I was doing a past OCR Further Pure 1 Paper from January 2011, but came across the following question that I could not solve, even with the help of the mark scheme:
Determine whether the simultaneous equations $$6x - 6y + z = 3k$$ $$3x + 6y + z = 0$$ $$4x + 2y + z = k$$
where $k$ is a non-zero constant, have a unique solution, no solution or an infinite number of solutions, justifying your answer.
Can anyone help me with the question?
I'm assuming here that the last equation should be $4x + 2y + z = k$.
We can solve this system using linear algebra. To do this, note that the system is equivalent to the matrix equation \begin{equation*} \begin{pmatrix} 6 & -6 & 1 \\ 3 & 6 & 1 \\ 4 & 2 & 1 \end{pmatrix}\begin{pmatrix} x\\y\\z \end{pmatrix} = \begin{pmatrix}3k\\0\\k\end{pmatrix}. \end{equation*}
Now use Gaussian elimination on the augmented matrix (I've skipped the details of how to do this): \begin{equation*} \left(\begin{array}{ccc|c} 6 & -6 & 1 & 3k \\ 3 & 6 & 1 & 0 \\ 4 & 2 & 1 & k \end{array}\right) \to \left(\begin{array}{ccc|c} 1 & 0 & \frac{2}{9} & \frac{k}{3}\\ 0 & 1 & \frac{1}{18} & \frac{-k}{6} \\ 0 & 0 & 0 & 0\end{array}\right)\end{equation*}
This matrix is equivalent to the system
\begin{eqnarray*} x +\frac{2}{9}z&=& \frac{k}{3} \\ y + \frac{1}{18}z &=& \frac{-k}{6}\end{eqnarray*} Now we see that given any value of $z$, we find a solution by setting \begin{eqnarray*} x &=& \frac{k}{3} - \frac{2}{9}z\\ y &=& \frac{-k}{6} - \frac{1}{18}z \end{eqnarray*}
Therefore, for any $k$, the system has infinitely many solutions. The solution set is a line in $\mathbb{R}^3$, with parametrization \begin{equation*} \left(\frac{k}{3},\frac{-k}{6}, 0\right) + t\left(\frac{-2}{9}, \frac{-1}{18}, 1\right).\end{equation*}