The system of equations
\(x-3y-8z=-10\)
\(2x+5y + \lambda z = 13\)
\(3x + y -4z = 0\)
has infinite number of solutions if the value of \(\lambda\) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 6
For infinitely many solutions, the coefficient determinant must be zero (otherwise the solution is unique). Expanding along the first row:
\( \displaystyle \begin{vmatrix}1 & -3 & -8\\ 2 & 5 & \lambda\\ 3 & 1 & -4\end{vmatrix} = 1(-20 - \lambda) + 3(-8 - 3\lambda) - 8(2 - 15) \)
\(= -20 - \lambda - 24 - 9\lambda + 104\)
\(= 60 - 10\lambda\)
\(60 - 10\lambda = 0,\quad \lambda = 6\)
A zero determinant alone allows no solution too, so the system must also be consistent. With \(\lambda = 6\), ten times the second equation is a combination of the other two:
\(-13(x - 3y - 8z) + 11(3x + y - 4z) = 20x + 50y + 60z = 10(2x + 5y + 6z)\)
\(-13(-10) + 11(0) = 130 = 10 \times 13\)
The right sides agree as well, so the system reduces to two equations in three unknowns and has infinitely many solutions.