Question

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.

Answer 1

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,

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.