Question

Consider the following statements. The system of equations
$\begin{align}
  & 2x-y=4 \\
 & px-y=q \\
\end{align}$
$1.$ has a unique solution if $p\ne 2$
$2.$ has infinitely many solutions if $p=2,q=4$ of these statements
A. $1$ alone is correct
B. $2$ alone is correct
C. $1$ and $2$ are correct
D. $1$ and $2$ are false

Answer 1

Hint: We first write it in another form as
$\begin{align}
  & 2x-y-4=0 \\
 & px-y-q=0 \\
\end{align}$ .
We then write the values of three determinants $\Delta =\left| \begin{matrix}
   2 & -1 \\
   p & -1 \\
\end{matrix} \right|$ , ${{\Delta }_{1}}=\left| \begin{matrix}
   -1 & -4 \\
   -1 & -q \\
\end{matrix} \right|$ , ${{\Delta }_{2}}=\left| \begin{matrix}
   -4 & 2 \\
   -q & p \\
\end{matrix} \right|$ . There are two cases which may arise:
$1.$ $\Delta \ne 0$ : The system of equations has unique solution
$2.$ $\Delta =0$ and either of ${{\Delta }_{1}},{{\Delta }_{2}}$ are non-zero: The system has no solution
$3.$ $\Delta =0$ and both ${{\Delta }_{1}},{{\Delta }_{2}}$ are zero: The system has infinite number of solutions
Solving accordingly, we get the correct option.

Complete step by step solution:
We can also write the system of equations $\begin{align}
  & 2x-y=4 \\
 & px-y=q \\
\end{align}$ as,
$\begin{align}
  & 2x-y-4=0 \\
 & px-y-q=0 \\
\end{align}$
Now, we can assume three determinants which are $\Delta ,{{\Delta }_{1}},{{\Delta }_{2}}$ . The formula for the determinants will be,
$\Delta =\left| \begin{matrix}
   2 & -1 \\
   p & -1 \\
\end{matrix} \right|$ , ${{\Delta }_{1}}=\left| \begin{matrix}
   -1 & -4 \\
   -1 & -q \\
\end{matrix} \right|$ , ${{\Delta }_{2}}=\left| \begin{matrix}
   -4 & 2 \\
   -q & p \\
\end{matrix} \right|$
We know that the entire solvability of the system of linear equations is dependent on the value of $\Delta ,{{\Delta }_{1}},{{\Delta }_{2}}$ . There are three cases which may arise:
$1.$ $\Delta \ne 0$ : The system of equations has unique solution
$2.$ $\Delta =0$ and either of ${{\Delta }_{1}},{{\Delta }_{2}}$ are non-zero: The system has no solution
$3.$ $\Delta =0$ and both ${{\Delta }_{1}},{{\Delta }_{2}}$ are zero: The system has infinite number of solutions
This means that for unique solution,
$\begin{align}
  & \Delta =\left| \begin{matrix}
   2 & -1 \\
   p & -1 \\
\end{matrix} \right|\ne 0 \\
 & \Rightarrow -2+p\ne 0 \\
 & \Rightarrow p\ne 2 \\
\end{align}$
So, the necessary and sufficient condition for unique solution is $p\ne 2$ .
For infinite number of solutions,
$\begin{align}
  & {{\Delta }_{1}}=\left| \begin{matrix}
   -1 & -4 \\
   -1 & -q \\
\end{matrix} \right|=0 \\
 & \Rightarrow q-4=0 \\
 & \Rightarrow q=4 \\
\end{align}$ and $\begin{align}
  & \Delta =\left| \begin{matrix}
   2 & -1 \\
   p & -1 \\
\end{matrix} \right|=0 \\
 & \Rightarrow -2+p=0 \\
 & \Rightarrow p=2 \\
\end{align}$
So, the necessary and sufficient condition for infinitely many solutions is $p=2,q=4$ .
Thus, we can conclude that the correct option is Option C.

Note: We can also solve the problem by intuition. For infinitely many solutions, the two equations must be one and the same. For that, $p=2,q=4$ . And, for a unique solution, the two equations must not be equivalent. $p\ne 2$ alone satisfies this condition as it gives a second line which is neither parallel nor the same as that of the line $2x-y=4$ .