Question

Let $S$ be the set of all $3\times 3$ matrices having three entries equal to $1$ and six entries equal to $0.$ A matrix $M$ is picked uniformly at random from the set $S.$ Then which of the following statements is (are) CORRECT ?
(The trace of a square matrix is defined to be the sum of elements on the main diagonal)

• A. Probability that $M$ is non-singular is $\frac{1}{14}$
• B. Probability that $M$ is singular is $\frac{13}{14}$
• C. Probability that $M$ is identity matrix is $\frac{1}{14}$
• D. Probability that $M$ has trace equal to $0$ is $\frac{5}{81}$​

Answer 1

Answer: (A), (B), (D)

Probability that $M$ has trace equal to $0$ is $\frac{5}{81}$​

To calculate the total number of matrices in the sample space, we may place the three $1^{\prime }$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 ${}^9{C}_3=84$

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

$\therefore$ Probability that $M$ is singular is $\frac{13}{14}$

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

For $\text{trace}\left(M\right)=0$, $0^{\prime }$s are present on the principal diagonal. Hence, $1^{\prime }$s can be placed on any of the $6$ remaining entries.
Hence, probability is $\frac{{}^6{C}_3}{84}=\frac{5}{21}.$