Step 1: Understanding the Concept:
This question tests the conditions for the consistency and nature of solutions for a system of linear equations \( AX = B \), where \( A \) is a square matrix. The determinant of \( A \), \( |A| \), plays a crucial role.
Step 3: Detailed Explanation:
Let's analyze the conditions for the system of equations \( AX = B \).
Case 1: \( |A| \neq 0 \) (A is non-singular)
If the determinant of the coefficient matrix is non-zero, the matrix \( A \) is invertible. The system has a unique solution given by \( X = A^{-1}B \).
- Statement (B) says the system has a unique solution if \( |A| \neq 0 \). This is true.
- Statement (A) says the system has a unique solution if \( |A| = 0 \). This is false.
Case 2: \( |A| = 0 \) (A is singular)
If the determinant is zero, the system may have no solution or infinitely many solutions. To determine which, we calculate \( (\text{adj} A)B \). The solution is given by \( X = A^{-1}B = \frac{(\text{adj} A)B}{|A|} \).
If \( (\text{adj} A)B \neq 0 \) (the zero vector), then the system is inconsistent and has no solution.
If \( (\text{adj} A)B = 0 \) (the zero vector), then the system is consistent and has infinitely many solutions.
- Statement (C) says the system has no solution if \( |A| = 0 \) and \( (\text{adj} A)B \neq 0 \). This is true.
- Statement (D) says the system has infinitely many solutions if \( |A| = 0 \) and \( (\text{adj} A)B = 0 \). This is true.
Step 4: Final Answer:
The correct statements are (B), (C), and (D). Therefore, the correct option is (2).
Match List-I with List-II
List-I (Matrix) | List-II (Inverse of the Matrix) |
---|---|
(A) \(\begin{bmatrix} 1 & 7 \\ 4 & -2 \end{bmatrix}\) | (I) \(\begin{bmatrix} \tfrac{2}{15} & \tfrac{1}{10} \\[6pt] -\tfrac{1}{15} & \tfrac{1}{5} \end{bmatrix}\) |
(B) \(\begin{bmatrix} 6 & -3 \\ 2 & 4 \end{bmatrix}\) | (II) \(\begin{bmatrix} \tfrac{1}{5} & -\tfrac{2}{15} \\[6pt] -\tfrac{1}{10} & \tfrac{7}{30} \end{bmatrix}\) |
(C) \(\begin{bmatrix} 5 & 2 \\ -5 & 4 \end{bmatrix}\) | (III) \(\begin{bmatrix} \tfrac{1}{15} & \tfrac{7}{30} \\[6pt] \tfrac{2}{15} & -\tfrac{1}{30} \end{bmatrix}\) |
(D) \(\begin{bmatrix} 7 & 4 \\ 3 & 6 \end{bmatrix}\) | (IV) \(\begin{bmatrix} \tfrac{2}{15} & -\tfrac{1}{15} \\[6pt] \tfrac{1}{6} & \tfrac{1}{6} \end{bmatrix}\) |