OCR Further Pure Core 2 2023 June — Question 1 8 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2023
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMatrices
TypeMatrix conformability and dimensions
DifficultyEasy -1.3 This is a straightforward question testing basic matrix definitions and properties: dimensions, transpose, conformability for addition/multiplication, and a simple matrix equation. All parts require only direct recall and routine application of definitions with minimal calculation. Even for Further Maths, this represents foundational knowledge with no problem-solving or insight required.
Spec4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar
1
  1. The matrix \(\mathbf { P }\) is given by \(\mathbf { P } = \left( \begin{array} { l l l l } 1 & 0 & - 2 & 2 \\ 4 & 2 & - 2 & 3 \end{array} \right)\).
    1. Write down the dimensions of \(\mathbf { P }\).
    2. Write down the transpose of \(\mathbf { P }\).
  2. The matrices \(\mathbf { Q } , \mathbf { R }\) and \(\mathbf { S }\) are given by \(\mathbf { Q } = \left( \begin{array} { l l } 1 & 2 \end{array} \right) , \mathbf { R } = \left( \begin{array} { r r } 3 & - 4 \\ 2 & 3 \end{array} \right)\) and \(\mathbf { S } = \left( \begin{array} { l l } 3 & - 2 \end{array} \right)\). Write down the sum of the two of these matrices which are conformable for addition.
  3. The dimensions of matrix \(\mathbf { A }\) are 4 by 5. The matrices \(\mathbf { A }\) and \(\mathbf { B }\) are conformable for multiplication so that the matrix \(\mathbf { C } = \mathbf { B A }\) can be formed. The matrix \(\mathbf { C }\) has 6 rows.
    1. Write down the number of columns that \(\mathbf { C }\) has.
    2. Write down the dimensions of \(\mathbf { B }\).
    3. Explain whether the matrix \(\mathbf { A B }\) can be formed.
  4. Find the value of \(c\) for which \(\left( \begin{array} { r r } - 2 & 3 \\ 6 & 10 \end{array} \right) \left( \begin{array} { r r } c & 5 \\ 10 & 13 \end{array} \right) = \left( \begin{array} { r r } c & 5 \\ 10 & 13 \end{array} \right) \left( \begin{array} { r r } - 2 & 3 \\ 6 & 10 \end{array} \right)\).
Question 1:
AnswerMarks Guidance
1(a) (i)
[1]
AnswerMarks
(ii) 1 4 
0 2
( )
P T =
− 2 − 2
AnswerMarks Guidance
2 3B1 1.2
here
[1]
AnswerMarks Guidance
(b)(4 0) B1
[1]
AnswerMarks Guidance
(c)(i) 5
[1]
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(c)(ii) 6 by 4
[1]
AnswerMarks Guidance
(c)(iii) No because the number of columns in A ( is 5
which) is not equal to the number of rows in
matrix B(which is 6) (and for the matrices to be
AnswerMarks Guidance
conformable these have to be the same.)B1 2.4
If numbers used, must have a word
to imply comparison (eg “while”,
AnswerMarks
“but” rather than “and”)Accept (4×5)×(6×4) and “5 ≠
6”
numbers given must be correct
[1]
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(d) − 2 3   c 5   3 0 − 2 c 2 9 
= and
6 1 0 1 0 1 3 6 c + 1 0 0 1 6 0
 c 5   − 2 3   3 0 − 2 c 3 c + 5 0 
=
AnswerMarks Guidance
1 0 1 3 6 1 0 5 8 1 6 0M1 1.1
directions sufficient to obtain one
pair ofequivalent entries in trailing
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diagonal.Can be implied by correct linear
equation
AnswerMarks Guidance
6c + 100 = 58 or 3c + 50 = 29 =>c = –7A1 1.1
[2]
1
Using factor theorem to find
another factor and fully
factorising
Question 1:
1 | (a) | (i) | 2 by 4 or 24 | B1 | 1.2
[1]
(ii) |  1 4 
0 2
( )
P T =
− 2 − 2
2 3 | B1 | 1.2 | condone poor/omitted brackets just
here
[1]
(b) | (4 0) | B1 | 2.5 | Do not allow (4, 0)
[1]
(c) | (i) | 5 | B1 | 2.2a
[1]
(c) | (ii) | 6 by 4 | B1 | 2.2a
[1]
(c) | (iii) | No because the number of columns in A ( is 5
which) is not equal to the number of rows in
matrix B(which is 6) (and for the matrices to be
conformable these have to be the same.) | B1 | 2.4 | Must include “number of” oe
If numbers used, must have a word
to imply comparison (eg “while”,
“but” rather than “and”) | Accept (4×5)×(6×4) and “5 ≠
6”
numbers given must be correct
[1]
(d) |  − 2 3   c 5   3 0 − 2 c 2 9 
= and
6 1 0 1 0 1 3 6 c + 1 0 0 1 6 0
 c 5   − 2 3   3 0 − 2 c 3 c + 5 0 
=
1 0 1 3 6 1 0 5 8 1 6 0 | M1 | 1.1 | Attempt at multiplication in both
directions sufficient to obtain one
pair ofequivalent entries in trailing
diagonal. | Can be implied by correct linear
equation
6c + 100 = 58 or 3c + 50 = 29 =>c = –7 | A1 | 1.1 | ignore errors in unused elements
[2]
1
Using factor theorem to find
another factor and fully
factorising
1
\begin{enumerate}[label=(\alph*)]
\item The matrix $\mathbf { P }$ is given by $\mathbf { P } = \left( \begin{array} { l l l l } 1 & 0 & - 2 & 2 \\ 4 & 2 & - 2 & 3 \end{array} \right)$.
\begin{enumerate}[label=(\roman*)]
\item Write down the dimensions of $\mathbf { P }$.
\item Write down the transpose of $\mathbf { P }$.
\end{enumerate}\item The matrices $\mathbf { Q } , \mathbf { R }$ and $\mathbf { S }$ are given by $\mathbf { Q } = \left( \begin{array} { l l } 1 & 2 \end{array} \right) , \mathbf { R } = \left( \begin{array} { r r } 3 & - 4 \\ 2 & 3 \end{array} \right)$ and $\mathbf { S } = \left( \begin{array} { l l } 3 & - 2 \end{array} \right)$.

Write down the sum of the two of these matrices which are conformable for addition.
\item The dimensions of matrix $\mathbf { A }$ are 4 by 5. The matrices $\mathbf { A }$ and $\mathbf { B }$ are conformable for multiplication so that the matrix $\mathbf { C } = \mathbf { B A }$ can be formed. The matrix $\mathbf { C }$ has 6 rows.
\begin{enumerate}[label=(\roman*)]
\item Write down the number of columns that $\mathbf { C }$ has.
\item Write down the dimensions of $\mathbf { B }$.
\item Explain whether the matrix $\mathbf { A B }$ can be formed.
\end{enumerate}\item Find the value of $c$ for which $\left( \begin{array} { r r } - 2 & 3 \\ 6 & 10 \end{array} \right) \left( \begin{array} { r r } c & 5 \\ 10 & 13 \end{array} \right) = \left( \begin{array} { r r } c & 5 \\ 10 & 13 \end{array} \right) \left( \begin{array} { r r } - 2 & 3 \\ 6 & 10 \end{array} \right)$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2023 Q1 [8]}}
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