation

The following is in matrix form with determinant symbols around the matrix
x y 1
a$$_{1}$$ b$$_{1}$$ 1 = 0
a$$_{2}$$ b$$_{2}$$ 1

All a's and b's are constants.

My answer was that the graph would be a line going through the origin. Like y=2x or y=3x

Am I way off track???

I have tried to attach the question in the document. Not sure if it will show.

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2. Jul 10, 2008

FrogginTeach

Those 1's and 2's are supposed to be subscripts.

3. Jul 11, 2008

HallsofIvy

Staff Emeritus
No, that is not, in general, a line through the origin. What is that determinant? It not difficult to calculate.

4. Jul 11, 2008

FrogginTeach

The determinant is b1x - a2b1

a2b1 is a constant. So wait, I had that wrong. it is a straight line with a y-intercept which is some constant.

How do you say that though?

5. Jul 11, 2008

Defennder

The determinant is not that. But I don't know what this question is asking for. What can you tell from the equation of a plane/line?

6. Jul 11, 2008

FrogginTeach

Sorry, I forgot about it being a 3x3 matrix.

7. Jul 11, 2008

FrogginTeach

I'm not sure if the question is asking what the graph of the determinant would look like or the matrix?

8. Jul 11, 2008

Dick

If you expand the determinant you get an expression like f(x,y)=0. The question is what's the graph of that equation? Determinants and matrices don't have graphs. Equations have graphs.

9. Jul 11, 2008

rock.freak667

Just find the determinant of the 3x3 matrix and you'll see what it describes. You have an idea of what it describes though.

10. Jul 13, 2008

FrogginTeach

Okay. I found the determinant of this 3x3 matrix:

It is b1x + a2y + a1b1 - a2b1 - b2x - a1y

I simplified it: b1x - b2x + a2y - a1y - a1b1 - a2b1 = 0

I'm not sure if this will help, but I did this next.

b1x - b2x + a2y - a1y + C ; c = constant

a2y-a1y = C - b1x + b2x

y(a2 - a1) = C - x(b1 - b2)

y = (C - x(b1 - B2)) / (a2 - a1)

I know this is the equation of a line, but what more do I need to say?

11. Jul 13, 2008

Dick

Yes, it's a line. I think we are just waiting for you to retract the 'through the origin' part.

12. Jul 14, 2008

rock.freak667

Could say what the gradient is and where it cuts the axes, but I think saying that it is a straight line should be sufficient.