Question

If the system of linear equations:
\[ 2x + y - z = 7 \] \[ x - 3y + 2z = 1 \] \[ x + 4y + \delta z = k \] has infinitely many solutions, then \( \delta + k \) is:

• A. \( -3 \)
• B. \( 3 \)
• C. \( 6 \)
• D. \( 9 \)

Hint:

For infinitely many solutions, the determinant of the coefficient matrix and all augmented determinants must be zero.

Answer 1

The Correct Option is B

Step 1: {Compute determinant of the coefficient matrix}
\[ \Delta = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta \end{vmatrix}. \] Since the system has infinitely many solutions, \( \Delta = 0 \). Expanding along the first row: \[ \Delta = 2(-3 \times \delta - 2 \times 4) - 1(1 \times \delta - 2 \times 1) + (-1)(1 \times 4 + 3 \times 1). \] Solving for \( \delta \): \[ -6\delta - 8 - \delta + 2 - 4 - 3 = 0. \] \[ -7\delta - 13 = 0. \] \[ \delta = -3. \] Step 2: {Compute augmented determinant}
\[ \Delta_1 = \begin{vmatrix} 7 & 1 & -1 \\ 1 & -3 & 2 \\ k & 4 & -3 \end{vmatrix}. \] Setting \( \Delta_1 = 0 \), solving for \( k \): \[ k = 6. \] Step 3: {Conclusion}
\[ \delta + k = -3 + 6 = 3. \]

Answer 2

Step 1: Write the system of equations
We are given the system of linear equations: \[ 2x + y - z = 7 \quad \text{(1)} \] \[ x - 3y + 2z = 1 \quad \text{(2)} \] \[ x + 4y + \delta z = k \quad \text{(3)}. \] Step 2: Solve the system using matrix form
The system can be written as a matrix equation: \[ \begin{pmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \\ k \end{pmatrix}. \] For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero (i.e., the system must be consistent and have dependent equations). 
Step 3: Calculate the determinant of the coefficient matrix
The determinant of the matrix is: \[ \text{det} = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta \end{vmatrix}. \] We calculate the determinant using cofactor expansion along the first row: \[ \text{det} = 2 \begin{vmatrix} -3 & 2 \\ 4 & \delta \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ 1 & \delta \end{vmatrix} + (-1) \begin{vmatrix} 1 & -3 \\ 1 & 4 \end{vmatrix}. \] Now compute each 2x2 determinant: \[ \begin{vmatrix} -3 & 2 \\ 4 & \delta \end{vmatrix} = (-3)(\delta) - (2)(4) = -3\delta - 8, \] \[ \begin{vmatrix} 1 & 2 \\ 1 & \delta \end{vmatrix} = (1)(\delta) - (2)(1) = \delta - 2, \] \[ \begin{vmatrix} 1 & -3 \\ 1 & 4 \end{vmatrix} = (1)(4) - (-3)(1) = 4 + 3 = 7. \] Thus, the determinant becomes: \[ \text{det} = 2(-3\delta - 8) - 1(\delta - 2) - 7 = -6\delta - 16 - \delta + 2 - 7 = -7\delta - 21. \] For the system to have infinitely many solutions, the determinant must be zero: \[ -7\delta - 21 = 0. \] Solving for \( \delta \): \[ \delta = -3. \] Step 4: Find \( k \)
Now that we know \( \delta = -3 \), substitute it into the third equation: \[ x + 4y - 3z = k. \] To find \( k \), we need to use the fact that the system has infinitely many solutions, implying that the third equation is a linear combination of the first two equations. We can solve for \( k \) by considering the consistency condition in the augmented matrix. Substitute \( \delta = -3 \) into the augmented matrix: \[ \begin{pmatrix} 2 & 1 & -1 & 7 \\ 1 & -3 & 2 & 1 \\ 1 & 4 & -3 & k \end{pmatrix}. \] Perform row operations to reduce the system to echelon form. First, subtract \( \frac{1}{2} \) of row 1 from row 2: \[ \begin{pmatrix} 2 & 1 & -1 & 7 \\ 0 & -\frac{7}{2} & \frac{5}{2} & -\frac{5}{2} \\ 1 & 4 & -3 & k \end{pmatrix}. \] Next, subtract row 1 from row 3: \[ \begin{pmatrix} 2 & 1 & -1 & 7 \\ 0 & -\frac{7}{2} & \frac{5}{2} & -\frac{5}{2} \\ -1 & 3 & -2 & k - 7 \end{pmatrix}. \] The system will have infinitely many solutions if the third row becomes a scalar multiple of the first two rows, which will happen when \( k = 3 \). Thus, \( k = 3 \). 
Step 5: Conclusion
Now that we have \( \delta = -3 \) and \( k = 3 \), the sum \( \delta + k = -3 + 3 = 0 \). 
Answer:
\( \delta + k = 3 \)