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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?

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    $\begingroup$ Does your third equation have a typo, or is there indeed no $z$ term in it? $\endgroup$
    – RandomUser
    May 15, 2014 at 20:53
  • $\begingroup$ @RandomUser Sorry, it was indeed a typo $\endgroup$ May 15, 2014 at 20:55
  • $\begingroup$ Are you familiar with the various techniques for solving systems of linear equations? purplemath.com/modules/systlin4.htm $\endgroup$
    – RandomUser
    May 15, 2014 at 21:15

1 Answer 1

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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*}

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  • $\begingroup$ I think you've made an error. Note that the determinant of the matrix is zero, so the system is degenerate. There is another solution: $(k/3,-k/6,0)$, and another: $(0,-k/4,3k/2)$. In fact, I believe the entire plane containing your solution and these 2 other points solves the system. $\endgroup$ May 15, 2014 at 22:23
  • $\begingroup$ Rick, thanks for the correction. All three solutions are collinear, so the solution space is a line. As the matrix is clearly not rank 1, we see that the nullspace, and hence the solution space, can't be dimension 2. $\endgroup$ May 16, 2014 at 0:09
  • $\begingroup$ Right. I was doing all the calculations in my head, so I didn't notice the rank when I wrote the comment. +1 for the corrected answer. $\endgroup$ May 16, 2014 at 1:34

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