Edexcel FP3 2016 June — Question 1 4 marks

Exam BoardEdexcel
ModuleFP3 (Further Pure Mathematics 3)
Year2016
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeDeterminant calculation and singularity
DifficultyModerate -0.3 This is a straightforward application of the singularity condition (det(A) = 0) for a 3×3 matrix. Students must calculate the determinant using cofactor expansion, set it equal to zero, and solve the resulting quadratic equation. While it involves more computation than a 2×2 case, it's a standard textbook exercise requiring only routine technique with no problem-solving insight, making it slightly easier than average.
Spec4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices

1. $$\mathbf { A } = \left( \begin{array} { r r r } - 2 & 1 & - 3 \\ k & 1 & 3 \\ 2 & - 1 & k \end{array} \right) \text {, where } k \text { is a constant }$$ Given that the matrix \(\mathbf { A }\) is singular, find the possible values of \(k\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\det \mathbf{A} = -2(k+3) - (k^2 - 6) - 3(-k-2)\) (row 1) or equivalent via other rows/columnsM1 Correct attempt at determinant (3 elements, may be implied if one is zero) with at least two elements correct. Various alternatives depending on choice of row or column.
Correct determinant in any formA1 e.g. \(\det \mathbf{A} = -2(k+3) - (k^2-6) - 3(-k-2)\)
\(-2(k+3) - (k^2-6) - 3(-k-2) = 0 \Rightarrow k = \ldots\)M1 Sets \(\det \mathbf{A} = 0\) and attempts to solve a 3-term quadratic. Correct quadratic is \(k^2 - k - 6 = 0\)
\((k+2)(k-3) = 0 \Rightarrow k = -2,\ 3\)A1 Both values correct
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\det \mathbf{A} = -2(k+3) - (k^2 - 6) - 3(-k-2)$ (row 1) or equivalent via other rows/columns | M1 | Correct attempt at determinant (3 elements, may be implied if one is zero) with at least two elements correct. Various alternatives depending on choice of row or column. |
| Correct determinant in any form | A1 | e.g. $\det \mathbf{A} = -2(k+3) - (k^2-6) - 3(-k-2)$ |
| $-2(k+3) - (k^2-6) - 3(-k-2) = 0 \Rightarrow k = \ldots$ | M1 | Sets $\det \mathbf{A} = 0$ and attempts to solve a 3-term quadratic. Correct quadratic is $k^2 - k - 6 = 0$ |
| $(k+2)(k-3) = 0 \Rightarrow k = -2,\ 3$ | A1 | Both values correct |

---
1.

$$\mathbf { A } = \left( \begin{array} { r r r } 
- 2 & 1 & - 3 \\
k & 1 & 3 \\
2 & - 1 & k
\end{array} \right) \text {, where } k \text { is a constant }$$

Given that the matrix $\mathbf { A }$ is singular, find the possible values of $k$.\\

\hfill \mbox{\textit{Edexcel FP3 2016 Q1 [4]}}