Question
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=0
3x+ky−2z=0
2x+4y−3z=0
Has a non-zero solution (x,y,z), then y2xz is equal to
A. −30
B. 30
C. −10
D. 10
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 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 thatΔ=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
⇒ 1(−3k+8)−k(−9+4)+3(12−2k)=0
⇒ −3k+8+9k−4k+36−6k=0
⇒ 4k=44
⇒ 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
⇒ 3x+11y−2z− 3x−33y−9z=0
⇒ 11(2y+z)=0
⇒ z=−2y −−−−−−−−(4)
Now we will eliminate x from (1) and (3)
Apply the operation(3)−2(1), we get
⇒ 2x+4y−3z−2x−22y−6z=0
⇒ −18y−9z=0
⇒ 2y+z=0
Hence we need another equation as this is the same one.
Adding (1) and (3),
⇒ (x+11y+3z)+(2x+4y−3z)=0
⇒ 3x+15y=0
⇒ x=−5y −−−−−(5)
Now we can use the equations (4) and (5) to get our desired result y2xz
So using the value of z from (4) and x from (5),
⇒ y2xz =y2−5y(−2y)=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.