Total Number of Matrices

The set $S$ contains all $3 \times 3$ matrices with three entries equal to $1$ and six entries equal to $0$. The total number of matrices is found by choosing the positions for the three $1$s from the $9$ available positions:

$|S| = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$

A matrix $M$ is chosen uniformly at random from $S$. We analyze the probabilities related to the given options.

Rank 3 Matrices (Nonsingular)

Nonsingular matrices have a non-zero determinant. For $3 \times 3$ matrices with entries $0$ or $1$, nonsingular implies rank $3$. The matrices in $S$ with rank $3$ are the $3! = 6$ permutation matrices. These are matrices where each row and each column contains exactly one $1$.

The probability of picking a nonsingular matrix is:

$P\{M \text{ is nonsingular}\} = P(\text{Rank } 3) = \frac{\text{Number of Rank 3 matrices}}{|S|} = \frac{6}{84} = \frac{1}{14}$

Rank 1 Matrices

Rank $1$ matrices in this context occur when the three $1$s are placed entirely within a single row or entirely within a single column.

• Matrices with three $1$s in a single row: There are $3$ rows, so $3$ such matrices exist.
• Matrices with three $1$s in a single column: There are $3$ columns, so $3$ such matrices exist.

The total number of rank $1$ matrices is $3 + 3 = 6$.

The probability of picking a rank $1$ matrix is:

$P\{M \text{ has rank } 1\} = \frac{\text{Number of Rank 1 matrices}}{|S|} = \frac{6}{84} = \frac{1}{14}$

Identity Matrix Probability

The identity matrix $I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ contains exactly three $1$s, so it belongs to the set $S$. There is only $1$ such matrix.

The probability of picking the identity matrix is:

$P\{M \text{ is identity}\} = \frac{1}{|S|} = \frac{1}{84}$

This probability is not $\frac{1}{14}$.

Trace Equal to Zero Probability

The trace $\text{trace}(M)$ is the sum of the diagonal elements ($M_{11}, M_{22}, M_{33}$). For the trace to be $0$, all diagonal entries must be $0$. Consequently, the three $1$s must be placed in the $9 - 3 = 6$ off-diagonal positions.

The number of ways to place the three $1$s in the off-diagonal positions is $\binom{6}{3}$.

Number of matrices with trace $0 = \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

The probability is:

$P\{\text{trace}(M) = 0\} = \frac{20}{84} = \frac{5}{21}$

This probability is not $\frac{1}{14}$.

Summary of Probabilities

The calculated probabilities for the events are:

• $P\{M \text{ is nonsingular}\} = \frac{1}{14}$
• $P\{M \text{ has rank } 1\} = \frac{1}{14}$
• $P\{M \text{ is identity}\} = \frac{1}{84}$
• $P\{\text{trace}(M) = 0\} = \frac{5}{21}$

The probabilities that equal $\frac{1}{14}$ correspond to the events that $M$ is nonsingular and that $M$ has rank $1$.