Question-98

Consider the given system of linear equations for some real \(\displaystyle k\).

\[ \begin{aligned} x+ky & =1\\ kx+y & =-1 \end{aligned} \]

Which of the following is/are true?

• There is exactly one value of 𝑘 for which the above system of equations has no solution.
• There exist an infinite number of values of 𝑘 for which the system of equations has no solution.
• There exists exactly one value of 𝑘 for which the system of equations has exactly one solution.
• There exists exactly one value of 𝑘 for which the system of equations has an infinite number of solutions.

NoteAnswer

• There is exactly one value of 𝑘 for which the above system of equations has no solution.
• There exist an infinite number of values of 𝑘 for which the system of equations has no solution.
• There exists exactly one value of 𝑘 for which the system of equations has exactly one solution.
• There exists exactly one value of 𝑘 for which the system of equations has an infinite number of solutions.

NoteSolution

• If \(\displaystyle k=0\), the system has exactly one solution, with \(\displaystyle x=1,y=-1\).

• If \(\displaystyle k\neq 0\), then the system will have a unique solution if \(\displaystyle \frac{1}{k} \neq \frac{k}{1} \Longrightarrow k^{2} \neq 1\Longrightarrow k\neq \pm 1\). So the system has a unique solution whenever \(\displaystyle k=0\) or \(\displaystyle k\neq \pm 1\).

• The system has infinitely many solutions when \(\displaystyle \frac{1}{k} =\frac{k}{1} =-1\). This happens when \(\displaystyle k=-1\).

• The system has no solution when \(\displaystyle \frac{1}{k} =\frac{k}{1} \neq -1\). This happens when \(\displaystyle k=1\).