Question: For the value(s) of k, if any, will the following system have (a) no...

daktielti

daktielti

Answered

2022-07-12

Question: For the value(s) of k, if any, will the following system have (a) no solution, (b) a unique solution, (c) infinity many solutons:
x + y + k z = 1 , x + k y + z = 1 , x k + y + z = 2.

Answer & Explanation

lywiau63

lywiau63

Expert

2022-07-13Added 13 answers

Let's calculate this determinant
det ( 1 1 k 1 k 1 k 1 1 ) = det ( k + 2 1 k k + 2 k 1 k + 2 1 1 ) = ( k + 2 1 k 0 k 1 1 k 0 0 1 k ) = ( k + 2 ) ( k 1 ) 2
hence
- If k 1 and k 2 there's exactly one solution.
- If k = 1 we see easily that the 3 equations aren't compatible so there is no solution.
- If k = 2 we can see also easily that there's infinitely many solutions.
Frank Day

Frank Day

Expert

2022-07-14Added 4 answers

Just add. This gives ( k + 1 ) ( x + y + z ) = 0. If k = 1, then there are an infinite number of solutions.
Then if k = 0... and so on.

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