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:
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:
\[ 2x + y - z = 7 \] \[ x - 3y + 2z = 1 \] \[ x + 4y + \delta z = k \] has infinitely many solutions, then \( \delta + k \) is:
Show Hint
For infinitely many solutions, the determinant of the coefficient matrix and all augmented determinants must be zero.
Updated On: Feb 4, 2025
- \( -3 \)
- \( 3 \)
- \( 6 \)
- \( 9 \)
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The Correct Option is B
Solution and Explanation
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}
\[ \delta + k = -3 + 6 = 3. \]
\[ \Delta = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta \end{vmatrix}. \] 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}
\[ \delta + k = -3 + 6 = 3. \]
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