OCR Further Pure Core AS 2023 June — Question 7 6 marks

Exam BoardOCR
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2023
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeDeterminant calculation and singularity
DifficultyStandard +0.8 This question requires computing a 3×3 determinant with algebraic entries, setting up an equation where det(A) equals its value at a=2, then solving the resulting cubic equation. While the determinant calculation is standard Further Maths content, the algebraic manipulation and solving a cubic (likely requiring factorization after finding one root) elevates this beyond routine practice to moderate-challenging problem-solving.
Spec4.03j Determinant 3x3: calculation
7 In this question you must show detailed reasoning. Matrix \(\mathbf { A }\) is given by \(\mathbf { A } = \left( \begin{array} { c c c } a & - 6 & a - 3 \\ a + 9 & a & 4 \\ 0 & - 13 & a - 1 \end{array} \right)\) where \(a\) is a constant.
Find all possible values of \(a\) for which \(\operatorname { det } \mathbf { A }\) has the same value as it has when \(a = 2\).
Question 7:
AnswerMarks Guidance
AnswerMarks Guidance
\(\det\mathbf{A} = a(a(a-1) - 4\times(-13)) - (-6)((a+9)(a-1)-0) + (a-3)((a+9)(-13)-0)\)M1 Attempt to expand determinant using standard method, at least two terms, at least one comprising \(\pm\) a number multiplied by residual determinant
\(= a^3 - 8a^2 + 22a + 297\)A1
\(a^3 - 8a^2 + 22a + 297 = 2^3 - 8\times2^2 + 22\times2 + 297\) (or \(= 317\))M1 Setting up equation equating determinant to specific value for \(a=2\)
\(a^3 - 8a^2 + 22a - 20 = 0\); \(a^2(a-2) - 6a(a-2) + 10(a-2) = 0\); \(a^2 - 6a + 10 = 0\)M1 Rearranging to \(=0\) and use of factor theorem to derive quadratic in \(a\) by dividing by \((a-2)\)
\((a-3)^2 - 9 + 10 = 0\); \((a-3)^2 = -1\); \(a - 3 = \pm\mathrm{i}\)M1 Attempt to solve quadratic involving \(\sqrt{-1} = \mathrm{i}\)
\(a = 3 \pm \mathrm{i}\) (and \(a=2\))A1 Both. No need to mention \(a=2\) explicitly 
# Question 7:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\det\mathbf{A} = a(a(a-1) - 4\times(-13)) - (-6)((a+9)(a-1)-0) + (a-3)((a+9)(-13)-0)$ | **M1** | Attempt to expand determinant using standard method, at least two terms, at least one comprising $\pm$ a number multiplied by residual determinant |
| $= a^3 - 8a^2 + 22a + 297$ | **A1** | |
| $a^3 - 8a^2 + 22a + 297 = 2^3 - 8\times2^2 + 22\times2 + 297$ (or $= 317$) | **M1** | Setting up equation equating determinant to specific value for $a=2$ |
| $a^3 - 8a^2 + 22a - 20 = 0$; $a^2(a-2) - 6a(a-2) + 10(a-2) = 0$; $a^2 - 6a + 10 = 0$ | **M1** | Rearranging to $=0$ and use of factor theorem to derive quadratic in $a$ by dividing by $(a-2)$ |
| $(a-3)^2 - 9 + 10 = 0$; $(a-3)^2 = -1$; $a - 3 = \pm\mathrm{i}$ | **M1** | Attempt to solve quadratic involving $\sqrt{-1} = \mathrm{i}$ |
| $a = 3 \pm \mathrm{i}$ (and $a=2$) | **A1** | Both. No need to mention $a=2$ explicitly |

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7 In this question you must show detailed reasoning.
Matrix $\mathbf { A }$ is given by $\mathbf { A } = \left( \begin{array} { c c c } a & - 6 & a - 3 \\ a + 9 & a & 4 \\ 0 & - 13 & a - 1 \end{array} \right)$ where $a$ is a constant.\\
Find all possible values of $a$ for which $\operatorname { det } \mathbf { A }$ has the same value as it has when $a = 2$.

\hfill \mbox{\textit{OCR Further Pure Core AS 2023 Q7 [6]}}
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