| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Find inverse then solve system |
| Difficulty | Standard +0.8 This is a substantial Further Maths question requiring determinant calculation, matrix inversion using cofactors, solving systems via inverse matrices, and analyzing consistency conditions. While the techniques are standard FP1 content, the multi-part structure (5 parts), the parameter analysis in parts (iv)-(v), and the need to recognize when systems are singular elevates this above routine exercises. It's harder than typical A-level but standard for Further Maths. |
| Spec | 4.03j Determinant 3x3: calculation4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix4.03s Consistent/inconsistent: systems of equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\det M = a(3-1) - 2(-(2)) - 1(-2-6) = 2a\) | M1, A1 | For correct expansion process / For showing given answer correctly |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(M^{-1} = \frac{1}{2a}\begin{pmatrix} 2 & -1 & 1 \\ -4 & a+2 & a-2 \\ -8 & a+4 & 3a-4 \end{pmatrix}\) | M1, A1, B1, A1 | For correct process for adjoint entries / For at least 4 correct entries in adjoint / For dividing by the determinant / For completely correct inverse |
| Answer | Marks | Guidance |
|---|---|---|
| So \(x = 0, y = 1, z = 1\) | B1, M1, A1 | For correct statement involving inverse / For carrying out the correct multiplication / For all three correct values |
| Answer | Marks | Guidance |
|---|---|---|
| So for consistency with 1st equn, \(k = 1\) | M1, M1, A1 | For eliminating \(x\) from 2nd and 3rd equns / For comparing two \(y\)-\(z\) equations / For correct value of \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| These values check in \(2y - x = 1\), so soln exists | M1, A1 | For using \(x = z\) to solve a pair of equns / For a completely correct demonstration |
**(i)** $\det M = a(3-1) - 2(-(2)) - 1(-2-6) = 2a$ | M1, A1 | For correct expansion process / For showing given answer correctly
**Total: 2 marks**
**(ii)** $M^{-1} = \frac{1}{2a}\begin{pmatrix} 2 & -1 & 1 \\ -4 & a+2 & a-2 \\ -8 & a+4 & 3a-4 \end{pmatrix}$ | M1, A1, B1, A1 | For correct process for adjoint entries / For at least 4 correct entries in adjoint / For dividing by the determinant / For completely correct inverse
**Total: 4 marks**
**(iii)** $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = M^{-1}\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$, with $a=1$
So $x = 0, y = 1, z = 1$ | B1, M1, A1 | For correct statement involving inverse / For carrying out the correct multiplication / For all three correct values
**Total: 3 marks**
**(iv)** Eliminating $y$ gives $4y - 2z = 2$
So for consistency with 1st equn, $k = 1$ | M1, M1, A1 | For eliminating $x$ from 2nd and 3rd equns / For comparing two $y$-$z$ equations / For correct value of $k$
**Total: 3 marks**
**(v)** Solving $x + 3y = 2, 3x - y = 0$ gives $x = \frac{3}{5}, y = \frac{7}{15}$
These values check in $2y - x = 1$, so soln exists | M1, A1 | For using $x = z$ to solve a pair of equns / For a completely correct demonstration
**Total: 2 marks**
**Question 8 Total: 14 marks**8 The matrix $\mathbf { M }$ is given by $\mathbf { M } = \left( \begin{array} { r r r } a & 2 & - 1 \\ 2 & 3 & - 1 \\ 2 & - 1 & 1 \end{array} \right)$, where $a$ is a constant.\\
(i) Show that the determinant of $\mathbf { M }$ is $2 a$.\\
(ii) Given that $a \neq 0$, find the inverse matrix $\mathbf { M } ^ { - 1 }$.\\
(iii) Hence or otherwise solve the simultaneous equations
$$\begin{array} { r }
x + 2 y - z = 1 \\
2 x + 3 y - z = 2 \\
2 x - y + z = 0
\end{array}$$
(iv) Find the value of $k$ for which the simultaneous equations
$$\begin{array} { r }
2 y - z = k \\
2 x + 3 y - z = 2 \\
2 x - y + z = 0
\end{array}$$
have solutions.\\
(v) Do the equations in part (iv), with the value of $k$ found, have a solution for which $x = z$ ? Justify your answer.
\hfill \mbox{\textit{OCR FP1 Q8 [14]}}