| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Matrix properties verification |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question on 3×3 matrices requiring calculation of a determinant (showing it's non-zero for all real k) and using the matrix inverse property AA^(-1) = I to find k. Both parts involve standard techniques with no novel insight required, making it slightly easier than average even for Further Maths content. |
| Spec | 4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(k\begin{vmatrix}5 & 2\\3 & -k\end{vmatrix} - \begin{vmatrix}6 & 2\\-1 & -k\end{vmatrix} = k(-5k-6)-(-6k+2) = -5k^2-2\) | M1 A1 | Evaluates determinant, forms quadratic expression. Allow 1 slip in the calculation of the determinant for M1 only. |
| No real value of \(k \Rightarrow\) Non-singular | A1 | Convincing conclusion using the discriminant or determinant. |
| Answer | Marks | Guidance |
|---|---|---|
| \(3k+1=1\) or \(-\dfrac{1}{2} = -\dfrac{1}{5k^2+2}\) | M1 | Uses \(\mathbf{A}\mathbf{A}^{-1} = \mathbf{I}\) or \(\det(\mathbf{A}^{-1}) = (\det\mathbf{A})^{-1}\) to find equation in \(k\). Could also use a minor determinant e.g. \(-k \times 1 - 3 \times 0 = 0\) |
| \(k = 0\) | A1 | Only \(k=0\), A0 for any additional solutions. |
## Question 1:
### Part (a):
$k\begin{vmatrix}5 & 2\\3 & -k\end{vmatrix} - \begin{vmatrix}6 & 2\\-1 & -k\end{vmatrix} = k(-5k-6)-(-6k+2) = -5k^2-2$ | **M1 A1** | Evaluates determinant, forms quadratic expression. Allow 1 slip in the calculation of the determinant for M1 only.
No real value of $k \Rightarrow$ Non-singular | **A1** | Convincing conclusion using the discriminant or determinant.
**Total: 3 marks**
---
### Part (b):
$3k+1=1$ or $-\dfrac{1}{2} = -\dfrac{1}{5k^2+2}$ | **M1** | Uses $\mathbf{A}\mathbf{A}^{-1} = \mathbf{I}$ or $\det(\mathbf{A}^{-1}) = (\det\mathbf{A})^{-1}$ to find equation in $k$. Could also use a minor determinant e.g. $-k \times 1 - 3 \times 0 = 0$
$k = 0$ | **A1** | Only $k=0$, A0 for any additional solutions.
**Total: 2 marks**1 The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { c c c }
k & 1 & 0 \\
6 & 5 & 2 \\
- 1 & 3 & - k
\end{array} \right)$$
where $k$ is a real constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathbf { A }$ is non-singular.
\item Given that $\mathbf { A } ^ { - 1 } = \left( \begin{array} { c c c } 3 & 0 & - 1 \\ 1 & 0 & 0 \\ - \frac { 23 } { 2 } & \frac { 1 } { 2 } & 3 \end{array} \right)$, find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q1 [5]}}