Question-6

Consider a system of linear equations:

\[ \begin{aligned} 2x_{1}+3x_{2} & =6\\ -2x_{1}+kx_{2} & =d\\ 4x_{1}+6x_{2} & =12 \end{aligned} \]

Come up with all possible values that \(k\) and \(d\) can take so that the system has:

Let us take one more look at the equations. Dividing the third equation by \(2\), we get:

\[ \begin{aligned} 2x_{1}+3x_{2} & =6\\ -2x_{1}+kx_{2} & =d\\ 2x_{1}+3x_{2} & =6 \end{aligned} \]

The first and last equations are essentially the same. Therefore we only have two equations:

\[ \begin{aligned} 2x_{1}+3x_{2} & =6\\ -2x_{1}+kx_{2} & =d \end{aligned} \]

It is now convenient to think about this geometrically. For the system to have no solution, these two equations should correspond to two distinct parallel lines. They have the same slope but different intercepts. This happens when:

\[ \cfrac{2}{-2}=\cfrac{3}{k}\neq\cfrac{6}{d} \]

This gives us \(k=-3\) and \(d\neq-6\). For the system to have infinitely many solutions, the lines should be identical, which happens when:

\[ \cfrac{2}{-2}=\cfrac{3}{k}=\cfrac{6}{d} \]

This gives \(k=-3\) and \(d=-6\). For the system to have a unique solution, the slopes should be different:

\[ \cfrac{2}{-2}\neq\cfrac{3}{k}\implies k\neq-3 \]