| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Non-singular matrix proof |
| Difficulty | Moderate -0.8 This is a straightforward Further Maths question requiring routine application of determinant formula (2×2 matrix), completing the square to show det A > 0, and finding a matrix inverse using the standard formula. All steps are mechanical with no problem-solving insight needed, making it easier than average despite being Further Maths content. |
| Spec | 4.03h Determinant 2x2: calculation4.03l Singular/non-singular matrices4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(\det \mathbf{A} = a(a+4) - (-5 \times 2) = a^2 + 4a + 10\) | M1 A1 (2) | Correct use of \(ad - bc\) for M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(a^2 + 4a + 10 = (a+2)^2 + 6\) | M1 A1ft | Attempt to complete the square for M1; Alt: attempt \(b^2-4ac\) for M1, correct \(-24\) for A1ft |
| Positive for all values of \(a\), so \(\mathbf{A}\) is non-singular | A1 cso (3) | Minimum value 6 for A1ft; "no real roots or equivalent" for final A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{A}^{-1} = \dfrac{1}{10}\begin{pmatrix} 4 & 5 \\ -2 & 0 \end{pmatrix}\) | B1 M1 A1 (3) | B1 for \(\dfrac{1}{10}\); swap leading diagonal and change sign of other diagonal for M1; correct matrix independent of \(\dfrac{1}{10}\) for final A1 |
## Question 5:
**(a)**
$\det \mathbf{A} = a(a+4) - (-5 \times 2) = a^2 + 4a + 10$ | M1 A1 (2) | Correct use of $ad - bc$ for M1
**(b)**
$a^2 + 4a + 10 = (a+2)^2 + 6$ | M1 A1ft | Attempt to complete the square for M1; Alt: attempt $b^2-4ac$ for M1, correct $-24$ for A1ft
Positive for all values of $a$, so $\mathbf{A}$ is non-singular | A1 cso (3) | Minimum value 6 for A1ft; "no real roots or equivalent" for final A1cso
**(c)**
$\mathbf{A}^{-1} = \dfrac{1}{10}\begin{pmatrix} 4 & 5 \\ -2 & 0 \end{pmatrix}$ | B1 M1 A1 (3) | B1 for $\dfrac{1}{10}$; swap leading diagonal and change sign of other diagonal for M1; correct matrix independent of $\dfrac{1}{10}$ for final A1
---5. $\mathbf { A } = \left( \begin{array} { c c } a & - 5 \\ 2 & a + 4 \end{array} \right)$, where $a$ is real.
\begin{enumerate}[label=(\alph*)]
\item Find $\operatorname { det } \mathbf { A }$ in terms of $a$.
\item Show that the matrix $\mathbf { A }$ is non-singular for all values of $a$.
Given that $a = 0$,
\item find $\mathbf { A } ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2010 Q5 [8]}}