# Homework Help: 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

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.