# Homework Help: Probability that determinant has 0 value

1. Oct 27, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
Entries of a 2X2 determinant are chosen from the set {-1,1}. The probability that determinant has zero value is

3. The attempt at a solution
Zero value can occur if all the elements of first row is 0. The remaining 2 places can be filled in 3*3=9 ways. Now other arrangements of -1 and 1 are also possible that will make it 0. I figured out all that arrangements manually and got a total of 7. So total possible arrangements = 7+9=16. the total no of ways by which this determinant can be constructed is 3*3*3*3=81. But the probability is incorrect. It's really tedious to manually figure out all the possibilities of zero and I'm not even sure if other possibilities also exist.

2. Oct 27, 2013

### Ray Vickson

The elements of the first row are not allowed to be 0, since all elements are either +1 or -1. Just write out the value of the determinant in terms of the four elements; you should be able to quickly list all the possibilities---there are not very many, and it is not tedious.

3. Oct 28, 2013

### utkarshakash

But the question says "from the set {-1,1}" This means 0 is also included.

4. Oct 28, 2013

### Dick

No, it doesn't. {a,b} is the set consisting of two elements a and b.

5. Oct 28, 2013

### HallsofIvy

You may be confusing the set notation {-1, 1} which, as Dick says, means "the set containing only -1 and 1", with the interval notation, (-1, 1), [-1, 1], (-1, 1], and [-1, 1].

6. Oct 29, 2013

### utkarshakash

Yes. I confused it with the interval notation. Thanks for pointing out.

7. Jan 3, 2014

Hey,even i am trying to solve this very same question...
but i m struggling with cases.???
were u able to crack it..?? if yes then plz tell me the cases and the solution

8. Jan 3, 2014

### Ray Vickson

He is not allowed to do that; it violates PF Rules. Also, do not use text-speak here; take the trouble to write things out properly. (You instead of u, I insted of i---i = sqrt(-1) here---please instead of plz, etc.)

9. Jan 3, 2014

### utkarshakash

I figured out all possible determinants which comes out to be 9. Out of them 6 determinants yield 0. But the probability 6/9 is incorrect.

10. Jan 4, 2014

### SteamKing

Staff Emeritus
You might want to check the total number of different determinant arrangements. I got a total which was greater than nine

11. Jan 4, 2014

### Saitama

How do you get 9? There are obviously $2^4$ possible determinants.

12. Jan 4, 2014

### utkarshakash

No. Some of them will be repeated twice if you interchange rows and columns. So effectively only 9 are possible.

13. Jan 4, 2014

### Mentallic

So for example, [1, -1; 1, 1] is considered the same as [1, 1; -1, 1]?

14. Jan 4, 2014

### utkarshakash

That is what I know. Interchanging rows and columns changes the matrix but not the determinant.

15. Jan 4, 2014

### SteamKing

Staff Emeritus
Yes, but the OP wanted to know what the probability was of getting a zero determinant. Although row interchange may not alter the value of the determinant, these are still valid matrices which have determinants, although coincidentally, their determinants are the same.

16. Jan 4, 2014

I m getting 8 possible cases,so can you please list your 9 cases.?

17. Jan 4, 2014

### lendav_rott

From what I understand in the assignment we have 4 spots to fill with either -1 or 1. There could be:
a)four 1s
b)four -1s
c)one 1 and three -1s
d)two 1s and two -1s
e)one -1 and three 1s.
How many different variations would we have with any of the given choices?
a)1 also = 0
b)1 also = 0
c)4 , none can be 0
d)6 , all are 0
e)4 , none can be 0
Total of 8 out of 16.

I am not too sure about the probability though. If we have a pool of balls each marked either as -1 or 1 and there are an equal number of both balls and reshuffled after the completion of a determinant what is the probability of any of those possible sets occurring?

so we see that we have to hit 3 out of 5 possible sets. What makes it simpler is that any variation within the 3 sets gives us a determinant of 0.

Let us say we have a jar with 8 balls, 4 white and 4 black. We pick 4 balls randomly and what we would like is to have: all blacks, all whites or 2 of either after a draw.
What is the probability of picking 3 whites and 1 black or 3 blacks and 1 white?

We have to pick 3 whites/blacks - 4/8 * 3/7 * 2/6, now we're left with 5 balls. To complete the set we have to pick an opposite coloured ball, but also we have 4 possible times when the opposite coloured ball could come out so:
2(either coloured set)* 4(order of picking) * 4/8 * 3/7 * 2/6 * 4/5(picking a set) = 0.4571 ~ 46% chance,

The probability of our desired sets occurring hence would be ~54%

I am not saying this is correct, but it is, atleast, how I would do it.

Last edited: Jan 4, 2014
18. Jan 4, 2014

### Ray Vickson

Wrong: interchanging rows (or columns) changes the sign of the determinant, but leaves its absolute value unchanged. (Of course, 0 = -0, so if the determinant = 0 it will remain unchanged.)

There are 16 independent matrices. One way to solve the problem is to list all the matrices and compute their determinants. There are 5 possible values of the determinant: -2, -1, 0, 1, 2.

19. Jan 4, 2014

### lendav_rott

As a way of rechecking I would say there are an N number of either coloured balls and N >= 4

To complete our 3:1 set the probability would be:
2*4 * (N/2N) * [ (N-1) / (2N -1) ] * [ (N-2) / (2N - 2) ] * [ N / (2N - 3) ]

after all that I get:
2( N³ - 3N² + 2N)
/
4N³ - 12N² + 11N - 3

think of the / as a division sign between the numerator and the demoninator.
With, say 10 balls of either the chance of our 3:1 set occurring is 49.5356%, but this is interesting, what happens if I increase the amount of either balls to, say, 50 -> The probability is now ALMOST 50%. What I can deduce is that the higher the N goes the more equal the chance of the determinant being 0 or not 0 becomes, however the chance of getting the 0 determinant set is always greater, if by a very small margin.

As N approaches infinity, the limit of that function would be 1/2 which is 50%, I don't know, why that is, though.

Last edited: Jan 4, 2014
20. Jan 4, 2014

### jbunniii

Suppose we write the general 2x2 matrix as
$$\left[ \begin{array}{cc} a & b\\ c & d\\ \end{array} \right]$$
The determinant of this matrix will be zero if and only if $[c \, d]$ is a multiple of $[a \, b]$. Consequently, I claim that I can choose $a,b,c$ however I like, and given this choice, there is exactly one value of $d$ which will result in a zero determinant. This should allow you to easily calculate the probability of a zero determinant.

21. Jan 4, 2014

### Ray Vickson

No: sometimes both values of d will result in a zero determinant. The determinant is
$$D = a d - b c$$
so if $a = 0$ and $bc = 0$, $d$ can be anything.

22. Jan 4, 2014

### jbunniii

But in this problem, $a,b,c,d \in \{-1, 1\}$.

23. Jan 4, 2014

### Ray Vickson

Oh, right.

24. Jan 4, 2014

### utkarshakash

The best method! Thank you!

25. Jan 5, 2014

### davidmoore63@y

The determinant (a b; c d) is zero iff ad=bc. The value of ad is either +1 (when a and d have the same sign), or -1 (when they have opposite sign). Each has a 50% probability. Likewise, bc is +1 with 50% probability or -1 with 50% probability, independently of what ad is. With similar reasoning, ad has a 50% chance of having the same sign as bc. That is precisely when the determinant is zero. So the answer is 50%.