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**Hint:**If a system of homogeneous equations has a trivial solution then the determinant of coefficients of x, y and z of the equations taken row-wise is equal to zero.

Proceeding to this we’ll obtain an equation in ‘k’. solving for ‘k’ we will obtain the required values of k.

**Complete step by step solution:**

Given: the system of equations having a trivial solution,

$

x - ky + z = 0 \\

kx + 3y - kz = 0 \\

3x + y - z = 0 \\

$

It is well known that, if the system of homogeneous equations say

$

{a_1}x + {b_1}y + {c_1}z = 0 \\

{a_2}x + {b_2}y + {c_2}z = 0 \\

{a_3}x + {b_3}y + {c_3}z = 0 \\

$

Have only a trivial solution then, it is said that

\[\left| {\begin{array}{*{20}{c}}

{{a_1}}&{{b_1}}&{{c_1}} \\

{{a_2}}&{{b_2}}&{{c_2}} \\

{{a_3}}&{{b_3}}&{{c_3}}

\end{array}} \right| = 0\]

Applying this to the systems of homogeneous equations that are given to us, we’ll get

\[

\left| {\begin{array}{*{20}{c}}

1&{ - k}&1 \\

k&3&{ - k} \\

3&1&{ - 1}

\end{array}} \right| = 0 \\

\Rightarrow 1( - 3 + k) + k( - k + 3k) + 1(k - 9) = 0 \\

\Rightarrow k - 3 + k(2k) + k - 9 = 0 \\

\Rightarrow 2k - 12 + 2{k^2} = 0 \\

\Rightarrow 2{k^2} + 2k - 12 = 0 \\

\]

Dividing the whole equation by 2, we’ll be left with

\[

{k^2} + k - 6 = 0 \\

\Rightarrow {k^2} + (3 - 2)k - 6 = 0 \\

\Rightarrow {k^2} + 3k - 2k - 6 = 0 \\

\Rightarrow k(k + 3) - 2(k + 3) = 0 \\

\]

Taking \[\left( {k + 3} \right)\] common from both the terms, we’ll have

\[

(k + 3)(k - 2) = 0 \\

i.e.{\text{ }}k + 3 = 0{\text{ }}or{\text{ }}k - 2 = 0 \\

\therefore k = - 3{\text{ }}or{\text{ }}k = 2 \\

\]

Therefore for ${\text{k = }}\left\{ {{\text{2, - 3}}} \right\}$ we fill obtain the trivial solution for the given system of homogeneous equations.

**(A) \[\left\{ {2, - 3} \right\}\] is the correct option.**

**Note:**Determinant can also be solved as taken in order of the first column, then we’ll obtain

\[

\left| {\begin{array}{*{20}{c}}

1&{ - k}&1 \\

k&3&{ - k} \\

3&1&{ - 1}

\end{array}} \right| = 0 \\

\Rightarrow 1( - 3 + k) - k(k - 1) + 3({k^2} - 3) = 0 \\

\Rightarrow k - 3 - {k^2} + k + 3{k^2} - 9 = 0 \\

\Rightarrow 2k - 12 + 2{k^2} = 0 \\

\Rightarrow 2{k^2} + 2k - 12 = 0 \\

\]

Dividing the whole equation by 2, we’ll be left with

\[

{k^2} + k - 6 = 0 \\

\Rightarrow {k^2} + (3 - 2)k - 6 = 0 \\

\Rightarrow {k^2} + 3k - 2k - 6 = 0 \\

\Rightarrow k(k + 3) - 2(k + 3) = 0 \\

\]

Taking \[\left( {k + 3} \right)\] common from both the terms

\[

(k + 3)(k - 2) = 0 \\

i.e.{\text{ }}k + 3 = 0{\text{ }}or{\text{ }}k - 2 = 0 \\

\therefore k = - 3{\text{ }}or{\text{ }}k = 2 \\

\]

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