OCR MEI Further Pure Core AS 2019 June — Question 3 6 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2019
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMatrices
TypeSolving matrix equations for unknown matrix
DifficultyModerate -0.3 This is a straightforward application of matrix inverse properties. Part (a) requires recognizing that (AB)^{-1} = B^{-1}A^{-1}, so AB is simply the inverse of the given matrix. Part (b) involves substituting the given A and solving AB = result from (a) using standard matrix multiplication. While it requires understanding of inverse properties and matrix operations, these are core Further Maths techniques with no novel problem-solving required, making it slightly easier than average.
Spec4.03o Inverse 3x3 matrix4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1)
3 In this question you must show detailed reasoning. \(\mathbf { A }\) and \(\mathbf { B }\) are matrices such that \(\mathbf { B } ^ { - 1 } \mathbf { A } ^ { - 1 } = \left( \begin{array} { r r } 2 & 1 \\ - 1 & 1 \end{array} \right)\).
  1. Find \(\mathbf { A B }\).
  2. Given that \(\mathbf { A } = \left( \begin{array} { l l } \frac { 1 } { 3 } & 1 \\ 0 & 1 \end{array} \right)\), find \(\mathbf { B }\).
Question 3:
AnswerMarks Guidance
3(a) DR
 2 1  2 1
(AB)−1 =B−1A−1 =   so AB=   −1
−1 1 −1 1
2 1
=3
−1 1
1 −1
⇒AB= 1
 
AnswerMarks
3 1 2M1
B1
A1cao
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1 2 1
AB= −1
 
−1 1
soi
1 −1
or  3 3 
1 2
 
3 3
AnswerMarks Guidance
3(b) DR
1 −1
A−1 =3 
0 1 
3
1 −11 −1
B=A−1AB=3×1
 
3 0 1 1 2
3
0 −3
= 
1 2
 
AnswerMarks
3 3B1
M1
AnswerMarks
A1cao1.1
3.1a
AnswerMarks
1.1−1
pre-multiply their AB by A
0 −9
Or 1  
3 1 2
Alternative solution
 2 11 1
B−1 =B−1A−1A= 3
  
AnswerMarks Guidance
−1 10 1M1 −1A −1
post-multiply B by A
 2 3
= 3 
 −1 0
AnswerMarks
3A1
0 −3
⇒B= 
1 2
 
AnswerMarks Guidance
3 3A1cao 0 −9
Or 1  
AnswerMarks
3 1 2`
Alternative solution
1 1a b 1 −1 
 3   = 3 3 
 
0 1c d 1 2
 
3 3
1 1 1 1
⇒ a+c= , b+d =−
AnswerMarks Guidance
3 3 3 3M1 forming equations (any two)
1 2
c= , d =
AnswerMarks
3 3A1
0 −3
⇒ a = 0, b = −3 ⇒B= 
1 2
 
AnswerMarks
3 3A1 
Question 3:
3 | (a) | DR
 2 1  2 1
(AB)−1 =B−1A−1 =   so AB=   −1
−1 1 −1 1
2 1
=3
−1 1
1 −1
⇒AB= 1
 
3 1 2 | M1
B1
A1cao
[3] | 3.1a
1.1
1.1 |  2 1
AB= −1
 
−1 1
soi
1 −1
or  3 3 
1 2
 
3 3
3 | (b) | DR
1 −1
A−1 =3 
0 1 
3
1 −11 −1
B=A−1AB=3×1
 
3 0 1 1 2
3
0 −3
= 
1 2
 
3 3 | B1
M1
A1cao | 1.1
3.1a
1.1 | −1
pre-multiply their AB by A
0 −9
Or 1  
3 1 2
Alternative solution
 2 11 1
B−1 =B−1A−1A= 3
  
−1 10 1 | M1 | −1A −1
post-multiply B by A
 2 3
= 3 
 −1 0
3 | A1
0 −3
⇒B= 
1 2
 
3 3 | A1cao | 0 −9
Or 1  
3 1 2 | `
Alternative solution
1 1a b 1 −1 
 3   = 3 3 
 
0 1c d 1 2
 
3 3
1 1 1 1
⇒ a+c= , b+d =−
3 3 3 3 | M1 | forming equations (any two) | ft their AB
1 2
c= , d =
3 3 | A1
0 −3
⇒ a = 0, b = −3 ⇒B= 
1 2
 
3 3 | A1
3 In this question you must show detailed reasoning.\\
$\mathbf { A }$ and $\mathbf { B }$ are matrices such that $\mathbf { B } ^ { - 1 } \mathbf { A } ^ { - 1 } = \left( \begin{array} { r r } 2 & 1 \\ - 1 & 1 \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { A B }$.
\item Given that $\mathbf { A } = \left( \begin{array} { l l } \frac { 1 } { 3 } & 1 \\ 0 & 1 \end{array} \right)$, find $\mathbf { B }$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2019 Q3 [6]}}
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