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# Find what value of k do the following system of equations possess a non-trivial solution over the set of rationals.$x + ky + 3z = 0,3x + ky - 2z = 0,2x + 3y - 4z = 0$Also, Find all the solutions of the system.

Last updated date: 19th Mar 2023
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Hint: To solve this question, first we will start with the condition for a non- trivial solution, converting these equations into a determinant of coefficients, find the value of k and then the solutions of the equations.

For a non-trivial solution D = 0 i.e. determinant = 0.
Determinant of the coefficients,
$\Rightarrow \left( {\begin{array}{*{20}{c}} 1&k&3 \\ 3&k&{ - 2} \\ 2&3&{ - 4} \end{array}} \right) = 0$
For making calculation easy we try to make the elements of determinant zero by applying ${R_2} - 3{R_1},{R_3} - 2{R_1}$(here R means row)
∴ $\Delta = \left( {\begin{array}{*{20}{c}} 1&k&3 \\ 0&{ - 2k}&{ - 11} \\ 0&{3 - 2k}&{ - 10} \end{array}} \right) = 0$
Or $20k + 11(3 - 2k) = 0$
Or 33-2k = 0
∴$k = \dfrac{{33}}{2}$
Putting t
the value of k, the equations are
$x + \dfrac{{33}}{2}y + 3z = 0..........(1)$
$3x + \dfrac{{33}}{2}y - 2z = 0..........(2)$
$2x + 3y - 4z = 0............(3)$
Multiply (1) by 3 and subtract from (2) and similarly multiply (1) by 2 and subtract from (3). Thus we get the equivalent system of equations as
$x + \dfrac{{33}}{2}y + 3z = 0$
$- 33y - 11z = 0$
$- 30y - 10z = 0$
From any of the last two, we get 3y = -z
Or$\dfrac{y}{1} = \dfrac{z}{{ - 3}} = \lambda$, say
∴ y =$\lambda$, z = -3$\lambda$
From (1), we get
$x + \dfrac{{33}}{2}y + 3z = 0 \\ x + \dfrac{{33}}{2}y - 9\lambda = 0 \\ \therefore x = \dfrac{{ - 15}}{2}\lambda \\ \therefore x:y:z = - \dfrac{{15}}{2}:1: - 3 \\$

Note: The system of equations in which the determinant of the coefficient is zero is called a non-trivial solution. And the system of equations in which the determinant of the coefficient matrix is not zero but the solutions are x=y=z=0 is called a trivial solution.