Question:
If the system of equations
\( 3x + y + 4z = 3 \)
\( 2x + \alpha y - z = -3 \)
\( x + 2y + z = 4 \)
has no solution, then the value of \( \alpha \) is equal to :
If the system of equations
\( 3x + y + 4z = 3 \)
\( 2x + \alpha y - z = -3 \)
\( x + 2y + z = 4 \)
has no solution, then the value of \( \alpha \) is equal to :
\( 3x + y + 4z = 3 \)
\( 2x + \alpha y - z = -3 \)
\( x + 2y + z = 4 \)
has no solution, then the value of \( \alpha \) is equal to :
Show Hint
For a \( 3 \times 3 \) system, \( \Delta = 0 \) is a necessary condition for both "no solution" and "infinitely many solutions".
Updated On: Mar 25, 2026
- 19
- 13
- 4
- 23
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
A system has no solution if the determinant of the coefficients (\( \Delta \)) is zero, but at least one of the Cramer's determinants (\( \Delta_x, \Delta_y, \Delta_z \)) is non-zero.
Step 2: Detailed Explanation:
Calculate \( \Delta \):
\[ \Delta = \begin{vmatrix} 3 & 1 & 4 \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 3(\alpha + 2) - 1(2 + 1) + 4(4 - \alpha) \]
\[ \Delta = 3\alpha + 6 - 3 + 16 - 4\alpha = 19 - \alpha \]
For no solution, set \( \Delta = 0 \):
\[ 19 - \alpha = 0 \implies \alpha = 19 \]
Verify for \( \alpha = 19 \):
\[ \Delta_x = \begin{vmatrix} 3 & 1 & 4 \\ -3 & 19 & -1 \\ 4 & 2 & 1 \end{vmatrix} = 3(19 + 2) - 1(-3 + 4) + 4(-6 - 76) = 63 - 1 - 328 \neq 0 \]
Since \( \Delta = 0 \) and \( \Delta_x \neq 0 \), there is no solution.
Step 3: Final Answer:
The value of \( \alpha \) is 19.
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