Let S be the set of all \[\lambda \] \[ \in \] R for which the system
of linear equations
2x -- y + 2z = 2
x -- 2y +
\[\lambda \]z = --4
x +
\[\lambda \]y + z = 4
has no solution. Then the set S :
Solutions · 1
Official solution
For no solution :
\[\Delta \] = 0 and \[\Delta \]1/\[\Delta \]2/\[\Delta \]3 \[ \ne \] 0
\[\Delta \] = \[\left| {211−1−2λ2λ1 } \right|\] = 0
\[ \Rightarrow \] 2(--2 -- \[\lambda \]2) + 1 (1 -- \[\lambda \]) + 2(\[\lambda \] + 2) = 0
\[ \Rightarrow \] --2\[\lambda \]2
+ \[\lambda \] + 1 = 0
\[ \Rightarrow \] \[\lambda \] = 1, \[ - {1 \over 2}\]
When \[\lambda \] = 1
2x -- y + 2z = 2 ...(1)
x -- 2y + z = --4 ...(2)
x + y + z = 4 ...(3)
Adding (2) and (3), we get
2x -- y + 2z = 0 (contradiction) hence no solution.
\[ \therefore \] \[\lambda \] = 1 belongs to set S.
When \[\lambda \] = \[ - {1 \over 2}\]
2x -- y + 2z = 2 ...(1)
x -- 2y \[ - {1 \over 2}\]z
= --4 ...(2)
x \[ - {1 \over 2}\]y
+ z = 4 ...(3)
(1) and (3) contradict each other, hence no
solution.
\[ \therefore \] \[\lambda \] = \[ - {1 \over 2}\] belongs to set S.