| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | General solution with parameters |
| Difficulty | Standard +0.3 This is a standard Further Maths matrix question testing routine techniques: finding a matrix inverse using cofactors/adjugate method, verifying a specific case, and solving linear systems. Part (iii) requires recognizing when a system is singular (a=4 case) and interpreting geometrically as three planes meeting in a line, which is slightly beyond pure mechanical calculation but still a well-practiced FM topic. The parameter work is straightforward substitution rather than requiring novel insight. |
| Spec | 4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Evaluating determinant: \(4-a\) | M1, A1 | |
| Finding at least four cofactors | M1 | |
| Six signed cofactors correct | A1 | |
| \(\mathbf{M}^{-1} = \dfrac{1}{4-a}\begin{pmatrix}2 & -2-2a & 2+a \\ 2 & 2-3a & 2a-2 \\ -1 & 5 & -3\end{pmatrix}\) | M1 | Transposing and dividing by det |
| When \(a=-1\), \(\mathbf{M}^{-1}=\dfrac{1}{5}\begin{pmatrix}2&0&1\\2&5&-4\\-1&5&-3\end{pmatrix}\) | A1 | \(\mathbf{M}^{-1}\) correct (in terms of \(a\)) and result for \(a=-1\) stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}x\\y\\z\end{pmatrix}=\dfrac{1}{5}\begin{pmatrix}2&0&1\\2&5&-4\\-1&5&-3\end{pmatrix}\begin{pmatrix}-2\\b\\1\end{pmatrix}\) | M2 | Attempting to multiply \((-2\ b\ 1)^T\) by given matrix (M0 if wrong order) |
| Multiplying out | M1 | |
| \(\Rightarrow x=-\dfrac{3}{5},\ y=b-\dfrac{8}{5},\ z=b-\dfrac{1}{5}\) | A2 | A1 for one correct |
| OR: \(4x+y=b-4\), \(x-y=1-b\) | M1 | Eliminating one unknown in 2 ways |
| Solve to obtain one value | M1 | Dep. on M1 above |
| \(\Rightarrow x=-\dfrac{3}{5}\) | A1 | One unknown correct |
| Finding the other two unknowns | M1 | |
| \(\Rightarrow y=b-\dfrac{8}{5},\ z=b-\dfrac{1}{5}\) | A1 | Both correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. \(3x-3y=2b+2\), \(5x-5y=4\) | M1, A1A1 | Eliminating one unknown in 2 ways; two correct equations |
| Consistent if \(\dfrac{2b+2}{3}=\dfrac{4}{5}\) | M1 | Attempting to find \(b\) |
| \(\Rightarrow b=\dfrac{1}{5}\) | A1 | |
| Solution is a line | B2 |
# Question 3:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Evaluating determinant: $4-a$ | M1, A1 | |
| Finding at least four cofactors | M1 | |
| Six signed cofactors correct | A1 | |
| $\mathbf{M}^{-1} = \dfrac{1}{4-a}\begin{pmatrix}2 & -2-2a & 2+a \\ 2 & 2-3a & 2a-2 \\ -1 & 5 & -3\end{pmatrix}$ | M1 | Transposing and dividing by det |
| When $a=-1$, $\mathbf{M}^{-1}=\dfrac{1}{5}\begin{pmatrix}2&0&1\\2&5&-4\\-1&5&-3\end{pmatrix}$ | A1 | $\mathbf{M}^{-1}$ correct (in terms of $a$) and result for $a=-1$ stated |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}x\\y\\z\end{pmatrix}=\dfrac{1}{5}\begin{pmatrix}2&0&1\\2&5&-4\\-1&5&-3\end{pmatrix}\begin{pmatrix}-2\\b\\1\end{pmatrix}$ | M2 | Attempting to multiply $(-2\ b\ 1)^T$ by given matrix (M0 if wrong order) |
| Multiplying out | M1 | |
| $\Rightarrow x=-\dfrac{3}{5},\ y=b-\dfrac{8}{5},\ z=b-\dfrac{1}{5}$ | A2 | A1 for one correct |
| OR: $4x+y=b-4$, $x-y=1-b$ | M1 | Eliminating one unknown in 2 ways |
| Solve to obtain one value | M1 | Dep. on M1 above |
| $\Rightarrow x=-\dfrac{3}{5}$ | A1 | One unknown correct |
| Finding the other two unknowns | M1 | |
| $\Rightarrow y=b-\dfrac{8}{5},\ z=b-\dfrac{1}{5}$ | A1 | Both correct |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. $3x-3y=2b+2$, $5x-5y=4$ | M1, A1A1 | Eliminating one unknown in 2 ways; two correct equations |
| Consistent if $\dfrac{2b+2}{3}=\dfrac{4}{5}$ | M1 | Attempting to find $b$ |
| $\Rightarrow b=\dfrac{1}{5}$ | A1 | |
| Solution is a line | B2 | |
---3 (i) Find the inverse of the matrix
$$\left( \begin{array} { r r r }
1 & 1 & a \\
2 & - 1 & 2 \\
3 & - 2 & 2
\end{array} \right)$$
where $a \neq 4$.\\
Show that when $a = - 1$ the inverse is
$$\frac { 1 } { 5 } \left( \begin{array} { r r r }
2 & 0 & 1 \\
2 & 5 & - 4 \\
- 1 & 5 & - 3
\end{array} \right)$$
(ii) Solve, in terms of $b$, the following system of equations.
$$\begin{aligned}
x + y - z & = - 2 \\
2 x - y + 2 z & = b \\
3 x - 2 y + 2 z & = 1
\end{aligned}$$
(iii) Find the value of $b$ for which the equations
$$\begin{aligned}
x + y + 4 z & = - 2 \\
2 x - y + 2 z & = b \\
3 x - 2 y + 2 z & = 1
\end{aligned}$$
have solutions. Give a geometrical interpretation of the solutions in this case.
Section B (18 marks)
\hfill \mbox{\textit{OCR MEI FP2 2010 Q3 [18]}}