Step-by-step Solution

1. Understanding the Sample Space and Events

   • S is the set of all 3x3 matrices with entries from {0, 1}. The total number of matrices in S is $2^9 = 512$.
   • E1 is the event that a chosen matrix A has a determinant of 0 (i.e., A is singular).
   • E2 is the event that the sum of the entries of matrix A is 7. Since entries are only 0 or 1, this means the matrix must have exactly seven 1s and two 0s.

2. Formulating the Conditional Probability We need to find the conditional probability P(E1 | E2). The formula for this is: $P(E1 | E2) = \frac{n(E1 \cap E2)}{n(E2)}$ where:

   • n(E2) is the number of matrices in event E2.
   • n(E1 ∩ E2) is the number of matrices that are in both E1 and E2 (i.e., matrices with seven 1s and two 0s, and also have a determinant of 0).

3. Calculating n(E2) n(E2) is the number of ways to form a 3x3 matrix with seven 1s and two 0s. This is equivalent to choosing 2 positions for the zeros out of the 9 available positions in the matrix. $n(E2) = \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36$ So, there are 36 matrices in our reduced sample space E2.

4. Calculating n(E1 ∩ E2) We need to find how many of these 36 matrices have a determinant of 0. A matrix has a determinant of 0 if its rows (or columns) are linearly dependent. Let's analyze the structure of a matrix A from E2.

   A matrix is singular if:

   • A row or column consists entirely of zeros. This is not possible for a matrix in E2, as it would require at least three 0s.
   • Two rows or two columns are identical.

   Let's consider the placement of the two 0s:

   • Case A: The two 0s are in the same row. If two 0s are in the same row, say row i, then the other two rows (row j and row k) will consist entirely of 1s. Thus, row j and row k are identical ([1, 1, 1]). This makes the rows linearly dependent, and det(A) = 0.

     • Number of ways to choose the row for the two 0s: 3 ways.
     • Number of ways to choose 2 positions for the 0s within that row: $\binom{3}{2} = 3$ ways.
     • Total matrices in this case: $3 \times 3 = 9$.

   • Case B: The two 0s are in the same column. Similarly, if two 0s are in the same column, say column i, then the other two columns (column j and column k) will be identical ($[1, 1, 1]^T$). This makes the columns linearly dependent, and det(A) = 0.

     • Number of ways to choose the column for the two 0s: 3 ways.
     • Number of ways to choose 2 positions for the 0s within that column: $\binom{3}{2} = 3$ ways.
     • Total matrices in this case: $3 \times 3 = 9$.

   The cases A and B are mutually exclusive since the two zeros cannot be in the same row and the same column simultaneously.

   • Case C: The two 0s are in different rows and different columns. Let's check if the determinant is zero in this case. Let the matrix A have zeros at positions $(i_1, j_1)$ and $(i_2, j_2)$ where $i_1 \neq i_2$ and $j_1 \neq j_2$. For example, let the zeros be at (1,1) and (2,2). $A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ The determinant is det(A) = 0(0-1) - 1(1-1) + 1(1-0) = 1. This is non-zero. In general, if the two zeros are in different rows and columns, the resulting matrix is non-singular. One can prove this by showing the rows are linearly independent. Thus, matrices in this case do not belong to E1.

   Therefore, the matrices in E1 ∩ E2 are only those from Case A and Case B. $n(E1 \cap E2) = \text{(Number of matrices from Case A)} + \text{(Number of matrices from Case B)}$ $n(E1 \cap E2) = 9 + 9 = 18$

5. Calculating the Final Probability Now we can substitute the values back into the conditional probability formula: $P(E1 | E2) = \frac{n(E1 \cap E2)}{n(E2)} = \frac{18}{36} = \frac{1}{2} = 0.5$

Final Answer

The conditional probability P(E1 | E2) is 0.5.