Conditional Probability of Singular Matrix Given Sum of Entries

To find the conditional probability $P(E_1|E_2)$, we use the definition of conditional probability: $P(E_1|E_2) = \frac{n(E_1 \cap E_2)}{n(E_2)}$

Step 1: Calculate $n(E_2)$

The set $E_2$ consists of all $3 \times 3$ matrices whose sum of entries is equal to $7$.

Since a $3 \times 3$ matrix has $9$ entries and each entry is either $0$ or $1$, a sum of $7$ means that exactly $7$ entries are $1$ and $2$ entries are $0$.

The number of ways to select $2$ positions out of $9$ to place the zeros is: $n(E_2) = \binom{9}{2} = \frac{9 \times 8}{2} = 36$

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Step 2: Calculate $n(E_1 \cap E_2)$

$E_1 \cap E_2$ is the set of matrices with $7$ ones and $2$ zeros such that $\det A = 0$.

We analyze all $36$ matrices in $E_2$ based on the relative positions of the two zeros:

1. Case 1: Both zeros are in the same row.

   • Choose $1$ row out of $3$ for the zeros: $\binom{3}{1} = 3$ ways.
   • Choose $2$ columns out of $3$ for the zeros: $\binom{3}{2} = 3$ ways.
   • Number of such matrices = $3 \times 3 = 9$.

   In these matrices, the other two rows contain no zeros, meaning all their entries are $1$. Since the matrix has two identical rows, its determinant is zero ($\det A = 0$). Thus, all $9$ matrices in this case belong to $E_1 \cap E_2$.

2. Case 2: Both zeros are in the same column.

   • Choose $1$ column out of $3$ for the zeros: $\binom{3}{1} = 3$ ways.
   • Choose $2$ rows out of $3$ for the zeros: $\binom{3}{2} = 3$ ways.
   • Number of such matrices = $3 \times 3 = 9$.

   In these matrices, the other two columns contain no zeros, meaning all their entries are $1$. Since the matrix has two identical columns, its determinant is zero ($\det A = 0$). Thus, all $9$ matrices in this case belong to $E_1 \cap E_2$.

3. Case 3: The two zeros are in different rows and different columns.

   • Total remaining matrices = $36 - 9 - 9 = 18$.
   • A representative matrix of this form is:

$$A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{pmatrix}$$

• Computing its determinant: $\det A = 0(0 - 1) - 1(1 - 1) + 1(1 - 0) = 1 \neq 0$

Since row and column permutations only affect the sign of the determinant, every matrix in this case has $\det A = \pm 1 \neq 0$. Thus, none of these $18$ matrices belong to $E_1 \cap E_2$.

Summing the valid matrices from all cases: $n(E_1 \cap E_2) = 9 + 9 = 18$

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Step 3: Calculate $P(E_1|E_2)$

Substituting the values into the conditional probability formula: $P(E_1|E_2) = \frac{18}{36} = \frac{1}{2} = 0.5$