OCR MEI Further Pure Core AS 2022 June — Question 1 6 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2022
SessionJune
Marks6
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Mark schemeDownload PDF ↗
TopicMatrices
TypeSolving linear systems using matrices
DifficultyModerate -0.3 Part (a) is a straightforward application of writing equations in matrix form and solving using standard techniques (likely inverse matrix or row reduction). Part (b) requires understanding that unique solutions exist when the determinant is non-zero, leading to a simple quadratic inequality. Both parts are routine Further Maths exercises with no novel problem-solving required, making this slightly easier than an average A-level question overall.
Spec4.03l Singular/non-singular matrices4.03r Solve simultaneous equations: using inverse matrix
1
    1. Write the following simultaneous equations as a matrix equation. $$\begin{aligned} x + y + 2 z & = 7 \\ 2 x - 4 y - 3 z & = - 5 \\ - 5 x + 3 y + 5 z & = 13 \end{aligned}$$
    2. Hence solve the equations.
  2. Determine the set of values of the constant \(k\) for which the matrix equation $$\left( \begin{array} { c c } k + 1 & 1 \\ 2 & k \end{array} \right) \binom { x } { y } = \binom { 23 } { - 17 }$$ has a unique solution.
Question 1:
AnswerMarks Guidance
1(a) (i)
2 − 4 − 3 y = − 5
AnswerMarks Guidance
− 5 3 5 z 1 3B1
[1]1.1
1(a) (ii)
 x   1 1 2   7 
y = 2 − 4 − 3 − 5
z − 5 3 5 1 3
1 3
x = , y = − , z = 4
AnswerMarks
2 2M1
A1
AnswerMarks
[2]1.1a
1.1(soi) allow M1 if order of matrices incorrect
BC. Accept in vector form. Give full marks for a correct answer.
AnswerMarks Guidance
1(b) k + 1 1
= k ( k + 1 ) − 2
2 k
k2 + k −2 = 0 when k = 1 or −2
AnswerMarks
so unique solution provided k ≠ 1 or −2B1
B1
B1
AnswerMarks
[3]1.1
2.1
AnswerMarks
2.2ax 1 ...
or  =  
y k(k+1)−2...
soi (e.g.from inequalities) or k = 1 or −2 found by inspection
oe, e.g. k < −2, −2 < k < 1, k > 1.
Question 1:
1 | (a) | (i) |  1 1 2   x   7 
2 − 4 − 3 y = − 5
− 5 3 5 z 1 3 | B1
[1] | 1.1
1 | (a) | (ii) | − 1
 x   1 1 2   7 
y = 2 − 4 − 3 − 5
z − 5 3 5 1 3
1 3
x = , y = − , z = 4
2 2 | M1
A1
[2] | 1.1a
1.1 | (soi) allow M1 if order of matrices incorrect
BC. Accept in vector form. Give full marks for a correct answer.
1 | (b) | k + 1 1
= k ( k + 1 ) − 2
2 k
k2 + k −2 = 0 when k = 1 or −2
so unique solution provided k ≠ 1 or −2 | B1
B1
B1
[3] | 1.1
2.1
2.2a | x 1 ...
or  =  
y k(k+1)−2...
soi (e.g.from inequalities) or k = 1 or −2 found by inspection
oe, e.g. k < −2, −2 < k < 1, k > 1.
1
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write the following simultaneous equations as a matrix equation.

$$\begin{aligned}
x + y + 2 z & = 7 \\
2 x - 4 y - 3 z & = - 5 \\
- 5 x + 3 y + 5 z & = 13
\end{aligned}$$
\item Hence solve the equations.
\end{enumerate}\item Determine the set of values of the constant $k$ for which the matrix equation

$$\left( \begin{array} { c c } 
k + 1 & 1 \\
2 & k
\end{array} \right) \binom { x } { y } = \binom { 23 } { - 17 }$$

has a unique solution.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2022 Q1 [6]}}
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Q1 6 Q2 7 Q3 5 Q4 6 Q5 5 Q6 10 Q7 9 Q8 12