# The system of equation$kx+y+z=1,x+ky+z=k$  and $x+y+kz=k^{^{2}}$ have resolution if k equals Option 1) 0 Option 2) 1 Option 3) -1 Option 4) -2

As learnt in

Solution of a system of equations -

$x_{1},x_{2},\cdot \cdot \cdot ,x_{n}$ satisfy the system of linear equations  $A_{x}=B$

- wherein

Determinant is $\begin{vmatrix} k &1 & 1\\ 1& k &1 \\ 1& 1 &k \end{vmatrix}$

on solving, $k(k^{2}-1)-1(k-1)+1(1-k)$

$=k^{3}-k-k+1+1-k$

$=k^{3}-3k+2$

For     $k=1$

$k^3-k^2+k^2-k-2k+2$

$\Rightarrow k^{2}(k-1)+k(k-1)-2(k-1)$

$\Rightarrow (k-1)[k^{2}+2k-k-2]$

$\Rightarrow (k-1)(k-1)(k+2)$

It has a solution when $k+2=0$

$k=-2$

and $\Delta_{1}, \Delta_{2}, \Delta_{3}$ are all zero for $k=-2$

Option 1)

0

This option is incorrect.

Option 2)

1

This option is incorrect.

Option 3)

-1

This option is incorrect.

Option 4)

-2

This option is correct.

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