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Question: If the system of the linear equations: \(x + ky + 3z = 0\) \(3x + ky - 2z = 0\) \(2x + 4y - 3z...

If the system of the linear equations:
x+ky+3z=0x + ky + 3z = 0
3x+ky2z=03x + ky - 2z = 0
2x+4y3z=02x + 4y - 3z = 0
Has a non-zero solution (x,y,z)(x,y,z), then xzy2\dfrac{{xz}}{{{y^2}}} is equal to
A. 30 - 30
B. 3030
C. 10 - 10
D. 1010

Explanation

Solution

At first, use the condition for the system of the linear equations to have the non-zero solution from which you will get the value of kk. Then, you will get all the three equations.
Solve these equations and get the answer.

Complete step-by-step answer:
As stated in the question, the system of equations has the non-zero solution. Hence, we need to use the condition thatΔ=0\Delta = 0. For the equations to have the non-zero solution:
\left| {\begin{array}{*{20}{c}} 1&k;&3 \\\ 3&k;&{ - 2} \\\ 2&4&{ - 3} \end{array}} \right| = 0
\Rightarrow 1(3k+8)k(9+4)+3(122k)=01( - 3k + 8) - k( - 9 + 4) + 3(12 - 2k) = 0
\Rightarrow 3k+8+9k4k+366k=0 - 3k + 8 + 9k - 4k + 36 - 6k = 0
\Rightarrow 4k=444k = 44
\Rightarrow k=11k = 11
Now the three equations become,
x+11y+3z=0x + 11y + 3z = 0 (1) - - - - - - - - (1)
3x+11y2z=03x + 11y - 2z = 0 (2) - - - - - - - (2)
2x+4y3z=02x + 4y - 3z = 0 (3) - - - - - - - - - (3)
Now we have to solve these equations and get the solution
Now applying the condition (2)3(1)(2) - 3(1)operation
\Rightarrow 3x+11y2z3x + 11y - 2z - 3x33y9z=03x - 33y - 9z = 0
\Rightarrow 11(2y+z)=011(2y + z) = 0
\Rightarrow z=2yz = - 2y (4) - - - - - - - - (4)
Now we will eliminate xx from (1) and (3)
Apply the operation(3)2(1)(3) - 2(1), we get
\Rightarrow 2x+4y3z2x22y6z=02x + 4y - 3z - 2x - 22y - 6z = 0
\Rightarrow 18y9z=0 - 18y - 9z = 0
\Rightarrow 2y+z=02y + z = 0
Hence we need another equation as this is the same one.
Adding (1) and (3),
\Rightarrow (x+11y+3z)+(2x+4y3z)=0(x + 11y + 3z) + (2x + 4y - 3z) = 0
\Rightarrow 3x+15y=03x + 15y = 0
\Rightarrow x=5yx = - 5y (5) - - - - - (5)
Now we can use the equations (4) and (5) to get our desired result xzy2\dfrac{{xz}}{{{y^2}}}
So using the value of zz from (4) and xx from (5),
\Rightarrow xzy2\dfrac{{xz}}{{{y^2}}} =5y(2y)y2=10 = \dfrac{{ - 5y( - 2y)}}{{{y^2}}} = 10

Hence option D is the correct answer.

Note: General form of a linear equation in one variable is ax+c=0.ax+c=0.
General form of a linear equation in two variables is ax+by=cax+by=c.
Linear equations come in handy when one variable is given and we have to find the other variable.