| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Geometric interpretation of systems |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring determinant calculation to find k values where planes don't meet uniquely, then analyzing geometric configurations. While it involves multiple steps (setting det=0, solving quadratic, checking consistency for each k), the techniques are standard for FM students: determinant of 3×3 matrix, factoring/solving quadratics, and row reduction to identify plane configurations. The conceptual demand is moderate—recognizing that non-unique solutions require det=0, then distinguishing between consistent (line/plane of intersection) and inconsistent (parallel) cases—but these are well-rehearsed FM topics without requiring novel insight. |
| Spec | 4.03r Solve simultaneous equations: using inverse matrix4.03s Consistent/inconsistent: systems of equations4.03t Plane intersection: geometric interpretation4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms |
| 7 |
|
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = (4k+1)[4(3-k)+6] + 3[4(k-1)-14] + (k-5)[-3(k-1)-7(3-k)]\) | B1 | Shows correctly that \(k = 4.5\) gives determinant of 0, or shows \(k = 4.5\) from solving det \(M = 0\) |
| \(0 = -12k^2 + 42k + 54\) | M1 | Correctly expands determinant of the matrix and equates to 0; condone misread of \(-3y\) as \(+3y\) |
| \(k = 4.5,\ k = -1\) | ||
| \(k = -1\) | A1 | Obtains \(k = -1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(k = 4.5\) matrix becomes: \(\begin{pmatrix}19 & -3 & -0.5 & 3\\ 3.5 & -1.5 & 2 & 1\\ 0 & 0 & 0 & 0\end{pmatrix}\). The system of equations is consistent. | B1 | When \(k = 4.5\), clearly shows or explains that the system of equations is consistent, using equations of planes or augmented matrix form. Must state the system is consistent. |
| Two planes are the same and intersect the third plane in a line. | B1 | States that two planes are the same and intersect the third plane |
| When \(k = -1\) matrix becomes: \(\begin{pmatrix}0 & 0 & 0 & 66\\ -2 & 4 & 2 & 1\\ 11 & -11 & 0 & 0\end{pmatrix}\). The system of equations is inconsistent. | M1 | When \(k = -1\), completes appropriate working to find the consistency of the system using their \(k\) |
| The three planes form a prism. | A1 | States that the system is inconsistent and that the three planes form a prism. CSO |
# Question 7:
## Part 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = (4k+1)[4(3-k)+6] + 3[4(k-1)-14] + (k-5)[-3(k-1)-7(3-k)]$ | B1 | Shows correctly that $k = 4.5$ gives determinant of 0, or shows $k = 4.5$ from solving det $M = 0$ |
| $0 = -12k^2 + 42k + 54$ | M1 | Correctly expands determinant of the matrix and equates to 0; condone misread of $-3y$ as $+3y$ |
| $k = 4.5,\ k = -1$ | | |
| $k = -1$ | A1 | Obtains $k = -1$ |
## Part 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $k = 4.5$ matrix becomes: $\begin{pmatrix}19 & -3 & -0.5 & 3\\ 3.5 & -1.5 & 2 & 1\\ 0 & 0 & 0 & 0\end{pmatrix}$. The system of equations is consistent. | B1 | When $k = 4.5$, clearly shows or explains that the system of equations is consistent, using equations of planes or augmented matrix form. Must state the system is consistent. |
| Two planes are the same and intersect the third plane in a line. | B1 | States that two planes are the same and intersect the third plane |
| When $k = -1$ matrix becomes: $\begin{pmatrix}0 & 0 & 0 & 66\\ -2 & 4 & 2 & 1\\ 11 & -11 & 0 & 0\end{pmatrix}$. The system of equations is inconsistent. | M1 | When $k = -1$, completes appropriate working to find the consistency of the system using their $k$ |
| The three planes form a prism. | A1 | States that the system is inconsistent and that the three planes form a prism. CSO |
---7 Three planes have equations
$$\begin{aligned}
( 4 k + 1 ) x - 3 y + ( k - 5 ) z & = 3 \\
( k - 1 ) x + ( 3 - k ) y + 2 z & = 1 \\
7 x - 3 y + 4 z & = 2
\end{aligned}$$
7
\begin{enumerate}[label=(\alph*)]
\item The planes do not meet at a unique point.\\
Show that $k = 4.5$ is one possible value of $k$, and find the other possible value of $k$.\\
\begin{center}
\begin{tabular}{|l|l|}
\hline
7
\item & \begin{tabular}{l}
For each value of $k$ found in part (a), identify the configuration of the given planes. \\
In each case fully justify your answer, stating whether or not the equations of the planes form a consistent system. \\[0pt]
[4 marks] $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ \\
\end{tabular} \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 2020 Q7 [7]}}