OCR Further Pure Core 2 2022 June — Question 7 13 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2022
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeFind inverse then solve system
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring determinant calculation with parameters, matrix inversion, geometric interpretation of singular systems, and understanding of volume scaling under transformations. While each individual technique is standard for FP2, the combination of algebraic manipulation, geometric reasoning about planes, and connecting determinants to orientation/volume scaling makes this moderately challenging, particularly parts (b)(ii) and (c).
Spec4.03j Determinant 3x3: calculation4.03o Inverse 3x3 matrix4.03s Consistent/inconsistent: systems of equations
7 You are given that \(a\) is a parameter which can take only real values.
The matrix \(\mathbf { A }\) is given by \(\mathbf { A } = \left( \begin{array} { c c r } 2 & 4 & - 6 \\ - 3 & 10 - 4 a & 9 \\ 7 & 4 & 4 \end{array} \right)\).
  1. Find an expression for the determinant of \(\mathbf { A }\) in terms of \(a\). You are given the following system of equations in \(x , y\) and \(z\). $$\begin{array} { r r } 2 x + & 4 y - 6 z = \\ - 3 x + & ( 10 - 4 a ) y + 9 z = \\ 7 x + & 4 y + 4 z = \\ 7 x + & 11 \end{array}$$ The system can be written in the form \(\mathbf { A } \left( \begin{array} { c } \mathrm { x } \\ \mathrm { y } \\ \mathrm { z } \end{array} \right) = \left( \begin{array} { r } 6 \\ - 9 \\ 11 \end{array} \right)\).
    1. In the case where \(\mathbf { A }\) is not singular, solve the given system of equations by using \(\mathbf { A } ^ { - 1 }\).
    2. In the case where \(\mathbf { A }\) is singular describe the configuration of the planes whose equations are the three equations of the system. The transformation represented by \(\mathbf { A }\) is denoted by T .
      A 3-D object of volume \(| 5 a - 20 |\) is transformed by T to a 3-D image.
    1. Determine the range of values of \(a\) for which the orientation of the image is the reverse of the orientation of the object.
    2. Determine the range of values of \(a\) for which the volume of the image is less than the volume of the object.
Question 7:
AnswerMarks Guidance
7(a) detA
=2((10−4a)4−94)−4(−34−97)
+(−6)(−34−(10−4a)7)
= − 2 ( 4 0 a 1 6 − − 3 6 ) 4 ( − 1 2 − 6 3 )
− − 6 ( 1 2 − 7 0 + a 2 8 )
= − 8 3 2 a + 3 + 0 0 4 9 2 − 1 6 8 a
AnswerMarks
= − 8 0 0 2 a 0 0 o r 2 0 0 ( 4 − a )M1
A1
AnswerMarks
[2]1.1
1.1Expanding the determinant
ISW once all a terms and all
AnswerMarks Guidance
number terms collectedCondone one calculation error
(b)(i)  4 − 1 6 a 7 5 2 8 a − 8 2 
− 4 0 5 0 2 0
9 6 − 2 4 a 0 3 2 − 8 a
 4 − 1 6 a − 4 0 9 6 − 2 4 a 
7 5 5 0 0
2 8 a − 8 2 2 0 3 2 − 8 a
 4 − 1 6 a − 4 0 9 6 − 2 4 a 
1
A -1 = 7 5 5 0 0
8 0 0 − 2 0 0 a
2 8 a − 8 2 2 0 3 2 − a 8
 6 
 
A−1 −9
 
 
11
 4−16a −40 96−24a 6 
1  
= 75 50 0 −9
  
  
28a−82 20 32−8a 11
1440−360a
1  
= 0
 
800−200a
AnswerMarks
 −320+80a *M1
dep*M
1
A1FT
AnswerMarks
M11.1
1.1
1.1
AnswerMarks
1.1Correctly finding at least 5
cofactors or minors (need not be
in matrix)
Transposing and changing signs
in correct way
FT their determinant
Forming correct product and
attempting multiplication,
resulting in a vector
 3 6 0 ( 4 − a  )
1 
= 0
2 0 0 ( 4 − a )
− 8 0 ( 4 − a ) 
 9 
  9  9 
5
4 0 ( 4 − a )  1
= 0 = 0 = 0

2 0 0 ( 4 − a ) 5
− 2  − 2 2
−
5
9 2
S o = x , y = 0 , z = −
AnswerMarks Guidance
5 5A1
[5]1.1 Simplification to correct
numerical solution (can be left as
AnswerMarks
an unmultiplied vector) 9 
 
5
 
i.e. A1 can be awarded for  0 
 
2
− 
 5
AnswerMarks Guidance
(b)(ii) Singular when detA = 0 => a = 4 so the
second equation is –3x – 6y + 9z = –9...
...which has the same normal (direction) as the
first equation and is consistent with it... oe
...so the first two planes are identical and the
AnswerMarks
third intersects them in a line.M1
M1
A1
AnswerMarks
[3]2.1
1.1
AnswerMarks
2.4Finding a from detA = 0 and
subbing in to 2nd equation.
Could just find the appropriate
normal
eg The second equation is a
multiple of the first.
AnswerMarks Guidance
(c)(i) Orientation reversed if detA < 0 so
200(4 – a) < 0 so a > 4B1FT
[1]3.1a Must be strict inequality
if linear function of a.
AnswerMarks Guidance
(c)(ii) Image volume smaller than object volume =>
–1 < detA < 1...
799 8 0 1
... but a 4 so a4or 4  a  oe
AnswerMarks
200 2 0 0M1
A1
AnswerMarks
[2]3.1a
3.2aUnderstanding that d e t A
represents the volume scale
factor and so d e t A  1 …
...and that𝑎 ≠ 4.
AnswerMarks
Must be strict inequalities7 9 9 8 0 1
 a  and a  4
2 0 0 2 0 0
Question 7:
7 | (a) | detA
=2((10−4a)4−94)−4(−34−97)
+(−6)(−34−(10−4a)7)
= − 2 ( 4 0 a 1 6 − − 3 6 ) 4 ( − 1 2 − 6 3 )
− − 6 ( 1 2 − 7 0 + a 2 8 )
= − 8 3 2 a + 3 + 0 0 4 9 2 − 1 6 8 a
= − 8 0 0 2 a 0 0 o r 2 0 0 ( 4 − a ) | M1
A1
[2] | 1.1
1.1 | Expanding the determinant
ISW once all a terms and all
number terms collected | Condone one calculation error
(b) | (i) |  4 − 1 6 a 7 5 2 8 a − 8 2 
− 4 0 5 0 2 0
9 6 − 2 4 a 0 3 2 − 8 a
 4 − 1 6 a − 4 0 9 6 − 2 4 a 
7 5 5 0 0
2 8 a − 8 2 2 0 3 2 − 8 a
 4 − 1 6 a − 4 0 9 6 − 2 4 a 
1
A -1 = 7 5 5 0 0
8 0 0 − 2 0 0 a
2 8 a − 8 2 2 0 3 2 − a 8
 6 
 
A−1 −9
 
 
11
 4−16a −40 96−24a 6 
1  
= 75 50 0 −9
  

  
28a−82 20 32−8a 11
1440−360a
1  
= 0
 
800−200a

 −320+80a  | *M1
dep*M
1
A1FT
M1 | 1.1
1.1
1.1
1.1 | Correctly finding at least 5
cofactors or minors (need not be
in matrix)
Transposing and changing signs
in correct way
FT their determinant
Forming correct product and
attempting multiplication,
resulting in a vector
 3 6 0 ( 4 − a  )
1 
= 0
2 0 0 ( 4 − a )
− 8 0 ( 4 − a ) 
 9 
  9  9 
5
4 0 ( 4 − a )  1
= 0 = 0 = 0

2 0 0 ( 4 − a ) 5
− 2  − 2 2
−
5
9 2
S o = x , y = 0 , z = −
5 5 | A1
[5] | 1.1 | Simplification to correct
numerical solution (can be left as
an unmultiplied vector) |  9 
 
5
 
i.e. A1 can be awarded for  0 
 
2
− 
 5
(b) | (ii) | Singular when detA = 0 => a = 4 so the
second equation is –3x – 6y + 9z = –9...
...which has the same normal (direction) as the
first equation and is consistent with it... oe
...so the first two planes are identical and the
third intersects them in a line. | M1
M1
A1
[3] | 2.1
1.1
2.4 | Finding a from detA = 0 and
subbing in to 2nd equation.
Could just find the appropriate
normal
eg The second equation is a
multiple of the first.
(c) | (i) | Orientation reversed if detA < 0 so
200(4 – a) < 0 so a > 4 | B1FT
[1] | 3.1a | Must be strict inequality | FT their expression for determinant
if linear function of a.
(c) | (ii) | Image volume smaller than object volume =>
–1 < detA < 1...
799 8 0 1
... but a 4 so a4or 4  a  oe
200 2 0 0 | M1
A1
[2] | 3.1a
3.2a | Understanding that d e t A
represents the volume scale
factor and so d e t A  1 …
...and that𝑎 ≠ 4.
Must be strict inequalities | 7 9 9 8 0 1
 a  and a  4
2 0 0 2 0 0
7 You are given that $a$ is a parameter which can take only real values.\\
The matrix $\mathbf { A }$ is given by $\mathbf { A } = \left( \begin{array} { c c r } 2 & 4 & - 6 \\ - 3 & 10 - 4 a & 9 \\ 7 & 4 & 4 \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the determinant of $\mathbf { A }$ in terms of $a$.

You are given the following system of equations in $x , y$ and $z$.

$$\begin{array} { r r } 
2 x + & 4 y - 6 z = \\
- 3 x + & ( 10 - 4 a ) y + 9 z = \\
7 x + & 4 y + 4 z = \\
7 x + & 11
\end{array}$$

The system can be written in the form $\mathbf { A } \left( \begin{array} { c } \mathrm { x } \\ \mathrm { y } \\ \mathrm { z } \end{array} \right) = \left( \begin{array} { r } 6 \\ - 9 \\ 11 \end{array} \right)$.
\item \begin{enumerate}[label=(\roman*)]
\item In the case where $\mathbf { A }$ is not singular, solve the given system of equations by using $\mathbf { A } ^ { - 1 }$.
\item In the case where $\mathbf { A }$ is singular describe the configuration of the planes whose equations are the three equations of the system.

The transformation represented by $\mathbf { A }$ is denoted by T .\\
A 3-D object of volume $| 5 a - 20 |$ is transformed by T to a 3-D image.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Determine the range of values of $a$ for which the orientation of the image is the reverse of the orientation of the object.
\item Determine the range of values of $a$ for which the volume of the image is less than the volume of the object.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2022 Q7 [13]}}
This paper (10 questions)
View full paper
Q1 6 Q2 5 Q3 6 Q4 4 Q5 7 Q6 10 Q7 13 Q8 7 Q9 9 Q10 8