Question

The greatest value of $c\in R$ for which the system of linear equation
  $\begin{align}
  & x-cy-cz=0 \\
 & cx-y+cz=0 \\
 & cx+cy-z=0 \\
\end{align}$
has a non-trivial solution, is:
(a) $\dfrac{1}{2}$
(b) $-1$
(c) $0$
(d) $2$

Answer 1

Hint: So here to solve the given question we have firstly find the single equation from $\left| \begin{matrix}
   1 & -c & -c \\
   c & -1 & c \\
   c & c & -1 \\
\end{matrix} \right|=0$ as we have given that this has non-trivial solution, and then we will get the equation after this we have to find the greatest value of the C
So For non-trivial solution, $D=0$ we have given that,
i.e., $\left| \begin{matrix}
   1 & -c & -c \\
   c & -1 & c \\
   c & c & -1 \\
\end{matrix} \right|=0$
Now, find the greatest value of $c$ .

Complete step-by-step solution
As you can see that is given in the question that the system of linear equations
 $\begin{align}
  & x-cy-cz=0 \\
 & cx-y+cz=0 \\
 & cx+cy-z=0 \\
\end{align}$
has a non-trivial solution.
Now, let us understand the condition for a system of linear equations to have a non-trivial solution.
If the system of linear equation
 $\begin{align}
  & px+qy+rz=0 \\
 & {{p}_{1}}x+{{q}_{1}}y+{{r}_{1}}z=0 \\
 & {{p}_{2}}x+{{q}_{2}}y+{{r}_{2}}z=0 \\
\end{align}$
 has a non-trivial solution, then
 $D=0$
i.e., $\left| \begin{matrix}
   p & q & r \\
   {{p}_{1}} & {{q}_{1}} & {{r}_{1}} \\
   {{p}_{2}} & {{q}_{2}} & {{r}_{2}} \\
\end{matrix} \right|=0$
 $\therefore $ We have
  $\left| \begin{matrix}
   1 & -c & -c \\
   c & -1 & c \\
   c & c & -1 \\
\end{matrix} \right|=0$
Expanding w.r.t. \[{{R}_{1}}\left( first\text{ }row \right)\] , we get
 $1\left| \begin{matrix}
   -1 & c \\
   c & -1 \\
\end{matrix} \right|-\left( -c \right)\left| \begin{matrix}
   c & c \\
   c & -1 \\
\end{matrix} \right|+\left( -c \right)\left| \begin{matrix}
   c & -1 \\
   c & c \\
\end{matrix} \right|=0$
 \[\begin{align}
  & \Rightarrow 1\left[ \left( -1 \right)\left( -1 \right)-\left( c \right)\left( c \right) \right]-\left( -c \right)\left[ \left( c \right)\left( -1 \right)-\left( c \right)\left( c \right) \right]+\left( -c \right)\left[ \left( c \right)\left( c \right)-\left( c \right)\left( -1 \right) \right]=0 \\
 & \Rightarrow 1\left( 1-{{c}^{2}} \right)+c\left( -c-{{c}^{2}} \right)-c\left( {{c}^{2}}+c \right)=0 \\
 & \Rightarrow 1-{{c}^{2}}-{{c}^{2}}-{{c}^{3}}-{{c}^{3}}-{{c}^{2}}=0 \\
 & \Rightarrow 1-3{{c}^{2}}-2{{c}^{3}}=0 \\
 & \Rightarrow 2{{c}^{3}}+3{{c}^{2}}-1=0 \\
\end{align}\]
Now we have a cubic equation. So we solve this cubic equation by remainder theorem. For this, we shall use trial and error methods.
Let us put \[c=-1\] . Then
 $\begin{align}
  & L.H.S.=2{{\left( -1 \right)}^{3}}+3{{\left( -1 \right)}^{2}}-1 \\
 & \text{ }=-2+3-1 \\
 & \text{ }=-3+3 \\
 & \text{ }=0=R.H.S. \\
\end{align}$
 \[\therefore \left( c+1 \right)\] is a root of the cubic equation $2{{c}^{3}}+3{{c}^{2}}-1=0$ .
Now we will divide the equation
 $2{{c}^{3}}+3{{c}^{2}}-1=0$ by \[\left( c+1 \right)\] .
 \[c+1\overset{2{{c}^{2}}+c-1}{\overline{\left){\dfrac{\begin{align}
  & 2{{c}^{3}}+3{{c}^{2}}-1 \\
 & 2{{c}^{3}}+2{{c}^{2}} \\
 & -\text{ }- \\
\end{align}}{\dfrac{\begin{align}
  & \text{ }{{c}^{2}}-1 \\
 & \text{ }{{c}^{2}}+c \\
 & \text{ }-\text{ }- \\
\end{align}}{\dfrac{\begin{align}
  & \text{ }-c-1 \\
 & \text{ }-c-1 \\
 & \text{ + +} \\
\end{align}}{\text{ }0}}}}\right.}}\]
 \[2{{c}^{3}}+3{{c}^{2}}-1=\left( c+1 \right)\left( 2{{c}^{2}}+c-1 \right)\]
Let us factorize the equation \[\left( 2{{c}^{2}}+c-1 \right)\] .
 \[\begin{align}
  & \therefore \left( 2{{c}^{2}}+c-1 \right)=2{{c}^{2}}+2c-c-1 \\
 & \text{ }=2c\left( c+1 \right)-\left( c+1 \right) \\
 & \text{ }=\left( 2c-1 \right)\left( c+1 \right) \\
\end{align}\]
Thus, we have
 \[\begin{align}
  & 2{{c}^{3}}+3{{c}^{2}}-1=\left( c+1 \right)\left( 2{{c}^{2}}+c-1 \right) \\
 & \text{ }=\left( c+1 \right)\left( 2c-1 \right)\left( c+1 \right) \\
 & \text{ }={{\left( c+1 \right)}^{2}}\left( 2c-1 \right) \\
\end{align}\]
 \[\begin{align}
  & \therefore 2{{c}^{3}}+3{{c}^{2}}-1=0 \\
 & \Rightarrow {{\left( c+1 \right)}^{2}}\left( 2c-1 \right)=0 \\
 & \Rightarrow {{\left( c+1 \right)}^{2}}=0\text{ }or\text{ }\left( 2c-1 \right)=0 \\
 & \Rightarrow c=-1,-1\text{ }or\text{ }c=\dfrac{1}{2} \\
\end{align}\]
Thus, the greatest value of $c\in R$ for which the system of linear equation
 $\begin{align}
  & x-cy-cz=0 \\
 & cx-y+cz=0 \\
 & cx+cy-z=0 \\
\end{align}$
has a non-trivial solution, is $\dfrac{1}{2}$ .
Hence, the correct option is (a).

Note: It is very important to remember the condition for a system of linear equations to have a non-trivial solution. One can expand the determinant with respect to any row but the calculation should be done properly in order to find the solution. We also have the shortcut we have given the condition $c\in R$ which means our answer should be a Rational number and we have a rational number in the option (a) only so this is the correct answer.