Question 2 - A-Level Maths

Edexcel F3 2024 January — Question 2 8 marks

Exam Board	Edexcel
Module	F3 (Further Pure Mathematics 3)
Year	2024
Session	January
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	3x3 Matrices
Type	Transformation mapping problems
Difficulty	Standard +0.8 This is a Further Maths question combining matrix inverse conditions with transformation of lines in 3D. Part (a) requires setting up and solving a system from TU=I (standard but computational). Part (b) requires finding the pre-image of a line under transformation T, involving applying T^(-1) to both a point and direction vector—this requires solid understanding of how linear transformations affect geometric objects and is non-routine for A-level.
Spec	4.03o Inverse 3x3 matrix 4.04a Line equations: 2D and 3D, cartesian and vector forms

2. $$\mathbf { T } = \left( \begin{array} { l l l } 2 & 3 & 7 \\ 3 & 2 & 6 \\ a & 4 & b \end{array} \right) \quad \mathbf { U } = \left( \begin{array} { r r r } 6 & - 1 & - 4 \\ 15 & c & - 9 \\ - 8 & a & 5 \end{array} \right)$$ where $a$, $b$ and $c$ are constants.
Given that $\mathbf { T U } = \mathbf { I }$

determine the value of $a$, the value of $b$ and the value of $c$ The transformation represented by the matrix $\mathbf { T }$ transforms the line $l _ { 1 }$ to the line $l _ { 2 }$ Given that $l _ { 2 }$ has equation $$\frac { x - 1 } { 3 } = \frac { y } { - 4 } = z + 2$$
determine a Cartesian equation for $l _ { 1 }$

Show mark scheme Show mark scheme source

Question 2(a):

Way 1 (TU = I):

Answer	Marks	Guidance
$\mathbf{TU} = \mathbf{I} \Rightarrow \begin{pmatrix}2&3&7\\3&2&6\\a&4&b\end{pmatrix}\begin{pmatrix}6&-1&-4\\15&c&-9\\-8&a&5\end{pmatrix} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$	M1	Obtains at least 2 equations with at least one correct. Condone column × row multiplication leading to Way 2 equations. Ignore errors in unused elements.
$6a + 60 - 8b = 0$, $-2 + 3c + 7a = 0$, $-4a - 36 + 5b = 1$, $-3 + 2c + 6a = 1$	dM1	Obtains values for two of $a$, $b$ and $c$. Do not need to check values. Sufficient to just write down values provided previous M mark scored.
$a = 2,\ b = 9,\ c = -4$	A1 A1	A1: Two correct values. A1: All three correct values and no extra values unless rejected.

Way 2 (UT = I):

Answer	Marks	Guidance
$12 - 3 - 4a = 1$, $42 - 6 - 4b = 0$, $[45 + 2c - 36 = 1]$	M1	Obtains at least 2 equations with at least one correct. Condone column × row multiplication. Ignore errors in unused elements.
$-4a = -8,\ -4b = -36\ [2c = -8] \Rightarrow a = \ldots,\ b = \ldots$	dM1	Obtains values for two of $a$, $b$ and $c$. Sufficient to just write down values provided previous M mark scored.

Way 3 (Inverses):

Answer	Marks	Guidance
$\mathbf{T}^{-1} = \mathbf{U} \Rightarrow \frac{1}{4a-5b+36}\begin{pmatrix}2b-24&-3b+28&4\\6a-3b&-7a+2b&9\\-2a+12&3a-8&-5\end{pmatrix} = \begin{pmatrix}6&-1&-4\\15&c&-9\\-8&a&5\end{pmatrix}$	M1	For $\mathbf{T}^{-1} = \frac{1}{f(a,b)}\mathbf{M}$ where M has at least 1 correct element and obtains 2 equations. No requirement to find all elements of M.
OR $\mathbf{U}^{-1} = \mathbf{T}$ similarly applied	M1	For $\mathbf{U}^{-1} = \frac{1}{f(a,c)}\mathbf{M}$ where M has at least 1 correct element and obtains 2 equations.

Question 2(b):

Answer	Marks	Guidance
$\frac{x-1}{3} = \frac{y}{-4} = z+2 \Rightarrow [l_2: \mathbf{r} =] \begin{pmatrix}1\\0\\-2\end{pmatrix} \pm \lambda\begin{pmatrix}3\\-4\\1\end{pmatrix}$	M1	Obtains parametric/vector form (allow one slip only) or clearly identifies position and direction vectors.
$\mathbf{U} \times$ parametric form or $\mathbf{U} \times$ both vectors or $\mathbf{U} \times$ 2 points on line and attempts direction	M1	Complete and correct method with their $b$ and $c$. Must be attempt to multiply correctly — clearly not row×row. Allow attempts using $\mathbf{T}^{-1}$ for $\mathbf{U}$ provided all elements are constants and it is a "changed" T.
$[l_1: \mathbf{r} =] \begin{pmatrix}14+18\lambda\\33+52\lambda\\-18-27\lambda\end{pmatrix} \Rightarrow \frac{x-14}{18} = \frac{y-33}{52} = \frac{z+18}{-27}$	dM1 A1	dM1: Correctly converts result into Cartesian equation. Requires previous method mark. A1: Correct Cartesian equation — allow equivalents e.g. $\frac{z-(-18)}{-27}$, $\frac{-z-18}{27}$

## Question 2(a):

**Way 1 (TU = I):**

$\mathbf{TU} = \mathbf{I} \Rightarrow \begin{pmatrix}2&3&7\\3&2&6\\a&4&b\end{pmatrix}\begin{pmatrix}6&-1&-4\\15&c&-9\\-8&a&5\end{pmatrix} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$ | M1 | Obtains at least 2 equations with at least one correct. Condone column × row multiplication leading to Way 2 equations. Ignore errors in unused elements.

$6a + 60 - 8b = 0$, $-2 + 3c + 7a = 0$, $-4a - 36 + 5b = 1$, $-3 + 2c + 6a = 1$ | dM1 | Obtains values for two of $a$, $b$ and $c$. Do **not** need to check values. Sufficient to just write down values provided previous M mark scored.

$a = 2,\ b = 9,\ c = -4$ | A1 A1 | A1: Two correct values. A1: All three correct values and no extra values unless rejected.

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**Way 2 (UT = I):**

$12 - 3 - 4a = 1$, $42 - 6 - 4b = 0$, $[45 + 2c - 36 = 1]$ | M1 | Obtains at least 2 equations with at least one correct. Condone column × row multiplication. Ignore errors in unused elements.

$-4a = -8,\ -4b = -36\ [2c = -8] \Rightarrow a = \ldots,\ b = \ldots$ | dM1 | Obtains values for two of $a$, $b$ and $c$. Sufficient to just write down values provided previous M mark scored.

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**Way 3 (Inverses):**

$\mathbf{T}^{-1} = \mathbf{U} \Rightarrow \frac{1}{4a-5b+36}\begin{pmatrix}2b-24&-3b+28&4\\6a-3b&-7a+2b&9\\-2a+12&3a-8&-5\end{pmatrix} = \begin{pmatrix}6&-1&-4\\15&c&-9\\-8&a&5\end{pmatrix}$ | M1 | For $\mathbf{T}^{-1} = \frac{1}{f(a,b)}\mathbf{M}$ where **M** has at least 1 correct element **and** obtains 2 equations. No requirement to find all elements of **M**.

OR $\mathbf{U}^{-1} = \mathbf{T}$ similarly applied | M1 | For $\mathbf{U}^{-1} = \frac{1}{f(a,c)}\mathbf{M}$ where **M** has at least 1 correct element **and** obtains 2 equations.

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## Question 2(b):

$\frac{x-1}{3} = \frac{y}{-4} = z+2 \Rightarrow [l_2: \mathbf{r} =] \begin{pmatrix}1\\0\\-2\end{pmatrix} \pm \lambda\begin{pmatrix}3\\-4\\1\end{pmatrix}$ | M1 | Obtains parametric/vector form (allow one slip only) or clearly identifies position and direction vectors.

$\mathbf{U} \times$ parametric form or $\mathbf{U} \times$ both vectors or $\mathbf{U} \times$ 2 points on line and attempts direction | M1 | Complete and correct method with their $b$ and $c$. Must be attempt to multiply correctly — clearly not row×row. Allow attempts using $\mathbf{T}^{-1}$ for $\mathbf{U}$ provided all elements are constants and it is a "changed" **T**.

$[l_1: \mathbf{r} =] \begin{pmatrix}14+18\lambda\\33+52\lambda\\-18-27\lambda\end{pmatrix} \Rightarrow \frac{x-14}{18} = \frac{y-33}{52} = \frac{z+18}{-27}$ | dM1 A1 | dM1: Correctly converts result into Cartesian equation. **Requires previous method mark.** A1: Correct Cartesian equation — allow equivalents e.g. $\frac{z-(-18)}{-27}$, $\frac{-z-18}{27}$

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Show LaTeX source

2.

$$\mathbf { T } = \left( \begin{array} { l l l } 
2 & 3 & 7 \\
3 & 2 & 6 \\
a & 4 & b
\end{array} \right) \quad \mathbf { U } = \left( \begin{array} { r r r } 
6 & - 1 & - 4 \\
15 & c & - 9 \\
- 8 & a & 5
\end{array} \right)$$

where $a$, $b$ and $c$ are constants.\\
Given that $\mathbf { T U } = \mathbf { I }$
\begin{enumerate}[label=(\alph*)]
\item determine the value of $a$, the value of $b$ and the value of $c$

The transformation represented by the matrix $\mathbf { T }$ transforms the line $l _ { 1 }$ to the line $l _ { 2 }$ Given that $l _ { 2 }$ has equation

$$\frac { x - 1 } { 3 } = \frac { y } { - 4 } = z + 2$$
\item determine a Cartesian equation for $l _ { 1 }$
\end{enumerate}

\hfill \mbox{\textit{Edexcel F3 2024 Q2 [8]}}

This paper (8 questions)

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Q1 7 Q2 8 Q3 11 Q4 8 Q5 11 Q6 12 Q7 9 Q8 9

Answer	Marks	Guidance
\(\mathbf{TU} = \mathbf{I} \Rightarrow \begin{pmatrix}2&3&7\\3&2&6\\a&4&b\end{pmatrix}\begin{pmatrix}6&-1&-4\\15&c&-9\\-8&a&5\end{pmatrix} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\)	M1	Obtains at least 2 equations with at least one correct. Condone column × row multiplication leading to Way 2 equations. Ignore errors in unused elements.
\(6a + 60 - 8b = 0\), \(-2 + 3c + 7a = 0\), \(-4a - 36 + 5b = 1\), \(-3 + 2c + 6a = 1\)	dM1	Obtains values for two of \(a\), \(b\) and \(c\). Do not need to check values. Sufficient to just write down values provided previous M mark scored.
\(a = 2,\ b = 9,\ c = -4\)	A1 A1	A1: Two correct values. A1: All three correct values and no extra values unless rejected.

Answer	Marks	Guidance
\(12 - 3 - 4a = 1\), \(42 - 6 - 4b = 0\), \([45 + 2c - 36 = 1]\)	M1	Obtains at least 2 equations with at least one correct. Condone column × row multiplication. Ignore errors in unused elements.
\(-4a = -8,\ -4b = -36\ [2c = -8] \Rightarrow a = \ldots,\ b = \ldots\)	dM1	Obtains values for two of \(a\), \(b\) and \(c\). Sufficient to just write down values provided previous M mark scored.

Answer	Marks	Guidance
\(\mathbf{T}^{-1} = \mathbf{U} \Rightarrow \frac{1}{4a-5b+36}\begin{pmatrix}2b-24&-3b+28&4\\6a-3b&-7a+2b&9\\-2a+12&3a-8&-5\end{pmatrix} = \begin{pmatrix}6&-1&-4\\15&c&-9\\-8&a&5\end{pmatrix}\)	M1	For \(\mathbf{T}^{-1} = \frac{1}{f(a,b)}\mathbf{M}\) where M has at least 1 correct element and obtains 2 equations. No requirement to find all elements of M.
OR \(\mathbf{U}^{-1} = \mathbf{T}\) similarly applied	M1	For \(\mathbf{U}^{-1} = \frac{1}{f(a,c)}\mathbf{M}\) where M has at least 1 correct element and obtains 2 equations.

Answer	Marks	Guidance
\(\frac{x-1}{3} = \frac{y}{-4} = z+2 \Rightarrow [l_2: \mathbf{r} =] \begin{pmatrix}1\\0\\-2\end{pmatrix} \pm \lambda\begin{pmatrix}3\\-4\\1\end{pmatrix}\)	M1	Obtains parametric/vector form (allow one slip only) or clearly identifies position and direction vectors.
\(\mathbf{U} \times\) parametric form or \(\mathbf{U} \times\) both vectors or \(\mathbf{U} \times\) 2 points on line and attempts direction	M1	Complete and correct method with their \(b\) and \(c\). Must be attempt to multiply correctly — clearly not row×row. Allow attempts using \(\mathbf{T}^{-1}\) for \(\mathbf{U}\) provided all elements are constants and it is a "changed" T.
\([l_1: \mathbf{r} =] \begin{pmatrix}14+18\lambda\\33+52\lambda\\-18-27\lambda\end{pmatrix} \Rightarrow \frac{x-14}{18} = \frac{y-33}{52} = \frac{z+18}{-27}\)	dM1 A1	dM1: Correctly converts result into Cartesian equation. Requires previous method mark. A1: Correct Cartesian equation — allow equivalents e.g. \(\frac{z-(-18)}{-27}\), \(\frac{-z-18}{27}\)