| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Consistency conditions for systems |
| Difficulty | Challenging +1.2 This is a structured multi-part question on singular matrices and consistency conditions. Part (i) requires finding determinant and matrix inverse (standard FP1/FP2 technique). Parts (ii)-(iv) guide students through discovering that k=3 makes M singular and exploring the null space and consistency conditions. While it involves several concepts, the question provides significant scaffolding and uses routine techniques throughout. It's moderately harder than average A-level due to being Further Maths content and requiring synthesis across parts, but doesn't demand novel insight. |
| Spec | 4.03j Determinant 3x3: calculation4.03o Inverse 3x3 matrix4.03s Consistent/inconsistent: systems of equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\det(\mathbf{M}) = 1(16-12)+1(20-18)+k(10-12) = 6-2k\) | M1, A1 | Obtaining \(\det(\mathbf{M})\) in terms of \(k\) |
| \(\Rightarrow\) no inverse if \(k=3\) | A1 | Accept \(k\neq 3\) after correct determinant |
| M1 | Evaluating at least four cofactors (including one involving \(k\)) | |
| \(\mathbf{M}^{-1} = \frac{1}{6-2k}\begin{pmatrix}4 & 4+2k & -6-4k\\-2 & 4-3k & 5k-6\\-2 & -5 & 9\end{pmatrix}\) | A1 | Six signed cofactors correct (including one involving \(k\)) |
| M1, A1 | Transposing and dividing by \(\det(\mathbf{M})\); dependent on previous M1M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix}1&-1&3\\5&4&6\\3&2&4\end{pmatrix}\begin{pmatrix}-3\\3\\1\end{pmatrix}=\begin{pmatrix}-3\\3\\1\end{pmatrix}\) | M1, A1 | Setting \(k=3\) and multiplying |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix}-3\\3\\1\end{pmatrix}\) is an eigenvector | B1 | For credit here, 2/2 scored in (ii); accept "invariant point" |
| corresponding to an eigenvalue of 1 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3x+6y=1-2t,\ x+2y=2,\ 5x+10y=-4t\) | M1 | Eliminating one variable in two different ways |
| (or \(9x+18z=4t+1,\ 5x+10z=2t,\ x+2z=-1\)) (or \(9y-9z=1-5t,\ 5y-5z=-3t,\ 2y-2z=3\)) | A1 | Two correct equations |
| For solutions, \(1-2t=3\times 2\) | M1 | Validly obtaining a value of \(t\) |
| \(\Rightarrow t = -\frac{5}{2}\) | A1 | |
| \(x=\lambda,\ y=1-\frac{1}{2}\lambda,\ z=-\frac{1}{2}-\frac{1}{2}\lambda\) | M1, A1 | Obtaining general solution by setting one unknown \(=\lambda\) and finding other two in terms of \(\lambda\) (accept unknown instead of \(\lambda\)) |
| Straight line | B1 | Accept "sheaf". Independent of all previous marks |
# Question 3:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\det(\mathbf{M}) = 1(16-12)+1(20-18)+k(10-12) = 6-2k$ | M1, A1 | Obtaining $\det(\mathbf{M})$ in terms of $k$ |
| $\Rightarrow$ no inverse if $k=3$ | A1 | Accept $k\neq 3$ after correct determinant |
| | M1 | Evaluating at least four cofactors (including one involving $k$) |
| $\mathbf{M}^{-1} = \frac{1}{6-2k}\begin{pmatrix}4 & 4+2k & -6-4k\\-2 & 4-3k & 5k-6\\-2 & -5 & 9\end{pmatrix}$ | A1 | Six signed cofactors correct (including one involving $k$) |
| | M1, A1 | Transposing and dividing by $\det(\mathbf{M})$; dependent on previous M1M1 |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix}1&-1&3\\5&4&6\\3&2&4\end{pmatrix}\begin{pmatrix}-3\\3\\1\end{pmatrix}=\begin{pmatrix}-3\\3\\1\end{pmatrix}$ | M1, A1 | Setting $k=3$ and multiplying |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix}-3\\3\\1\end{pmatrix}$ is an eigenvector | B1 | For credit here, 2/2 scored in (ii); accept "invariant point" |
| corresponding to an eigenvalue of 1 | B1 | |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $3x+6y=1-2t,\ x+2y=2,\ 5x+10y=-4t$ | M1 | Eliminating one variable in two different ways |
| (or $9x+18z=4t+1,\ 5x+10z=2t,\ x+2z=-1$) (or $9y-9z=1-5t,\ 5y-5z=-3t,\ 2y-2z=3$) | A1 | Two correct equations |
| For solutions, $1-2t=3\times 2$ | M1 | Validly obtaining a value of $t$ |
| $\Rightarrow t = -\frac{5}{2}$ | A1 | |
| $x=\lambda,\ y=1-\frac{1}{2}\lambda,\ z=-\frac{1}{2}-\frac{1}{2}\lambda$ | M1, A1 | Obtaining general solution by setting one unknown $=\lambda$ and finding other two in terms of $\lambda$ (accept unknown instead of $\lambda$) |
| Straight line | B1 | Accept "sheaf". Independent of all previous marks |
---3 (i) Find the value of $k$ for which the matrix
$$\mathbf { M } = \left( \begin{array} { r r r }
1 & - 1 & k \\
5 & 4 & 6 \\
3 & 2 & 4
\end{array} \right)$$
does not have an inverse.\\
Assuming that $k$ does not take this value, find the inverse of $\mathbf { M }$ in terms of $k$.\\
(ii) In the case $k = 3$, evaluate
$$\mathbf { M } \left( \begin{array} { r }
- 3 \\
3 \\
1
\end{array} \right)$$
(iii) State the significance of what you have found in part (ii).\\
(iv) Find the value of $t$ for which the system of equations
$$\begin{array} { r }
x - y + 3 z = t \\
5 x + 4 y + 6 z = 1 \\
3 x + 2 y + 4 z = 0
\end{array}$$
has solutions. Find the general solution in this case and describe the solution geometrically.
\hfill \mbox{\textit{OCR MEI FP2 2011 Q3 [18]}}