Question

The value of \[\lambda \] and \[\mu \] for which the system of equations \[x+y+z=6\] , \[x+2y+3z=10\] , \[x+2y+\lambda z=\mu \] has no solution, are?
A) \[\lambda =3\] , \[\mu \ne 10\]
B) \[\lambda \ne 3\] , \[\mu =10\]
C) \[\lambda \ne 3\] , \[\mu \ne 10\]
D) None of these

Answer 1

Hint: For a given system of linear equations, the condition for no solution is \[\Delta =0\] . Here, we have \[\Delta =\left| \begin{align}
  & \begin{matrix}
   1 & 1 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 2 & 3 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 2 & \lambda \\
\end{matrix} \\
\end{align} \right|\] and make \[\Delta =0\] . We can get the value of \[\lambda \] . Put the value of \[\lambda \] in the equation \[x+2y+\lambda z=\mu \] . The RHS of the equation \[x+2y+3z=10\] and \[x+2y+\lambda z=\mu \] must not be equal to each other.
Complete step-by-step answer:
According to the question we have three equations,
\[x+y+z=6\] ……………………(1)
\[x+2y+3z=10\] …………………………(2)
\[x+2y+\lambda z=\mu \] ………………………….(3)
For the system of linear equations having no solution, we know that \[\Delta =0\] .
Now, we can represent the equations in determinant form. This can be done by writing the coefficients of variables x, y and z as elements of the determinant. Then, after applying the above condition we get,
\[\Delta =\left| \begin{align}
  & \begin{matrix}
   1 & 1 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 2 & 3 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 2 & \lambda \\
\end{matrix} \\
\end{align} \right|=0\] ……………………..(4)
Now, expanding the determinant of equation (4), we get

\[\left| \begin{align}
  & \begin{matrix}
   1 & 1 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 2 & 3 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 2 & \lambda \\
\end{matrix} \\
\end{align} \right|=0\]
\[\begin{align}
  & \Rightarrow 1(2\lambda -6)-1(\lambda -3)+1(2-2)=0 \\
 & \Rightarrow 2\lambda -6-\lambda +3=0 \\
 & \Rightarrow \lambda -3=0 \\
 & \Rightarrow \lambda =3 \\
\end{align}\]
So, the value of \[\lambda \] is 3.
Now, putting the value of \[\lambda \] in equation (3), we get
\[x+2y+3z=\mu \] ………………..(5)
From equation (2) and equation (5), we have
\[x+2y+3z=10\]
\[x+2y+3z=\mu \]
Here, we can see that if \[\mu =10\] then equation (2) and equation (5) will be the same. So, \[\mu \ne 10\] .
Therefore, \[\lambda =3\] and \[\mu \ne 10\] .
Hence, the option (A) is correct.

Note: We can also solve this question in another way. We know that if we have to find the solution of three variables then we require three equations.
We can assume the given three equations like the equations of the planes. If three planes intersect at a point then we will have one solution. If any two equations are parallel then we will have no solution.
According to the question, we have
\[x+y+z=6\] ……………………(1)
\[x+2y+3z=10\] …………………………(2)
\[x+2y+\lambda z=\mu \] ………………………….(3)
From these three equations, only the equation of a plane from equation (2) and equation (3) can be parallel. For being two planes parallel, the coefficients of the x, y, and z must be the same while the constant at the RHS should not be equal to each other.
If we make the coefficients of the variables of equation (2) and equation (3) equal and the constant at the RHS not equal to each other, then we will have no solution.
Comparing the coefficients and constant terms of equation (2) and equation (3), we get
\[\lambda =3\] and \[\mu \ne 10\] .