The value of \( \lambda \) for which the following system of equations has no solution is
\( x+y+2z=4 \)
\( 2x+y+3z=5 \)
\( x+3y+\lambda z=7 \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4
A system of three equations can have no solution only when the determinant of the coefficients is zero.
\( \displaystyle \begin{vmatrix}1&1&2\\2&1&3\\1&3&\lambda\end{vmatrix}=1(\lambda-9)-1(2\lambda-3)+2(6-1) \)
\( \displaystyle =4-\lambda \)
So \( \lambda=4 \) is the only possible value. For \( \lambda=4 \) the equations must also be inconsistent; \( R_{2}-2R_{1} \) and \( R_{3}-R_{1} \) give
\( \displaystyle y+z=3 \)
\( \displaystyle 2y+2z=3 \)
These contradict each other, so the system has no solution when \( \lambda=4 \).