OCR Further Pure Core 2 2024 June — Question 3 7 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2024
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeFind inverse then solve system
DifficultyStandard +0.3 This is a straightforward Further Pure question testing basic matrix operations: scalar multiplication, subtraction, identifying the identity matrix, determinant calculation, and using inverse matrices to solve simultaneous equations. All parts are routine applications of standard techniques with no problem-solving insight required. While it's Further Maths content, these are foundational matrix skills that are slightly easier than average A-level questions overall.
Spec4.03b Matrix operations: addition, multiplication, scalar4.03h Determinant 2x2: calculation4.03n Inverse 2x2 matrix4.03r Solve simultaneous equations: using inverse matrix
Matrices \(\mathbf{A}\) and \(\mathbf{B}\) are given by \(\mathbf{A} = \begin{pmatrix} 4 & -3 \\ -2 & 2 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 3 & -5 \\ 0 & 1 \end{pmatrix}\).
  1. Find \(2\mathbf{A} - 4\mathbf{B}\). [2]
  2. Write down the matrix \(\mathbf{C}\) such that \(\mathbf{A}\mathbf{C} = 2\mathbf{A}\). [1]
  3. Find the value of \(\det \mathbf{A}\). [1]
  4. In this question you must show detailed reasoning. Use \(\mathbf{A}^{-1}\) to solve the equations \(4x - 3y = 7\) and \(-2x + 2y = 9\). [3]
Question 3:
AnswerMarks Guidance
3(a)  4 −  3  3 − 5 
C = 2  −  4 
−  2 2  0  1
 8 − 6   1 2 − 2 0 
= −  
− 4 4 0   4
 − 8 1 2 − − − 6 2  0
= 
− − 4 0 − 4 4
 − 4 1 4 
=
AnswerMarks
− 4 0M1
A1
AnswerMarks
[2]1.1
1.1Sufficient working to
demonstrate knowledge of scalar
multiplication of a matrix and
subtraction of matrices. Can be
implied by 3 out of 4 entries
AnswerMarks
correct.Or BC
(b) 2 0 
( C = ) or 2I
AnswerMarks Guidance
0 2B1
[1]2.2a
(c)(detA = 42 – (–3)(–2) = 8 – 6 =) 2 B1
[1]1.1
(d) 4 − 3   x   7 
=
− 2 2 y 9
1  2 3   7   7 
or A -1 if A-1defined
2 2 4 9 9
4 1
s o x = , y = 2 5
AnswerMarks
2M1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1DR Expressing the system in
matrix form. Can be implied by
the next line.
Forming correct solution as
matrix/vector product with
inverse matrix.
AnswerMarks
20½ or 20.5Matrix method must be used. Any
other method 0/3.
FT their detA from (c). Except for
this, inverse must be correct.
 4 1 
 x 
Condone = 2 but not
y
2 5
x 141
 =  
y 250
AnswerMarks Guidance
Alternative method:M1 Correctly setting up S and
writing it as the difference
AnswerMarks
between two sums of squares.7 8 0
or r r 2 − ( 2 2 + 4 2 + 6 2 + ... + 7 8 0 2 )
= 1
S = 1 2 + 3 2 + 5 2 + ... + 7 7 2 9
7 7 9
= r r 2 − ( 2 2 + 4 2 + 6 2 + ... + 7 7 8 2 )
= 1
7 7 9 1
r r 2 =  7 7 9 ( 7 7 9 + 1 ) ( 2  7 7 9 + 1 )
6
= 1
AnswerMarks Guidance
( = 1 5 7 8 7 9 9 3 0 )B1 Correctly using the formula for
the sum of squares up to and
AnswerMarks
including 779 (or 780)or 158 488 330 up to 780
779
S =r2 −(22 +42 +62 +...+7782)
r=1
779
=r2 −4(12 +22 +32 +...+3892)
AnswerMarks Guidance
r=1M1 Bringing out a factor of 22 (or 4)
to convert the sum of even
squares to a sum of squares.
Could be separate.
3 8 9 1
 r 2 =  3 8 9 ( 3 8 9 + 1 ) ( 2  3 8 9 + 1 )
6
r = 1
AnswerMarks Guidance
( = 1 9 6 9 7 0 1 5 )B1 Correctly using the formula for
the sum of squares up to and
AnswerMarks
including 389 (or 390)or 19 849 115 up to 390
NB B1B1 can only be attained for
779 and 389 or 780 and 390
AnswerMarks Guidance
So S = 157 879 930 – 419697015A1 158 488 330 – 419 849 115
= 79 091 870
[5]
M1
Correctly setting up S and
writing it as the difference
between two sums of squares.
7 8 0
or r r 2 − ( 2 2 + 4 2 + 6 2 + ... + 7 8 0 2 )
= 1
Correctly using the formula for
the sum of squares up to and
including 779 (or 780)
Bringing out a factor of 22 (or 4)
to convert the sum of even
squares to a sum of squares.
Could be separate.
Correctly using the formula for
the sum of squares up to and
including 389 (or 390)
or 19 849 115 up to 390
NB B1B1 can only be attained for
779 and 389 or 780 and 390
Question 3:
3 | (a) |  4 −  3  3 − 5 
C = 2  −  4 
−  2 2  0  1
 8 − 6   1 2 − 2 0 
= −  
− 4 4 0   4
 − 8 1 2 − − − 6 2  0
= 
− − 4 0 − 4 4
 − 4 1 4 
=
− 4 0 | M1
A1
[2] | 1.1
1.1 | Sufficient working to
demonstrate knowledge of scalar
multiplication of a matrix and
subtraction of matrices. Can be
implied by 3 out of 4 entries
correct. | Or BC
(b) |  2 0 
( C = ) or 2I
0 2 | B1
[1] | 2.2a
(c) | (detA = 42 – (–3)(–2) = 8 – 6 =) 2 | B1
[1] | 1.1
(d) |  4 − 3   x   7 
=
− 2 2 y 9
1  2 3   7   7 
or A -1 if A-1defined
2 2 4 9 9
4 1
s o x = , y = 2 5
2 | M1
M1
A1
[3] | 1.1
1.1
1.1 | DR Expressing the system in
matrix form. Can be implied by
the next line.
Forming correct solution as
matrix/vector product with
inverse matrix.
20½ or 20.5 | Matrix method must be used. Any
other method 0/3.
FT their detA from (c). Except for
this, inverse must be correct.
 4 1 
 x 
Condone = 2 but not
y
2 5
x 141
 =  
y 250
Alternative method: | M1 | Correctly setting up S and
writing it as the difference
between two sums of squares. | 7 8 0
or r r 2 − ( 2 2 + 4 2 + 6 2 + ... + 7 8 0 2 )
= 1
S = 1 2 + 3 2 + 5 2 + ... + 7 7 2 9
7 7 9
= r r 2 − ( 2 2 + 4 2 + 6 2 + ... + 7 7 8 2 )
= 1
7 7 9 1
r r 2 =  7 7 9 ( 7 7 9 + 1 ) ( 2  7 7 9 + 1 )
6
= 1
( = 1 5 7 8 7 9 9 3 0 ) | B1 | Correctly using the formula for
the sum of squares up to and
including 779 (or 780) | or 158 488 330 up to 780
779
S =r2 −(22 +42 +62 +...+7782)
r=1
779
=r2 −4(12 +22 +32 +...+3892)
r=1 | M1 | Bringing out a factor of 22 (or 4)
to convert the sum of even
squares to a sum of squares.
Could be separate.
3 8 9 1
 r 2 =  3 8 9 ( 3 8 9 + 1 ) ( 2  3 8 9 + 1 )
6
r = 1
( = 1 9 6 9 7 0 1 5 ) | B1 | Correctly using the formula for
the sum of squares up to and
including 389 (or 390) | or 19 849 115 up to 390
NB B1B1 can only be attained for
779 and 389 or 780 and 390
So S = 157 879 930 – 419697015 | A1 | 158 488 330 – 419 849 115
= 79 091 870
[5]
M1
Correctly setting up S and
writing it as the difference
between two sums of squares.
7 8 0
or r r 2 − ( 2 2 + 4 2 + 6 2 + ... + 7 8 0 2 )
= 1
Correctly using the formula for
the sum of squares up to and
including 779 (or 780)
Bringing out a factor of 22 (or 4)
to convert the sum of even
squares to a sum of squares.
Could be separate.
Correctly using the formula for
the sum of squares up to and
including 389 (or 390)
or 19 849 115 up to 390
NB B1B1 can only be attained for
779 and 389 or 780 and 390
Matrices $\mathbf{A}$ and $\mathbf{B}$ are given by $\mathbf{A} = \begin{pmatrix} 4 & -3 \\ -2 & 2 \end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix} 3 & -5 \\ 0 & 1 \end{pmatrix}$.

\begin{enumerate}[label=(\alph*)]
\item Find $2\mathbf{A} - 4\mathbf{B}$. [2]

\item Write down the matrix $\mathbf{C}$ such that $\mathbf{A}\mathbf{C} = 2\mathbf{A}$. [1]

\item Find the value of $\det \mathbf{A}$. [1]

\item \textbf{In this question you must show detailed reasoning.}

Use $\mathbf{A}^{-1}$ to solve the equations $4x - 3y = 7$ and $-2x + 2y = 9$. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2024 Q3 [7]}}
This paper (9 questions)
View full paper
Q1 5 Q2 6 Q3 7 Q4 5 Q5 6 Q6 11 Q7 10 Q8 13 Q9 12