Question

# If the trivial solution is the only solution to the system of equations $x - ky + z = 0 \\ kx + 3y - kz = 0 \\ 3x + y - z = 0 \\$ Then the set of all values of k is:A) $\left\{ {2, - 3} \right\}$B) $R - \left\{ {2, - 3} \right\}$C) $R - \left\{ 2 \right\}$D) $R - \left\{ { - 3} \right\}$

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
$(k + 3)(k - 2) = 0 \\ i.e.{\text{ }}k + 3 = 0{\text{ }}or{\text{ }}k - 2 = 0 \\ \therefore k = - 3{\text{ }}or{\text{ }}k = 2 \\$