| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Solving linear systems using matrices |
| Difficulty | Standard +0.3 This is a structured multi-part question testing standard matrix operations (multiplication, inverse, solving systems) with algebraic parameters. Part (a) requires matrix multiplication and comparing coefficients, (b) uses the determinant condition, (c) applies the formula A^{-1} = B/(3k+c), and (d) is routine application. While it involves multiple steps and parameter k, each part follows standard procedures without requiring novel insight or complex problem-solving—slightly easier than average due to the scaffolded structure. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-4k + 2 + 20 - 21 + 7k\) or \(3 + 3k - 2\) or \(7k - 4 - 19 + 24 - 4k\) or \(3k+1\) | M1 | 1.1b - Calculates one element of leading diagonal of AB, condone sign slips |
| \(\{1 + 3k = 3k + c\} \Rightarrow c = 1\) | A1 | 1.1b - Sets diagonal \(= 3k + c\), deduces correct value for \(c\). Award for sight of \(3k+1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3k + 1 = 0 \Rightarrow k = \ldots\) or attempts determinant and sets \(= 0\) | M1 | 1.1b - Attempts to solve \(3k + \text{"1"} = 0\) or attempts determinant, condone sign slips in minors |
| \(\Rightarrow k = -\frac{1}{3}\) | A1ft | 1.1b - Correct value or follow through on \(c\), allow \(k = -\frac{c}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\{A^{-1}\} = \frac{1}{3k+1}\begin{pmatrix} 4k-2 & 1 & 7k-4 \\ -10 & 3 & -19 \\ 3-k & -1 & 6-k \end{pmatrix}\) | B1ft | 2.2a - Deduces correct inverse matrix. Follow through on \(c\), allow \(\frac{1}{3k+c}\mathbf{B}\) or \(\frac{1}{\text{their det}}\mathbf{B}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{3k+1}\begin{pmatrix} 4k-2 & 1 & 7k-4 \\ -10 & 3 & -19 \\ 3-k & -1 & 6-k \end{pmatrix}\begin{pmatrix} 10 \\ 3 \\ 1 \end{pmatrix} = \ldots\) | M1 | 1.2 - Complete method to find values of \(x\), \(y\) and \(z\) using inverse matrix |
| \(\left(\frac{47k-21}{3k+1}, -\frac{110}{3k+1}, \frac{33-11k}{3k+1}\right)\) | A1, A1 | 1.1b - At least one correct coordinate (A1); all coordinates correct and simplified (A1). Condone as column vector |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-4k + 2 + 20 - 21 + 7k$ or $3 + 3k - 2$ or $7k - 4 - 19 + 24 - 4k$ or $3k+1$ | M1 | 1.1b - Calculates one element of leading diagonal of **AB**, condone sign slips |
| $\{1 + 3k = 3k + c\} \Rightarrow c = 1$ | A1 | 1.1b - Sets diagonal $= 3k + c$, deduces correct value for $c$. Award for sight of $3k+1$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3k + 1 = 0 \Rightarrow k = \ldots$ or attempts determinant and sets $= 0$ | M1 | 1.1b - Attempts to solve $3k + \text{"1"} = 0$ or attempts determinant, condone sign slips in minors |
| $\Rightarrow k = -\frac{1}{3}$ | A1ft | 1.1b - Correct value or follow through on $c$, allow $k = -\frac{c}{3}$ |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\{A^{-1}\} = \frac{1}{3k+1}\begin{pmatrix} 4k-2 & 1 & 7k-4 \\ -10 & 3 & -19 \\ 3-k & -1 & 6-k \end{pmatrix}$ | B1ft | 2.2a - Deduces correct inverse matrix. Follow through on $c$, allow $\frac{1}{3k+c}\mathbf{B}$ or $\frac{1}{\text{their det}}\mathbf{B}$ |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{3k+1}\begin{pmatrix} 4k-2 & 1 & 7k-4 \\ -10 & 3 & -19 \\ 3-k & -1 & 6-k \end{pmatrix}\begin{pmatrix} 10 \\ 3 \\ 1 \end{pmatrix} = \ldots$ | M1 | 1.2 - Complete method to find values of $x$, $y$ and $z$ using inverse matrix |
| $\left(\frac{47k-21}{3k+1}, -\frac{110}{3k+1}, \frac{33-11k}{3k+1}\right)$ | A1, A1 | 1.1b - At least one correct coordinate (A1); all coordinates correct and simplified (A1). Condone as column vector |
---4.
$$\mathbf { A } = \left( \begin{array} { r r r }
- 1 & - 2 & - 7 \\
3 & k & 2 \\
1 & 1 & 4
\end{array} \right) \quad \mathbf { B } = \left( \begin{array} { c c c }
4 k - 2 & 1 & 7 k - 4 \\
- 10 & 3 & - 19 \\
3 - k & - 1 & 6 - k
\end{array} \right)$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Determine the value of the constant $c$ for which
$$\mathbf { A B } = ( 3 k + c ) \mathbf { I }$$
\item Hence determine the value of $k$ for which $\mathbf { A } ^ { - 1 }$ does not exist.
Given that $\mathbf { A } ^ { - 1 }$ does exist,
\item write down $\mathbf { A } ^ { - 1 }$ in terms of $k$.
\item Use the answer to part (c) to solve the simultaneous equations
$$\begin{aligned}
- x - 2 y - 7 z & = 10 \\
3 x + k y + 2 z & = 3 \\
x + y + 4 z & = 1
\end{aligned}$$
giving the values of $x , y$ and $z$ in simplest form in terms of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP AS 2024 Q4 [8]}}