Edexcel FP1 2012 June — Question 2 6 marks

Exam BoardEdexcel
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMatrices
TypeSingular matrix conditions
DifficultyModerate -0.8 Part (a) is straightforward matrix multiplication requiring only careful arithmetic. Part (b) requires finding when a matrix is singular (determinant = 0), which is a standard technique in FP1. Both parts are routine applications of basic further maths matrix operations with no problem-solving insight needed.
Spec4.03b Matrix operations: addition, multiplication, scalar4.03l Singular/non-singular matrices

2. (a) Given that $$\mathbf { A } = \left( \begin{array} { l l l } 3 & 1 & 3 \\ 4 & 5 & 5 \end{array} \right) \quad \text { and } \quad \mathbf { B } = \left( \begin{array} { r r } 1 & 1 \\ 1 & 2 \\ 0 & - 1 \end{array} \right)$$ find \(\mathbf { A B }\).
(b) Given that $$\mathbf { C } = \left( \begin{array} { l l } 3 & 2 \\ 8 & 6 \end{array} \right) , \quad \mathbf { D } = \left( \begin{array} { r r } 5 & 2 k \\ 4 & k \end{array} \right) , \text { where } k \text { is a constant }$$ and $$\mathbf { E } = \mathbf { C } + \mathbf { D }$$ find the value of \(k\) for which \(\mathbf { E }\) has no inverse.

Question 2:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{AB} = \begin{pmatrix} 3 & 1 & 3 \\ 4 & 5 & 5 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 2 \\ 0 & -1 \end{pmatrix}\)
\(= \begin{pmatrix} 3+1+0 & 3+2-3 \\ 4+5+0 & 4+10-5 \end{pmatrix}\)M1 Correct method to multiply two matrices. Implied by two out of four correct (unsimplified) elements in a dimensionally correct 2×2 matrix with a number or calculation at each corner
\(= \begin{pmatrix} 4 & 2 \\ 9 & 9 \end{pmatrix}\)A1 Correct answer. A correct answer with no working can score both marks
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{C} + \mathbf{D} = \begin{pmatrix} 3 & 2 \\ 8 & 6 \end{pmatrix} + \begin{pmatrix} 5 & 2k \\ 4 & k \end{pmatrix} = \begin{pmatrix} 8 & 2k+2 \\ 12 & 6+k \end{pmatrix}\)M1 Attempt to add C to D. Implied by two out of four correct (unsimplified) elements in dimensionally correct matrix
\(\mathbf{E}\) does not have an inverse \(\Rightarrow \det \mathbf{E} = 0\)
\(8(6+k) - 12(2k+2)\)M1 Applies \(ad - bc\) to E where E is a 2×2 matrix
\(8(6+k) - 12(2k+2) = 0\)M1 States or applies \(\det(\mathbf{E}) = 0\) where \(\det(\mathbf{E}) = ad - bc\) or \(ad + bc\) only and E is a 2×2 matrix
\(48 + 8k = 24k + 24\), \(24 = 16k\) Note \(8(6+k) - 12(2k+2) = 0\) or \(8(6+k) = 12(2k+2)\) could score both M's
\(k = \frac{3}{2}\)A1 oe
## Question 2:

### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{AB} = \begin{pmatrix} 3 & 1 & 3 \\ 4 & 5 & 5 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 2 \\ 0 & -1 \end{pmatrix}$ | | |
| $= \begin{pmatrix} 3+1+0 & 3+2-3 \\ 4+5+0 & 4+10-5 \end{pmatrix}$ | M1 | Correct method to multiply two matrices. Implied by two out of four correct (unsimplified) elements in a dimensionally correct 2×2 matrix with a number or calculation at each corner |
| $= \begin{pmatrix} 4 & 2 \\ 9 & 9 \end{pmatrix}$ | A1 | Correct answer. A correct answer with no working can score both marks |

### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{C} + \mathbf{D} = \begin{pmatrix} 3 & 2 \\ 8 & 6 \end{pmatrix} + \begin{pmatrix} 5 & 2k \\ 4 & k \end{pmatrix} = \begin{pmatrix} 8 & 2k+2 \\ 12 & 6+k \end{pmatrix}$ | M1 | Attempt to add C to D. Implied by two out of four correct (unsimplified) elements in dimensionally correct matrix |
| $\mathbf{E}$ does not have an inverse $\Rightarrow \det \mathbf{E} = 0$ | | |
| $8(6+k) - 12(2k+2)$ | M1 | Applies $ad - bc$ to **E** where **E** is a 2×2 matrix |
| $8(6+k) - 12(2k+2) = 0$ | M1 | States or applies $\det(\mathbf{E}) = 0$ where $\det(\mathbf{E}) = ad - bc$ or $ad + bc$ only and **E** is a 2×2 matrix |
| $48 + 8k = 24k + 24$, $24 = 16k$ | | Note $8(6+k) - 12(2k+2) = 0$ or $8(6+k) = 12(2k+2)$ could score both M's |
| $k = \frac{3}{2}$ | A1 oe | |

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2. (a) Given that

$$\mathbf { A } = \left( \begin{array} { l l l } 
3 & 1 & 3 \\
4 & 5 & 5
\end{array} \right) \quad \text { and } \quad \mathbf { B } = \left( \begin{array} { r r } 
1 & 1 \\
1 & 2 \\
0 & - 1
\end{array} \right)$$

find $\mathbf { A B }$.\\
(b) Given that

$$\mathbf { C } = \left( \begin{array} { l l } 
3 & 2 \\
8 & 6
\end{array} \right) , \quad \mathbf { D } = \left( \begin{array} { r r } 
5 & 2 k \\
4 & k
\end{array} \right) , \text { where } k \text { is a constant }$$

and

$$\mathbf { E } = \mathbf { C } + \mathbf { D }$$

find the value of $k$ for which $\mathbf { E }$ has no inverse.\\

\hfill \mbox{\textit{Edexcel FP1 2012 Q2 [6]}}