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If the system of equations $x-ky-z=0,kx-y-z=0$ , $x+y-z=0$ has a non trivial solution then possible values of k are

Option 1)
-2

Option 2)
0

Option 3)
-1

Option 4)
2

Answers (1)

best_answer

As we have learned

Solution of a homogeneous system of linear equations -

Let $Ax=0$

If $A$ is singular then the system of equations will have infinitely many solutions

-

$x=ky+z$ ..(1)

$kx=y+z$ …(2)

$x+y=z$ …(3)

(1) & (3) $\Rightarrow$ $k=-1$ or $y=0$

Case 1: $k=-1$

Then $x=z-y$ and $x=-(y+z)$

$z=0$ $x=-y$

Case 2: $y=0$

$x=z$ and $kx=z$

$\therefore k=1$ for non trivial solution.

Option 1)

-2

Option 2)

0

Option 3)

-1

Option 4)

2

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