# Homework Help: Tricks for finding determinates

1. Nov 12, 2009

### thomas49th

1. The problem statement, all variables and given/known data
Consider the following matrices
1 2 3
4 5 6
7 8 9

y+z z + x x+ y
x y z
1 1 1

1 1 1
a b c
a² b² c²

Keeping direct expansion to a minimum obtain the value of the determinates

I know the rules, but what give away pointers should I look out for? Do I want to aim for for rows with 0s in by adding and subtracting columns or rows?

Thanks
Tom

2. Nov 12, 2009

### foxjwill

Try looking for symmetry.

3. Nov 12, 2009

### thomas49th

okay, but I cannot see any in the first matrix

4. Nov 12, 2009

### foxjwill

Well, there's definitely a pattern (try finding the sum of each diagonal). I'm not sure off the top of my head how that'd help you find the determinant, though.

5. Nov 13, 2009

### LCKurtz

What if you subtract the 2nd row from the third and... what next?

6. Nov 14, 2009

### HallsofIvy

A "row operation", such as "subtract the first row from the second" as LKurtz suggested, does not change the value of the determinant.

7. Nov 16, 2009

### thomas49th

okay so if I subtract 4 5 6 from7 8 9 I get:

1 2 3
4 5 6
3 3 3

is the correct notation for this

(-R2 + R3)

okay I have all 3 elements in the third row the same? Can I now do

(R1+ R3)

then

(-R2 + R3)

you have 3 0s in row 3 and hence the determinate is 0?

Is that right?

8. Nov 16, 2009

### LCKurtz

Yes. You could even stop earlier when you have two rows identical. That's enough isn't it?

9. Nov 16, 2009

### thomas49th

You mean I could also of done

1 2 3
4 5 6
7 8 9

(-R2 + R3) as before ----- (btw if i wrote this as (R3 - R2), does that means take R3 from R2 or R2 take R3?)

then do (-R1 + R2) to give 3 3 3 as well

If so, good!

So for the next question

y + z, z + x, x + y
x , y , z
1, 1, 1

okay any clues?? :)

Those ones look like they need to be multiplied by something :S.

Thanks
Tom

P.S Is there a rule where I can multiply one row by another and the determinate will remain the same?

10. Nov 16, 2009

### LCKurtz

I guess it means whatever you want it to mean.
Hint: You can do column operations just as well as row ops.
What do you think? Try it with a small 2 by 2 determinant and see.

11. Nov 16, 2009

### thomas49th

So there is NO standard notation for row/column manipulation

Doesn't work. Didn't think so, not listed as a determinant rule or property.

So back to the question

y + z, z + x, x+y
x y z
1 1 1

so (-C2 + C3)

y + z, z + x, y - z
x y y - z
1 1 0

so (-C1 + C2)

y + z, x - y, y - z
x ,x - y, y - z
1 0 0

then find the det of the last row

(x-y)(y-z) - (x-y)(y-z)

= 0!

Is my logic correct?? :D This is fun

Right last one

1 1 1
a b c
a² b² c²

Tricky tricky.

I can see a way of getting only one one in the top row, but that would leave a nasty expansion :S

What should I look for?

Thanks
Tom

12. Nov 16, 2009

### thomas49th

getting down to 0 1 0 on the top row means I get the expression

ac² - ab² - bc² - ca² + cb² + ba²

is that of help at all?

13. Nov 16, 2009

### LCKurtz

The determinant

$$\left | \begin{array}{ccc} 1&1&1\\ a&b&c\\ a^2&b^2&c^2 \end{array}\right |$$

is indeed tricky. I'm going to lead you through a few steps you probably wouldn't think of. First let's consider the function P(x) gotten by substituting x in for the a:

$$P(x) =\left | \begin{array}{ccc} 1&1&1\\ x&b&c\\ x^2&b^2&c^2 \end{array}\right |$$

Now some questions for you to think about.

1. Do you see that P(x) is a second degree polynomial in x? (Think about what would happen if you expanded it by minors of the first column, but don't actually expand it.)

2. Can you see immediately what P(b) and P(c) would be?

3. Can you see what the coefficient of x2 is?

4. Can you see what the factors of P(x) must be and what the factored form of P(x) must look like?

5. Finally what would P(a) be in the factored form, and note that P(a) is your original determinant.

Last edited: Nov 16, 2009
14. Nov 17, 2009

### thomas49th

Okay I can see P(x) has a degree of 2 because of row 3 column 1. The minors of column 1 would be along the lines of bc² - cb².

P(b), does that means replace x by b?

$$P(x) =\left | \begin{array}{ccc} 1&1&1\\ b&b&c\\ b^2&b^2&c^2 \end{array}\right |$$

is that right so far?

Thanks :D
Tom

15. Nov 17, 2009

### LCKurtz

Yes, and what is the value of a determinant with two columns or rows equal?

16. Nov 18, 2009

### thomas49th

0! but you can't say a = b. that's just ludicrous!

i can see the coefficient of x² is 1

so

4. Can you see what the factors of P(x) must be and what the factored form of P(x) must look like?

nope :( unless i can write it like x² + x + 1, but you get imaginary as the discriminant is -ve and you can't just start writing equations like that out of thin air!

Thanks!
Tom

17. Nov 18, 2009

### LCKurtz

I'm trying to show you something that you might like if you stick with it. I said to consider the function of x that you get if a is replaced by x:

$$P(x) =\left | \begin{array}{ccc} 1&1&1\\ x&b&c\\ x^2&b^2&c^2 \end{array}\right |$$

There is nothing "ludicrous" about it. This is a second degree polynomial in x as you would see if you expanded it by minors of the first column.
No, it isn't 1. What would x2 be multiplied by when you expand the determinant by the first column? It isn't 1.
One thing at a time. There was a typo in my first post which I don't know if you saw. In any case, can you see why P(b) and P(c) would be 0?

Are you familiar with the factor theorem for polynomials which says if a polynomial has a root of x = r then it must have a factor of (x - r)? The polynomial P(x) has roots of x = b and x = c. We will continue from here...

18. Nov 22, 2009

### thomas49th

Hi LCKurtz! Sorry I've not replied... been away!

Ok, I can see P(x) is what you said. I will atempt to expand it:

If I expand by the first column by it's minors I get:

bc² - cb²
x(c²-cb)
x²(c - b²)

Is that correct?

So the coefficient of x² is (c - b²)?

Yup I remember (mostly) the factor theorom, from my Core or "Pure" maths 3. How do you know the polynomial P(x) has roots of x = b and x = c? In other words I cannot see how P(b) and P(c) = 0... unless because 2 columns are equal in the matrix?

Thanks alot for all the help so far!
Thomas :)

19. Nov 22, 2009

### LCKurtz

Look again at your coefficients for x and x2. Remember, to get the coefficient of each, you cross out the row and column it is in and take the remaining 2x2 determinant with appropriate sign.

That is exactly why as you noted earlier. A determinant with two rows or columns equal has a value of zero. (You could subtract one from the other to get a row or column of zeros).

So right now, you have a second degree polynomial P(x). We are almost done with what I want to show you. Forgetting for a moment where P(x) came from, if you know it has roots of x = b and x = c, what must P(x) have as factors? What must the factored form of P(x) look like?

20. Nov 23, 2009

### thomas49th

$$P(x) =\left | \begin{array}{ccc} 1&1&1\\ x&b&c\\ x^2&b^2&c^2 \end{array}\right |$$

the determinant for the first one is:
$$P(x) =\left | \begin{array}{cc} b&c\\ b^2&c^2 \end{array}\right |$$

+(bc² - cb²)

x is:

$$P(x) =\left | \begin{array}{cc} 1&1\\ b^2&c^2 \end{array}\right |$$
-(c²-b²)

and x²:

$$P(x) =\left | \begin{array}{cc} 1&1\\ b&c \end{array}\right |$$

+(c-b)

That okay?

If P(x) is a second degree polynominal then (x-b)(x-c)

Everything seem okay so far?
Thanks :)
Tom

21. Nov 23, 2009

### LCKurtz

That is the constant coefficient of the polynomial but let's not call it P(x). We have already called the top one P(x).
Again, this is the coefficient of x, but don't call it P(x).
Yes, the coefficient of x2 is (c-b). Again, not P(x) though.
Good. You are almost there. Since P(b) and P(c) are zero P(x) must have factors (x-b)(x-c). But that doesn't completely determine P(x) because P(x) might be a constant times that and it would still have those roots. So at this point you know:

$$P(x) = A(x-b)(x-c)$$

for some constant A. Notice that that constant A would be the coefficient of x2 if you multiply it out. But you know from above that the coefficient of x2 in P(x) is (c-b). So A must equal (c-b). This tells you that

$$P(x) = (c-b)(x-b)(x-c)$$

So now we have an identity:

$$P(x) =\left | \begin{array}{ccc} 1&1&1\\ x&b&c\\ x^2&b^2&c^2 \end{array}\right | = (c-b)(x-b)(x-c)$$

Since this is an identity it is true if we put x = a in both sides:

$$P(a) =\left | \begin{array}{ccc} 1&1&1\\ a&b&c\\ a^2&b^2&c^2 \end{array}\right | = (c-b)(a-b)(a-c)$$

That last determinant is what you started with and we have shown what its value is. If you multiply out the determinant the long way you might have difficulty showing it factors nicely like this. Like I said when we started, I wanted to show you a way to expand this determinant that you likely wouldn't think of. It's pretty nifty and I hope you like it.

22. Nov 23, 2009

### thomas49th

A-ha! That is ingenious! Thank you very much. I hope to use this more often. Does this method go by a name at all

Thanks for all the help
Thomas

23. Nov 23, 2009

### LCKurtz

This determinant is called a Vandermonde determinant. The method I showed you works because of the special form it has. There is lots of stuff on the internet about Vandermonde determinants, much of which might be too advanced for you right now. Type it into Google and you will see lots of hits.