Question

The value of $ \lambda $ for which the system of equation $ 2x-y-z=12 $ , $ x-2y+z=-4 $ and $ x+y+\lambda z=4 $ has no solution.
a.) 3
b.) -3
c.) 2
d.) -2

Answer 1

Hint: To solve the above system non-linear homogenous equation we will use the determinant method. Since we know that for an equation having three variables then the values of those are given by: $ x=\dfrac{{{D}_{1}}}{D} $ , $ y=\dfrac{{{D}_{2}}}{D} $ and $ z=\dfrac{{{D}_{3}}}{D} $ , where D is the determinant formed by the coefficient of x, y, z and $ {{D}_{1}},{{D}_{2}},{{D}_{3}} $ are the determinant formed by replacing first, the second and third column of the coefficient determinant by the constant of each equation respectively.
The given system’s non-linear homogenous equation will have no solution when D = 0, which means the determinant formed by the coefficient of the equations should have zero value.

Complete step by step answer:
Since we can see that the given equations are systems of non-linear homogenous equations. So, we will solve it by using the determinant method.
From the determinant method we know that value of x, y, z is given by:
 $ x=\dfrac{{{D}_{1}}}{D} $ , $ y=\dfrac{{{D}_{2}}}{D} $ , $ z=\dfrac{{{D}_{3}}}{D} $ , where D is the determinant formed by coefficient of x, y, z and $ {{D}_{1}},{{D}_{2}},{{D}_{3}} $ are the determinant formed by replacing first, second and third column of the coefficient determinant by the constant of each equations respectively.
And, from the we know that the equations are: $ 2x-y-z=12 $ , $ x-2y+z=-4 $ and $ x+y+\lambda z=4 $
So, $ D=\left| \begin{matrix}
   2 & -1 & -1 \\
   1 & -2 & 1 \\
   1 & 1 & \lambda \\
\end{matrix} \right| $
And, we know that the given system of the linear equation will have no solution if and only if D = 0.
So, we will equation the determinant formed by the coefficient of the equations to zero.
So, $ D=\left| \begin{matrix}
   2 & -1 & -1 \\
   1 & -2 & 1 \\
   1 & 1 & \lambda \\
\end{matrix} \right|=0 $
 $ \Rightarrow D=2\left( -2\lambda -1 \right)+\left( \lambda -1 \right)-\left( 1+2 \right)=0 $
 $ \Rightarrow D=-3\lambda -6=0 $
 $ \therefore \lambda =-2 $
This is our required solution.
Hence, option (d) is the right answer.

Note:
Students are required to note that when we are given a system of the non-linear homogenous equation we will always use the determinant method to solve them and we must memorize the unique solution will exist only when the determinant formed by the coefficient of the equations is not equal to zero and when it becomes equal to zero then the equation will have no solution or infinite solution.