| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Topic | 3x3 Matrices |
| Type | Consistency conditions for systems |
| Difficulty | Challenging +1.2 This is a standard Further Maths linear algebra question requiring determinant calculation to find when the system has no unique solution, then checking consistency via row reduction or substitution. While it involves multiple steps and requires understanding of when systems are consistent/inconsistent, the techniques are routine for Further Maths students and follow a predictable pattern. The conceptual demand is moderate—understanding det=0 for non-unique solutions and distinguishing consistent from inconsistent cases—but the execution is algorithmic. |
| Spec | 4.03s Consistent/inconsistent: systems of equations |
| Answer | Marks |
|---|---|
| (i) No unique soln. \(\Leftrightarrow \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & k \\ 1 & k & 6 \end{vmatrix} = 0\) | M1 |
| Gaining and solving a quadratic eqn. in \(k\) | M1 |
| \(0 = -(k^2 - 8k + 15) = -(k-3)(k-5) \Rightarrow k = 3, 5\) | A1 |
| [3] | |
| (ii) For \(k=3\): | |
| \(x + 2y + 3z = 4\) | M1 |
| \(2 \times (1) - (2) \Rightarrow y + 3z = -1\) | A1 |
| \((3) - (1) \Rightarrow y + 3z = -3\) | |
| Subst'. back and eliminating one variable; inconsistency correctly shown | M1, A1 |
| [4] | |
| For \(k=5\): | |
| \(x + 2y + 3z = 4\) | |
| \(2 \times (1) - (2) \Rightarrow y + z = -1\) | M1 |
| \((3) - (1) \Rightarrow 3y + 3z = -3\) | A1 |
| Subst'. back and eliminating one variable; consistency correctly shown | M1, A1 |
| [4] | |
| ALTERNATIVE (whole qn.) Row reduction with convincing attempt at Gaussian elimination | M1 |
| \(\begin{vmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & k & 9 \\ 1 & k & 6 & 1 \end{vmatrix} \to \begin{vmatrix} 0 & -1 & k-6 & 1 \\ 0 & k-2 & 3 & -3 \end{vmatrix}\) or by Cramer's Rule | M1 |
| \(\to \begin{vmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 6-k & -1 \\ 0 & k-2 & 3 & -3 \end{vmatrix}\) | |
| \(\to \begin{vmatrix} 0 & 1 & 6-k & -1 \\ 0 & k-2 & 3 & -3 \end{vmatrix}\) | |
| Final \(R_2\) | A1 |
| Final \(R_3\) | A1 |
| \(0 \quad k^2 - 8k + 15 \mid k-5\) \(R_3' = R_3 - (k-2)R_2\) | M1 |
| \(k^2 - 8k + 15 = 0\) for no unique solution | M1, A1 |
| \(k = 3, 5\) | A1, B1 |
| Noting \(k = 3 \Rightarrow R_3 = 0\ 0\ 0 \mid -2\) giving inconsistency | B1 |
| Noting \(k = 5 \Rightarrow R_3 = 0\ 0\ 0 \mid 0\) giving consistency | B1 |
| [7] |
**(i)** No unique soln. $\Leftrightarrow \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & k \\ 1 & k & 6 \end{vmatrix} = 0$ | M1 |
Gaining and solving a quadratic eqn. in $k$ | M1 |
$0 = -(k^2 - 8k + 15) = -(k-3)(k-5) \Rightarrow k = 3, 5$ | A1 |
| [3] |
**(ii)** For $k=3$: | |
$x + 2y + 3z = 4$ | M1 |
$2 \times (1) - (2) \Rightarrow y + 3z = -1$ | A1 |
$(3) - (1) \Rightarrow y + 3z = -3$ | |
Subst'. back and eliminating one variable; inconsistency correctly shown | M1, A1 |
| [4] |
For $k=5$: | |
$x + 2y + 3z = 4$ | |
$2 \times (1) - (2) \Rightarrow y + z = -1$ | M1 |
$(3) - (1) \Rightarrow 3y + 3z = -3$ | A1 |
Subst'. back and eliminating one variable; consistency correctly shown | M1, A1 |
| [4] |
**ALTERNATIVE (whole qn.)** Row reduction with convincing attempt at Gaussian elimination | M1 |
$\begin{vmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & k & 9 \\ 1 & k & 6 & 1 \end{vmatrix} \to \begin{vmatrix} 0 & -1 & k-6 & 1 \\ 0 & k-2 & 3 & -3 \end{vmatrix}$ or by Cramer's Rule | M1 |
$\to \begin{vmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 6-k & -1 \\ 0 & k-2 & 3 & -3 \end{vmatrix}$ | |
$\to \begin{vmatrix} 0 & 1 & 6-k & -1 \\ 0 & k-2 & 3 & -3 \end{vmatrix}$ | |
Final $R_2$ | A1 |
Final $R_3$ | A1 |
$0 \quad k^2 - 8k + 15 \mid k-5$ $R_3' = R_3 - (k-2)R_2$ | M1 |
$k^2 - 8k + 15 = 0$ for no unique solution | M1, A1 |
$k = 3, 5$ | A1, B1 |
Noting $k = 3 \Rightarrow R_3 = 0\ 0\ 0 \mid -2$ giving inconsistency | B1 |
Noting $k = 5 \Rightarrow R_3 = 0\ 0\ 0 \mid 0$ giving consistency | B1 |
| [7] |
---\begin{enumerate}[label=(\roman*)]
\item Determine the two values of $k$ for which the system of equations
\begin{align}
x + 2y + 3z &= 4 \\
2x + 3y + kz &= 9 \\
x + ky + 6z &= 1
\end{align}
has no unique solution. [3]
\item Show that the system is consistent for one of these values of $k$ and inconsistent for the other. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2011 Q8 [7]}}