Question

The value of $k$ for which the following system of equations possess a non-zero solution is:
$4x + ky + 2z = 0$
$kx + 4y + z = 0$
$2x + 2y + z = 0$
$\,$
$A) \,3 \\\ B) \,2 \\\ C) \,1 \\\ D) \,4 \\\$

Answer 1

Form a $3 \times 3$ determinant matrix for these linear equations by using their coefficients such as $x$ coefficient in first $y$ coefficient in second and $z$ coefficient in third. Then, find the values of $k$ for which determinant is zero.

Complete step-by-step answer:
The linear equations are
$4x + ky + 2z = 0 \\\ kx + 4y + z = 0 \\\ 2x + 2y + z = 0 \\\$
The condition for these linear equations to possess a non-zero solution is that the determinant formed by their coefficients is equal to zero i.e.
\left| {\begin{array}{*{20}{c}} 4&k;&2 \\\ k&4&1 \\\ 2&2&1 \end{array}} \right| = 0
Opening the determinant
$\Rightarrow 4(4 - 2) - k(k - 2) + 2(2k - 8) = 0 \\\ \Rightarrow {k^2} - 6k + 8 = 0 \\\$
Sum is $- 6$ and product is $8$
$\Rightarrow {k^2} - 4k - 2k + 8 = 0 \\\ \Rightarrow k(k - 4) - 2(k - 4) = 0 \\\ \Rightarrow (k - 2)(k - 4) = 0 \\\ \Rightarrow k - 2 = 0\,\,,\,\,k - 4 = 0 \\\ \Rightarrow k = 2\,\,,\,\,k = 4 \\\$
Hence $k$ has two values $4\,\&\,2$ for which these linear equations possess a non-zero solution.

So, the correct answer is “Option A”.

Note: The condition for which equations possess no solution is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ and for for infinite many solution is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.This is correct but for two linear equations.
The condition for three equations is:
$\Delta = 0\,$ and one of ${\Delta _1},{\Delta _2},{\Delta _3}$ is zero for no solution.
$\Delta = 0\,$ and one of ${\Delta _1} = {\Delta _2} = {\Delta _3} = 0$ is zero for an infinite solution.
Where,
\Delta = \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| , {\Delta _1} = \left| {\begin{array}{*{20}{c}} {{d_1}}&{{b_1}}&{{c_1}} \\\ {{d_2}}&{{b_2}}&{{c_2}} \\\ {{d_3}}&{{b_3}}&{{c_3}} \end{array}} \right| , {\Delta _2} =\left| {\begin{array}{*{20}{c}} {{a_1}}&{{d_1}}&{{c_1}} \\\ {{a_2}}&{{d_2}}&{{c_2}} \\\ {{a_3}}&{{d_3}}&{{c_3}} \end{array}} \right| , {\Delta _3} = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{d_1}} \\\ {{a_2}}&{{b_2}}&{{d_2}} \\\ {{a_3}}&{{b_3}}&{{d_3}} \end{array}} \right|