Let `[{:(,a,o,b),(,1,e,1),(,c,o,d):}]=[{:(,0),(,0),(,0):}]` where a,b,c,d,e `in (0,1)`
then number of such matrix A which system of equationa AX=0 have unique solution.

Options

A. 16
B. 6
C. 5
D. none of these

Answer

To solve the problem, we need to determine the number of matrices \( A \) such that the system of equations \( AX = 0 \) has a unique solution. For this to happen, the determinant of matrix \( A \) must be non-zero.

Given the matrix:

\[ A = \begin{pmatrix} a & 0 & b \\ 1 & e & 1 \\ c & 0 & d \end{pmatrix} \]

where \( a, b, c, d, e \in (0, 1) \).

### Step 1: Calculate the Determinant of Matrix \( A \)

The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:

\[ \text{det}(A) = a \cdot (e \cdot d - 1 \cdot 0) - 0 \cdot (1 \cdot d - 1 \cdot c) + b \cdot (1 \cdot 0 - e \cdot c) \]

This simplifies to:

\[ \text{det}(A) = a \cdot (e \cdot d) - b \cdot (e \cdot c) \]

So we have:

\[ \text{det}(A) = ae d - be c \]

### Step 2: Set the Determinant Not Equal to Zero

For the system \( AX = 0 \) to have a unique solution, we need:

\[ ae d - be c \neq 0 \]

This implies:

\[ ae d \neq be c \]

### Step 3: Analyze the Conditions

Since \( a, b, c, d, e \) are all in the interval \( (0, 1) \), we can analyze the conditions under which \( ae d \) is not equal to \( be c \).

1. **Case 1:** If \( e \neq 0 \), then we can divide both sides by \( e \) (since \( e > 0 \)):
\[ ad \neq bc \]

2. **Case 2:** If \( e = 0 \), then \( ae d = 0 \) and \( be c = 0 \), which means the determinant is zero, leading to no unique solution.

Thus, we must ensure \( e \neq 0 \) and \( ad \neq bc \).

### Step 4: Count the Valid Combinations

Now we need to find combinations of \( a, b, c, d \) such that \( ad \neq bc \).

- Possible values for \( a, b, c, d \) are \( 0 \) or \( 1 \).
- We can create a table of combinations for \( a, b, c, d \) and check which satisfy \( ad \neq bc \).

The valid combinations are:

1. \( a = 1, b = 0, c = 0, d = 1 \) → \( ad = 1 \), \( bc = 0 \) → Valid
2. \( a = 0, b = 1, c = 1, d = 0 \) → \( ad = 0 \), \( bc = 1 \) → Valid
3. \( a = 1, b = 1, c = 0, d = 0 \) → \( ad = 0 \), \( bc = 0 \) → Invalid
4. \( a = 0, b = 0, c = 1, d = 1 \) → \( ad = 0 \), \( bc = 0 \) → Invalid
5. \( a = 1, b = 0, c = 1, d = 0 \) → \( ad = 0 \), \( bc = 0 \) → Invalid
6. \( a = 0, b = 1, c = 0, d = 1 \) → \( ad = 0 \), \( bc = 0 \) → Invalid

After checking all combinations, we find that there are **6 valid combinations** where \( ad \neq bc \).

### Conclusion

The number of such matrices \( A \) for which the system of equations \( AX = 0 \) has a unique solution is **6**.