| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | General solution with parameters |
| Difficulty | Standard +0.8 This is a substantial Further Maths question requiring matrix inversion with a parameter, solving systems using the inverse, and analyzing consistency conditions for a singular case. Part (iii) requires recognizing when the system has no unique solution and finding the general solution with a free parameter, which goes beyond routine matrix manipulation. The multi-part structure and need to handle both non-singular and singular cases elevates this above standard exercises. |
| Spec | 4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix4.03s Consistent/inconsistent: systems of equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Finding cofactors matrix | M1 | |
| \(\det = 4(26-25)-1(39-40)+k(15-16)\) \(= 4+1-k = 5-k\) | M1 A1 | |
| Transpose of cofactor matrix divided by determinant | M1 | |
| Correct inverse matrix expressed with \((5-k)\) in denominator | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Writing in matrix form; using inverse with \(k=7\) | M1 | |
| \(x = \frac{f(m)}{5-k}\), \(y = \frac{g(m)}{5-k}\), \(z = \frac{h(m)}{5-k}\) | M1 | |
| \(x = \frac{13m-26}{-2}\), \(y=...\), \(z=...\) (correct expressions in \(m\)) | A1 A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For solutions to exist with \(k=5\): need consistent equations | M1 | |
| \(p = \frac{m\text{-value giving consistency}}{...}\); \(p=6\) | M1 A1 | |
| General solution with one free parameter \(\lambda\) | M1 | |
| Correct general solution stated | A1 A1 A1 |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Finding cofactors matrix | M1 | |
| $\det = 4(26-25)-1(39-40)+k(15-16)$ $= 4+1-k = 5-k$ | M1 A1 | |
| Transpose of cofactor matrix divided by determinant | M1 | |
| Correct inverse matrix expressed with $(5-k)$ in denominator | A1 A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Writing in matrix form; using inverse with $k=7$ | M1 | |
| $x = \frac{f(m)}{5-k}$, $y = \frac{g(m)}{5-k}$, $z = \frac{h(m)}{5-k}$ | M1 | |
| $x = \frac{13m-26}{-2}$, $y=...$, $z=...$ (correct expressions in $m$) | A1 A1 A1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| For solutions to exist with $k=5$: need consistent equations | M1 | |
| $p = \frac{m\text{-value giving consistency}}{...}$; $p=6$ | M1 A1 | |
| General solution with one free parameter $\lambda$ | M1 | |
| Correct general solution stated | A1 A1 A1 | |
---3 (i) Find the inverse of the matrix $\left( \begin{array} { r r r } 4 & 1 & k \\ 3 & 2 & 5 \\ 8 & 5 & 13 \end{array} \right)$, where $k \neq 5$.\\
(ii) Solve the simultaneous equations
$$\begin{aligned}
& 4 x + y + 7 z = 12 \\
& 3 x + 2 y + 5 z = m \\
& 8 x + 5 y + 13 z = 0
\end{aligned}$$
giving $x , y$ and $z$ in terms of $m$.\\
(iii) Find the value of $p$ for which the simultaneous equations
$$\begin{aligned}
& 4 x + y + 5 z = 12 \\
& 3 x + 2 y + 5 z = p \\
& 8 x + 5 y + 13 z = 0
\end{aligned}$$
have solutions, and find the general solution in this case.
\hfill \mbox{\textit{OCR MEI FP2 2006 Q3 [18]}}