| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Parameter values for unique solution |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring determinant calculation with parameters, identifying singularity conditions, and crucially recognizing that the system in part (c) corresponds to matrix A with t = b² + 1. The connection between parts requires insight beyond routine calculation, and the parameter substitution adds conceptual complexity typical of Further Pure questions. |
| Spec | 4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((t-1)(6-t(2-2t)) -(t-1)((1-t)-t(2-2t)) +(t-1)((1-t)(2-2t)-6(2-2t))\) | M1 | 1.1 — Correct process for expanding determinant. Fully expanded form: \(2t^3+7t^2-14t+5\) |
| \((t-1)[(6-t(2-2t))-((1-t)-t(2-2t))+((1-t)(2-2t)-6(2-2t))]\) | M1 | 1.1 — Bringing \((t-1)\) or \((t+5)\) or \((2t-1)\) out as factor of the entire expression. Factors may appear BC from no working |
| \((t-1)(6-2t+2t^2-1+t+2t-2t^2+2-4t+2t^2-12+12t) = (t-1)(2t^2+9t-5) = (t-1)(2t-1)(t+5)\) | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-5,\ \frac{1}{2},\ 1\) | B1 | 1.1 — FT their complete factorisation of determinant into 3 linear factors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t = b^2+2\) so that the system is \(\mathbf{Ar}=\mathbf{c}\) | M1 | 2.1 |
| and so \(t\geq2\) so cannot be \(-5\), \(\frac{1}{2}\) or \(1\), therefore \(\mathbf{A}^{-1}\) will exist (for all values of \(b\)) and so there will be a unique solution to the system for all values of \(b\). | A1 | 2.4 — Complete reasoning must be seen. Could test \(t=1\), \(\frac{1}{2}\), \(-5\) in \(b^2=t-2\), and show that these do not give real values of \(b\) |
# Question 8:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(t-1)(6-t(2-2t)) -(t-1)((1-t)-t(2-2t)) +(t-1)((1-t)(2-2t)-6(2-2t))$ | M1 | 1.1 — Correct process for expanding determinant. Fully expanded form: $2t^3+7t^2-14t+5$ |
| $(t-1)[(6-t(2-2t))-((1-t)-t(2-2t))+((1-t)(2-2t)-6(2-2t))]$ | M1 | 1.1 — Bringing $(t-1)$ or $(t+5)$ or $(2t-1)$ out as factor of the entire expression. Factors may appear BC from no working |
| $(t-1)(6-2t+2t^2-1+t+2t-2t^2+2-4t+2t^2-12+12t) = (t-1)(2t^2+9t-5) = (t-1)(2t-1)(t+5)$ | A1 | 1.1 |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-5,\ \frac{1}{2},\ 1$ | B1 | 1.1 — FT their complete factorisation of determinant into 3 linear factors |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t = b^2+2$ so that the system is $\mathbf{Ar}=\mathbf{c}$ | M1 | 2.1 |
| and so $t\geq2$ so cannot be $-5$, $\frac{1}{2}$ or $1$, therefore $\mathbf{A}^{-1}$ will exist (for all values of $b$) and so there will be a unique solution to the system for all values of $b$. | A1 | 2.4 — Complete reasoning must be seen. Could test $t=1$, $\frac{1}{2}$, $-5$ in $b^2=t-2$, and show that these do not give real values of $b$ |
---8 The matrix $\mathbf { A }$ is given by $\mathbf { A } = \left( \begin{array} { c c c } t - 1 & t - 1 & t - 1 \\ 1 - t & 6 & t \\ 2 - 2 t & 2 - 2 t & 1 \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find, in fully factorised form, an expression for $\operatorname { det } \mathbf { A }$ in terms of $t$.
\item State the values of $t$ for which $\mathbf { A }$ is singular.
You are given the following system of equations in $x , y$ and $z$, where $b$ is a real number.
$$\begin{aligned}
\left( b ^ { 2 } + 1 \right) x + \left( b ^ { 2 } + 1 \right) y + \left( b ^ { 2 } + 1 \right) z & = 5 \\
\left( - b ^ { 2 } - 1 \right) x + \quad 6 y + \left( b ^ { 2 } + 2 \right) z & = 10 \\
\left( - 2 b ^ { 2 } - 2 \right) x + \left( - 2 b ^ { 2 } - 2 \right) y + \quad z & = 15
\end{aligned}$$
\item Determine which one of the following statements about the solution of the equations is true.
\begin{itemize}
\item There is a unique solution for all values of $b$.
\item There is a unique solution for some, but not all, values of $b$.
\item There is no unique solution for any value of $b$.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core AS 2021 Q8 [6]}}