Question

If the system of linear equations
 \[\begin{align}
  & x+y+z=5 \\
 & x+2y+2z=6 \\
 & x+3y+\lambda z=\mu \left( \lambda ,\mu \in R \right) \\
\end{align}\]
has infinitely many solutions, then the value of $\lambda +\mu $ is : \[\]
A.12\[\]
B.10\[\]
C.9\[\]
D.7\[\]

Answer 1

Hint: We find the determinants from Kramer’s rule $\Delta ,{{\Delta }_{a}},{{\Delta }_{b}},{{\Delta }_{c}}$ where $\Delta $ is the determinant with entries as coefficients. ${{\Delta }_{a}},{{\Delta }_{b}},{{\Delta }_{c}}$ is $\Delta $ with the first, second and third column replaced with column with constant terms respectively. The necessary condition for infinite solutions is $\Delta ={{\Delta }_{a}}={{\Delta }_{b}}={{\Delta }_{c}}=0$. We find $\lambda $ from $\Delta =0$ and $\mu $ from ${{\Delta }_{a}}={{\Delta }_{b}}={{\Delta }_{c}}=0$. \[\]

Complete step by step answer:
We know the $3\times 3$ linear system of equations with three unknowns $x,y,z$ .
\[\begin{align}
  & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}} \\
 & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}} \\
 & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}} \\
\end{align}\]
If we want to solve the system of equations using Kramer’s rule we need the value of four determinants. The first determinant is the determinant of matrix A which we denote as $\Delta $. So we have
\[\Delta =\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\]
The second one is the determinant where the coefficients of $x$ are replaced with the constant terms in each equation. We denote it as ${{\Delta }_{a}}$and get
\[{{\Delta }_{a}}=\left| \begin{matrix}
   {{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\]

The third one is the determinant where the coefficients of $y$ are replaced with the constant terms in each equation. We denote it as ${{\Delta }_{b}}$and get
\[{{\Delta }_{b}}=\left| \begin{matrix}
   {{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\]
The fourth one is the determinant where the coefficients of $x$ are replaced with the constant terms in each equation. We denote it as ${{\Delta }_{c}}$and get
\[{{\Delta }_{c}}=\left| \begin{matrix}
   {{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\]

We know that if the system of equation has infinitely many solutions then $\Delta ={{\Delta }_{a}}={{\Delta }_{b}}={{\Delta }_{c}}=0$. The given system of equation is
\[\begin{align}
  & x+y+z=5 \\
 & x+2y+2z=6 \\
 & x+3y+\lambda z=\mu \left( \lambda ,\mu \in R \right) \\
\end{align}\]
We are given that the system of equation of equations has finitely many solutions.
So if we use the condition $\Delta =0$ then we can find the value of $\lambda $. Now we have,
\[\Delta =\left| \begin{matrix}
   1 & 1 & 1 \\
   1 & 2 & 2 \\
   1 & 3 & \lambda \\
\end{matrix} \right|=0\]
We can expand by first column and get
\[\begin{align}
  & 1\left( 2\lambda -6 \right)-\left( \lambda -2 \right)+1\left( 3-2 \right)=0 \\
 & \Rightarrow \lambda =3 \\
\end{align}\]
We observe that if we want to find the value of $\mu $ using the condition ${{\Delta }_{a}}=0$ we are getting two same columns ${{\left[ 1,2,3 \right]}^{T}}$. Then we cannot find the value of $\mu $. So either we use ${{\Delta }_{b}}=0$ or ${{\Delta }_{c}}=0$. There may be two possible values of $\mu $ . If we take ${{\Delta }_{b}}=0$ we have,
\[\Delta =\left| \begin{matrix}
   1 & 5 & 1 \\
   1 & 6 & 2 \\
   1 & \mu & \lambda \\
\end{matrix} \right|=0\]
We put $\lambda =3$ and expand by first column to get

\[\begin{align}
  & \Delta =\left| \begin{matrix}
   1 & 5 & 1 \\
   1 & 6 & 2 \\
   1 & \mu & 3 \\
\end{matrix} \right|=0 \\
 & \Rightarrow 1\left( 18-2\mu \right)-1\left( 15-\mu \right)+1\left( 10-6 \right)=0 \\
 & \Rightarrow \mu =7 \\
\end{align}\]
Similarly if we take ${{\Delta }_{c}}=0$we get,
\[\begin{align}
  & \Delta =\left| \begin{matrix}
   1 & 1 & 5 \\
   1 & 2 & 6 \\
   1 & 3 & \mu \\
\end{matrix} \right|=0 \\
 & \Rightarrow \left( 2\mu -18 \right)-\left( \mu -15 \right)+\left( 6-10 \right)=0 \\
 & \Rightarrow \mu =7 \\
\end{align}\]
So the value of $\mu $ is unique and the required value from the question is $\lambda +\mu =3+7=10.$So the correct option is B. \[\]

Note:
We note that the converse of the statement “ the system of equation has infinitely many solutions then $\Delta ={{\Delta }_{a}}={{\Delta }_{b}}={{\Delta }_{c}}=0$.” is not true. The sufficient condition for the system to have infinitely many solutions is $\Delta ={{\Delta }_{a}}={{\Delta }_{b}}={{\Delta }_{c}}=0$ and one of ${{d}_{1}},{{d}_{2}},{{d}_{3}}$ is non- zero.