Question:
An ordered pair \((\alpha, \beta)\) for which the system of linear equations
\((1 + \alpha)x + \beta y + z = 2\)
\(\alpha x + (1 + \beta)y + z = 3\)
\(\alpha x + \beta y + 2z = 2\)
has a unique solution, is :
An ordered pair \((\alpha, \beta)\) for which the system of linear equations
\((1 + \alpha)x + \beta y + z = 2\)
\(\alpha x + (1 + \beta)y + z = 3\)
\(\alpha x + \beta y + 2z = 2\)
has a unique solution, is :
\((1 + \alpha)x + \beta y + z = 2\)
\(\alpha x + (1 + \beta)y + z = 3\)
\(\alpha x + \beta y + 2z = 2\)
has a unique solution, is :
Show Hint
A unique solution needs the coefficient determinant to be non-zero. Simplify the determinant to a function of \(\alpha+\beta\).
Updated On: Nov 5, 2026
- \((-3, 1)\)
- \((1, -3)\)
- \((-4, 2)\)
- \((2, 4)\)
Show Solution
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The Correct Option is D
Solution and Explanation
Coefficient determinant \(\Delta = \begin{vmatrix} 1+\alpha & \beta & 1\\ \alpha & 1+\beta & 1\\ \alpha & \beta & 2\end{vmatrix}\).
Apply \(R_1 \to R_1 - R_2\): \(R_1 = (1, -1, 0)\).
\(\Delta = 1\cdot[2(1+\beta) - \beta] + 1\cdot[2\alpha - \alpha] = 2 + \beta + \alpha\).
Unique solution iff \(\Delta \ne 0\), i.e. \(\alpha + \beta \ne -2\).
(1) \(-3+1 = -2\) no; (2) \(1-3 = -2\) no; (3) \(-4+2 = -2\) no; (4) \(2+4 = 6 \ne -2\) yes.
Answer: (4) \((2, 4)\)
Apply \(R_1 \to R_1 - R_2\): \(R_1 = (1, -1, 0)\).
\(\Delta = 1\cdot[2(1+\beta) - \beta] + 1\cdot[2\alpha - \alpha] = 2 + \beta + \alpha\).
Unique solution iff \(\Delta \ne 0\), i.e. \(\alpha + \beta \ne -2\).
(1) \(-3+1 = -2\) no; (2) \(1-3 = -2\) no; (3) \(-4+2 = -2\) no; (4) \(2+4 = 6 \ne -2\) yes.
Answer: (4) \((2, 4)\)
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