Question

The system of linear equations \[ x + 2y + 3z = 4, \quad 2x - y - 2z = a^2, \quad -x - 7y - 11z = a \] has a solution if the values of \( a \) are:

• A. \( -1 \text{ and } 5 \)
• B. \( -2 \text{ and } 3 \)
• C. \( -5 \text{ and } 1 \)
• D. \( -3 \text{ and } 4 \)

Hint:

To verify consistency in a linear system, always compute the determinant of the coefficient matrix and analyze the augmented matrix for specific values.

Answer 1

The Correct Option is D

Step 1: Represent the system in matrix form. 
The given system of equations can be written in matrix form as: \[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & -2 \\ -1 & -7 & -11 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 \\ a^2 \\ a \end{bmatrix}. \] Let the coefficient matrix be \( A \): \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & -2 \\ -1 & -7 & -11 \end{bmatrix}. \] 

Step 2: Compute the determinant of \( A \). 
The determinant of the coefficient matrix \( A \) is: \[ \det(A) = \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & -2 \\ -1 & -7 & -11 \end{vmatrix}. \] Expanding along the first row: \[ \det(A) = 1 \cdot \begin{vmatrix} -1 & -2 \\ -7 & -11 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & -2 \\ -1 & -11 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & -1 \\ -1 & -7 \end{vmatrix}. \] Now, calculating each 2x2 determinant: \[ \det(A) = 1 \cdot \left( (-1) \cdot (-11) - (-7) \cdot (-2) \right) - 2 \cdot \left( 2 \cdot (-11) - (-2) \cdot (-1) \right) + 3 \cdot \left( 2 \cdot (-7) - (-1) \cdot (-1) \right). \] \[ \det(A) = 1 \cdot (11 - 14) - 2 \cdot (-22 - 2) + 3 \cdot (-14 - 1). \] \[ \det(A) = 1 \cdot (-3) - 2 \cdot (-24) + 3 \cdot (-15). \] \[ \det(A) = -3 + 48 - 45 = 0. \] Since \( \det(A) = 0 \), the coefficient matrix is **singular**, and the system may or may not have a solution depending on the values of \( a \). Step 3: Analyze the augmented matrix. 
For the system to have a solution, the augmented matrix must satisfy the consistency conditions. Specifically, we examine the augmented matrix: \[ \left[ \begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 2 & -1 & -2 & a^2 \\ -1 & -7 & -11 & a \\ \end{array} \right]. \] For certain values of \( a \), the system may be consistent and have a solution. Specifically, testing for \( a = -3 \) and \( a = 4 \) makes the system solvable, while other values might lead to inconsistency. Conclusion: The system has a solution if \( a = -3 \) and \( a = 4 \).