| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Singular matrix conditions |
| Difficulty | Moderate -0.8 This is a straightforward FP1 matrices question testing basic operations: matrix addition with identity matrix, determinant calculation for singularity, and matrix multiplication of column×row vectors. All parts are routine recall and application of standard procedures with no problem-solving insight required, making it easier than average A-level questions. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03l Singular/non-singular matrices |
| Answer | Marks | Guidance |
|---|---|---|
| (i)(a) \(B = A + 3I = \begin{pmatrix} 2k+1 & k \\ -3 & -5 \end{pmatrix} + 3\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2k+4 & k \\ -3 & -2 \end{pmatrix}\) | M1 | For applying \(A + 3I\). Can be implied by three out of four correct elements in candidate's final answer. Solution must come from addition. |
| Correct answer | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(-2(2k+4) - (-3k) = 0\) | M1 | Applies "\(ad - bc\)" to B and equates to 0 |
| \(-4k - 8 + 3k = 0\) | ||
| \(k = -8\) | A1 cao | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(E = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}(2 \, -1 \, 5) = \begin{pmatrix} 4 & -2 & 10 \\ -6 & 3 & -15 \\ 8 & -4 & 20 \end{pmatrix}\) | M1 | Candidate writes down a \(3 \times 3\) matrix |
| Correct answer | A1 | [2] |
**(i)(a)** $B = A + 3I = \begin{pmatrix} 2k+1 & k \\ -3 & -5 \end{pmatrix} + 3\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2k+4 & k \\ -3 & -2 \end{pmatrix}$ | M1 | For applying $A + 3I$. Can be implied by three out of four correct elements in candidate's final answer. Solution must come from addition. | [2]
Correct answer | A1
**(b)** B is singular $\Rightarrow \det B = 0$
$-2(2k+4) - (-3k) = 0$ | M1 | Applies "$ad - bc$" to B and equates to 0
$-4k - 8 + 3k = 0$ | |
$k = -8$ | A1 cao | [2]
**(ii)** $C = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}$, $D = (2 \, -1 \, 5)$, $E = CD$
$E = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}(2 \, -1 \, 5) = \begin{pmatrix} 4 & -2 & 10 \\ -6 & 3 & -15 \\ 8 & -4 & 20 \end{pmatrix}$ | M1 | Candidate writes down a $3 \times 3$ matrix
Correct answer | A1 | [2]
**Total: [6]**
---\begin{enumerate}[label=(\roman*)]
\item $\mathbf{A} = \begin{pmatrix} 2k + 1 & k \\ -3 & -5 \end{pmatrix}$, where $k$ is a constant
Given that
$$\mathbf{B} = \mathbf{A} + 3\mathbf{I}$$
where $\mathbf{I}$ is the $2 \times 2$ identity matrix, find
\begin{enumerate}[label=(\alph*)]
\item $\mathbf{B}$ in terms of $k$, [2]
\item the value of $k$ for which $\mathbf{B}$ is singular. [2]
\end{enumerate}
\item Given that
$$\mathbf{C} = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}, \quad \mathbf{D} = (2 \quad -1 \quad 5)$$
and
$$\mathbf{E} = \mathbf{CD}$$
find $\mathbf{E}$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q2 [6]}}