What is the probability that the determinant lies in the interval $[-9, 9]$?

Question

Given the $3 \times 3$ matrix

$$A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{pmatrix}$$

where every $a_{ij} \in {\Bbb R}$ is randomly selected from the interval $\left[3-3i-j,3i+j-3\right]$. What is the probability that $|A|\in[-9,9]$?

I have no idea on how to solve this problem in a purely mathematical way without any software. I used Microsoft Excel just to get a rough estimate. In columns A, B, and C, I created random $3 \times 3$ matrices with the specified criterion. In column D, I evaluated the determinant. Then, using Excel functions (COUNT, COUNTIF, SUM,...). I generated $2\cdot 10^5$ such matrices and the required probability turned out to be $\displaystyle \color{red}{\approx}\frac{35}{99}$. How can we evaluate this probability without any software?

Correction after seeing @JimB answer:

I used a slightly different way for the formulae in EXCEL, that is:

(-1)^RANDBETWEEN(0,1)*(RANDBETWEEN(0,0)+RAND()) in A2

(-1)^RANDBETWEEN(0,1)*(RANDBETWEEN(0,1)+RAND()) in B2

(-1)^RANDBETWEEN(0,1)*(RANDBETWEEN(0,2)+RAND()) in C2

and so on until

(-1)^RANDBETWEEN(0,1)*(RANDBETWEEN(0,8)+RAND()) in I2

With J2 being the determinant:

A2*(E2*I2-F2*H2)-B2*(D2*I2-F2*G2)+C2*(D2*H2-E2*G2)

With COUNTIF function for the criterion:

COUNTIFS(J2:J1000001,">-9",J2:J1000001,"<9")

I got $\approx 0.273902$.

I am looking for a purely mathematical way. Is there any?

Your help would be appreciated.

Columns are usually numbered. Bad to refer to a column with the same letter as the matrix name. Have you figured out what those intervals are? — Thomas Andrews, Jan 28 '25 at 08:12
@Pranay No, I have not. But that is the same idea as $3 \times 3$, I think. — Hussain-Alqatari, Jan 28 '25 at 08:13
What is the source of this problem? For example, are you assured by the source that there is a mathematical non-software way to solve this problem? — Sarvesh Ravichandran Iyer, Jan 29 '25 at 06:30
@SarveshRavichandranIyer I am not assured that there is a mathematical non-software way to solve this problem. There is no specific source. When studying relevant topics, (probability and integration) and (linear algebra and integration), this question popped into my mind. — Hussain-Alqatari, Jan 29 '25 at 06:55
I think the answer is closer to 0.274648 rather than 35/99 which is approximately 0.353535. — JimB, Jan 31 '25 at 06:08
@JimB could you please clarify how you got this? And to be sure, did you use only integers or real numbers as mentioned in the problem? — Hussain-Alqatari, Feb 02 '25 at 18:37
From the rule of Sarrus, the determinant of a $3 \times 3$ matrix is a $\pm 1$ combination of trilinear monomials (e.g., $a_{11} a_{22} a_{33}$). Thus, you start by finding the distribution of the monomials, then worry about the distribution of their $\pm 1 $ combination (which is found via convolution). Take a look at this — Rodrigo de Azevedo, Feb 03 '25 at 10:46

score 1 · Answer 1 · answered Feb 03 '25 at 02:58

This is just an extended comment that responds to your request for me to justify my suggestion that the answer is closer to 0.274648. I show that is roughly the answer using Excel, R, and Mathematica.

Excel

The 0.2797 is what I get with 10,000 simulations. Cell J3 is

 =-C3*E3*G3+B3*F3*G3+C3*D3*H3-A3*F3*H3-B3*D3*I3+A3*E3*I3Cell K3 is

Cell K3 is

 =(COUNTIF(J3:J10002,">=-9")-COUNTIF(J3:J10002,">9"))/10000

R

set.seed(12345)
a11 <- -1 +  2*runif(10000)
a12 <- -2 +  4*runif(10000)
a13 <- -3 +  6*runif(10000)
a21 <- -4 +  8*runif(10000)
a22 <- -5 + 10*runif(10000)
a23 <- -6 + 12*runif(10000)
a31 <- -7 + 14*runif(10000)
a32 <- -8 + 16*runif(10000)
a33 <- -9 + 18*runif(10000)
d <- -a13a22a31 + a12a23a31 + a13a21a32 - a11a23a32 - a12a21a33 + a11a22a33
length(d[d >= -9 & d <= 9])/10000
0.2743

Mathematica

A = Table[a[i, j], {i, 1, 3}, {j, 1, 3}];
uniform = Flatten[Table[a[i, j] \[Distributed] 
    UniformDistribution[{3 - 3  i - j, 3  i + j - 3}], {i, 1, 3}, {j, 1, 3}], 1]
dist = TransformedDistribution[Det[A], uniform];
SeedRandom[12345];
data = RandomVariate[dist, 1000000];
(*Proportion of observations between-9 and 9*)
Length[Select[data, -9 <= # <= 9 &]]/1000000 // N
(* 0.274421 *)

You're right that my approx was not good. I used a slightly different way for the formulae in EXCEL, that is:
(-1)^RANDBETWEEN(0,1)*(RANDBETWEEN(0,0)+RAND()) in A2

(-1)^RANDBETWEEN(0,1)*(RANDBETWEEN(0,1)+RAND()) in B2

(-1)^RANDBETWEEN(0,1)*(RANDBETWEEN(0,2)+RAND()) in C2

and so on until

(-1)^RANDBETWEEN(0,1)*(RANDBETWEEN(0,8)+RAND()) in I2

With J2 being the determinant:

A2(E2I2-F2H2)-B2(D2I2-F2G2)+C2(D2H2-E2*G2)

With COUNTIF function for the criterion:

COUNTIFS(J2:J1000001,">-9",J2:J1000001,"<9")

I got similar to you: 0.273902. Any purely mathematical way? — Hussain-Alqatari, Feb 10 '25 at 13:10
As asked in a comment, going through the process for a 2x2 matrix will result in a "mathematical" solution and get you ready for the resulting complexity of a solution for your 3x3 matrix. — JimB, Apr 11 '25 at 18:40

What is the probability that the determinant lies in the interval $[-9, 9]$?

1 Answers1

0.2743