Let A be a 3 by 3 matrix with entries chosen from the digits 0,1,2, ..., 9. Digits may appear more than once. (One way to get such a matrix A is to enter the digits of your social security number.) What is the probability that the columns of A are linearly dependent?

Solution

The Solution: The probability is 0.019945864.

There are 10^9 = 1 billion possible matrices. The columns of a square matrix are linearly dependent iff the determinant of the matrix is 0. (We say the matrix is singular in this case.) How many of these 1 billion matrices have a determinant of 0? It took my computer about an hour to count all the possible cases: 19,945,864. That means the probability that a given matrix of digits will have linearly dependent columns is 19,945,864/1,000,000,000 = 0.019945864. That's just a little less than .02 = 1/50. That means that roughly 1 person in 50 will have a Social Security Number that yields a singular matrix. I don't know how they assign those numbers, but they probably don't allow numbers like 000-00-000 or even 666-66-6666, so that will change the probability slightly.