DeterminantHard
Question
If the system of equation x - ky - z = 0, kx - y - z = o, x + y - z = 0 has a non-zero solution, then possible values of k are
Options
A.- 1, 2
B.1, 2
C.0, 1
D.-1, 1
Solution
Since, the given system has non-zero solution.
∴
Applying C1 → C1 - C2, C2 → C2 + C3
⇒
⇒ 2(k + 1) - (k + 1)2 = 0
⇒ (k + 1)(2 - k - 1) = 0
⇒ k =
1
Note There is a golden rule in determinant that n one, s ⇒ (n -1) zero′s or n (constant) ⇒ (n - 1) zero′s for all constant should be in a single row or a single column.
∴
Applying C1 → C1 - C2, C2 → C2 + C3
⇒
⇒ 2(k + 1) - (k + 1)2 = 0
⇒ (k + 1)(2 - k - 1) = 0
⇒ k =
1 Note There is a golden rule in determinant that n one, s ⇒ (n -1) zero′s or n (constant) ⇒ (n - 1) zero′s for all constant should be in a single row or a single column.
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