The value of k for which the system of equations
\(x+y+z=1\)
\(x-ky+z=1\)
\(x-y+z=1\)
has more than one solutions is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → any real number
Subtracting the third equation from the first:
\(2y = 0\)
\(y = 0\)
With \(y = 0\), the term \(-ky\) vanishes whatever the value of k, and all three equations become
\(x + z = 1\)
So for every real k the solutions are
\(x = t,\quad y = 0,\quad z = 1 - t\) (t any real number)
which are infinitely many solutions. The values \(k = 1\) and \(k = -1\), which make the second equation identical to another one, are not the only ones.