If the system of equations
\(x + 2y + 3z = 10\)
\(-x + y + \lambda z = 20\)
\(2x + 3y + \lambda z = 0\)
does not possess a unique solution, then \(\lambda\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{15}{4}\)
A system \(AX = B\) has a unique solution exactly when \(|A| \ne 0\). So it does not possess a unique solution when \(|A| = 0\).
\( |A| = \begin{vmatrix}1&2&3\\-1&1&\lambda\\2&3&\lambda\end{vmatrix} \)
Expanding along the first row:
\( |A| = 1(\lambda - 3\lambda) - 2(-\lambda - 2\lambda) + 3(-3 - 2) \)
\(= -2\lambda + 6\lambda - 15 = 4\lambda - 15\)
Putting \(|A| = 0\):
\( \displaystyle 4\lambda - 15 = 0,\quad \lambda = \frac{15}{4} \)