If the system of the linear equations:
$x + ky + 3z = 0$
$3x + ky - 2z = 0$
$2x + 4y - 3z = 0$
Has a non-zero solution $(x,y,z)$, then $\dfrac{{xz}}{{{y^2}}}$ is equal to
A. $ - 30$
B. $30$
C. $ - 10$
D. $10$
Answer
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Hint: 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 $k$. 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 that$\Delta = 0$. 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$$1( - 3k + 8) - k( - 9 + 4) + 3(12 - 2k) = 0$
$\Rightarrow$$ - 3k + 8 + 9k - 4k + 36 - 6k = 0$
$\Rightarrow$$4k = 44$
$\Rightarrow$$k = 11$
Now the three equations become,
$x + 11y + 3z = 0$$ - - - - - - - - (1)$
$3x + 11y - 2z = 0$$ - - - - - - - (2)$
$2x + 4y - 3z = 0$$ - - - - - - - - - (3)$
Now we have to solve these equations and get the solution
Now applying the condition $(2) - 3(1)$operation
$\Rightarrow$$3x + 11y - 2z - $$3x - 33y - 9z = 0$
$\Rightarrow$$11(2y + z) = 0$
$\Rightarrow$$z = - 2y$$ - - - - - - - - (4)$
Now we will eliminate $x$ from (1) and (3)
Apply the operation$(3) - 2(1)$, we get
$\Rightarrow$$2x + 4y - 3z - 2x - 22y - 6z = 0$
$\Rightarrow$$ - 18y - 9z = 0$
$\Rightarrow$$2y + z = 0$
Hence we need another equation as this is the same one.
Adding (1) and (3),
$\Rightarrow$$(x + 11y + 3z) + (2x + 4y - 3z) = 0$
$\Rightarrow$$3x + 15y = 0$
$\Rightarrow$$x = - 5y$$ - - - - - (5)$
Now we can use the equations (4) and (5) to get our desired result $\dfrac{{xz}}{{{y^2}}}$
So using the value of $z$ from (4) and $x$ from (5),
$\Rightarrow$$\dfrac{{xz}}{{{y^2}}}$$ = \dfrac{{ - 5y( - 2y)}}{{{y^2}}} = 10$
Hence option D is the correct answer.
Note: General form of a linear equation in one variable is $ax+c=0.$
General form of a linear equation in two variables is $ax+by=c$.
Linear equations come in handy when one variable is given and we have to find the other variable.
Solve these equations and get the answer.
Complete step-by-step answer:
As stated in the question, the system of equations has the non-zero solution. Hence, we need to use the condition that$\Delta = 0$. For the equations to have the non-zero solution:
$\left| {\begin{array}{*{20}{c}}
1&k&3 \\
3&k&{ - 2} \\
2&4&{ - 3}
\end{array}} \right| = 0$
$\Rightarrow$$1( - 3k + 8) - k( - 9 + 4) + 3(12 - 2k) = 0$
$\Rightarrow$$ - 3k + 8 + 9k - 4k + 36 - 6k = 0$
$\Rightarrow$$4k = 44$
$\Rightarrow$$k = 11$
Now the three equations become,
$x + 11y + 3z = 0$$ - - - - - - - - (1)$
$3x + 11y - 2z = 0$$ - - - - - - - (2)$
$2x + 4y - 3z = 0$$ - - - - - - - - - (3)$
Now we have to solve these equations and get the solution
Now applying the condition $(2) - 3(1)$operation
$\Rightarrow$$3x + 11y - 2z - $$3x - 33y - 9z = 0$
$\Rightarrow$$11(2y + z) = 0$
$\Rightarrow$$z = - 2y$$ - - - - - - - - (4)$
Now we will eliminate $x$ from (1) and (3)
Apply the operation$(3) - 2(1)$, we get
$\Rightarrow$$2x + 4y - 3z - 2x - 22y - 6z = 0$
$\Rightarrow$$ - 18y - 9z = 0$
$\Rightarrow$$2y + z = 0$
Hence we need another equation as this is the same one.
Adding (1) and (3),
$\Rightarrow$$(x + 11y + 3z) + (2x + 4y - 3z) = 0$
$\Rightarrow$$3x + 15y = 0$
$\Rightarrow$$x = - 5y$$ - - - - - (5)$
Now we can use the equations (4) and (5) to get our desired result $\dfrac{{xz}}{{{y^2}}}$
So using the value of $z$ from (4) and $x$ from (5),
$\Rightarrow$$\dfrac{{xz}}{{{y^2}}}$$ = \dfrac{{ - 5y( - 2y)}}{{{y^2}}} = 10$
Hence option D is the correct answer.
Note: General form of a linear equation in one variable is $ax+c=0.$
General form of a linear equation in two variables is $ax+by=c$.
Linear equations come in handy when one variable is given and we have to find the other variable.
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