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If the system of equations x-k y-z=0, k ...

If the system of equations `x-k y-z=0, k x-y-z=0,x+y-z=0` has a nonzero solution, then the possible value of `k` are `-1,2` b. `1,2` c. `0,1` d. `-1,1`

A

`-1, 2`

B

1, 2

C

0, 1

D

`-1, 1`

Text Solution

Verified by Experts

The correct Answer is:
D
Since, the given system has non-zero solution.
`therefore |{:(1, -k, -1),(k, -1, -1), (1,1,-1):}|=0`
Applying `C_(1) to C_(1), -C_(2), C_(2) to C_(2) + C_(3)`
`rArr |{:(1+k, -k-1, -1),(1+k, -2, -1), (0,0,-1):}|=0`
`rArr 2(k+1)-(k+1)^(2) = 0`
`rArr (k+1)(2-k-1) =0 rArr k = +-1`
Note There is a golden rule in determinant that n one's `rArr (n-1)` zero's or n (constatnt)`rArr (n-1)` zero's for all constant should be in a single row or single column.
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