The set of all values of \(\alpha\) for which the system of linear equations
\(x + y + z = 1\)
\(x + 2y + 4z = \alpha\)
\(x + 4y + 10z = \alpha^2\)
is consistent; is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → {1, 2}
The determinant of the coefficients is
\( \displaystyle \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{vmatrix} = 1(20 - 16) - 1(10 - 4) + 1(4 - 2) \)
\(= 4 - 6 + 2 = 0\)
So there is no unique solution, and we must check when the equations do not contradict each other. Subtracting the first equation from the other two:
\(y + 3z = \alpha - 1 \qquad (1)\)
\(3y + 9z = \alpha^2 - 1 \qquad (2)\)
The left side of (2) is 3 times the left side of (1), so the system is consistent only if the right sides are in the same ratio:
\(\alpha^2 - 1 = 3(\alpha - 1)\)
\(\alpha^2 - 3\alpha + 2 = 0\)
\((\alpha - 1)(\alpha - 2) = 0\)
So \(\alpha = 1\) or 2 (the system then has infinitely many solutions), and the required set is {1, 2}.