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Question : 47 of 100
Marks:
+1,
-0
Solution:
4+(3−λ2)+6 =
4+2λ1+6 ⇒ 3 -
λ2 =
2λ1 ⇒
2λ1+λ2 = 3 ... (1)
Given
. = 0
⇒ 6 +
6λ1+3(λ3−1) = 0
⇒
2λ1+λ3 = - 1 ... (2)
Now
(λ1,λ2,λ3) =
(λ1,3−2λ1,−1−2λ1) Now check the options, option (2) is correct
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