Probability and Matrices

Firstly , we note that , to calculate the total number of matrices in the sample space , we may place the three $1's$ in any of the $9$ entries of $M$ and the remaining $6$ entries would be all $0$ . Hence , total number of matrices $M$ in the sample space is $\binom{9}{3} = 84$ .

For $M$ to be non - singular , all rows must be linearly independent so that $M$ has full rank. Hence , each row must have exactly one $1$ and no two $1's$ must be present on the same column. This can be done in $6$ ways. Hence , probability is $\dfrac{6}{84} = \dfrac{1}{14}$ .

For $\text{rank}(M) = 1$ , all $1's$ should be present on the same column or the same row. Hence , there are $6$ total choices. Probability is again $\dfrac{1}{14}$ .

$\text{Prob}(M = I_3) = \dfrac{1}{84}$ because all $1's$ need to be present on the principal diagonal and hence there is only one such $M$ .

For $\text{trace}(M) = 0$ , $0's$ are present on the principal diagonal . Hence , $1's$ can be placed on any of the $6$ remaining entries. Hence , probability is $\dfrac{\binom{6}{3}}{84} = \dfrac{5}{21}$ .