Question 10 - A-Level Maths

OCR MEI FP1 2009 January — Question 10 12 marks

Exam Board	OCR MEI
Module	FP1 (Further Pure Mathematics 1)
Year	2009
Session	January
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	3x3 Matrices
Type	Matrix equation solving (AB = C)
Difficulty	Standard +0.3 This is a structured multi-part question on matrix multiplication and inverse matrices. Part (i) requires routine matrix multiplication to verify given entries. Parts (ii)-(iii) involve substitution and recognizing that AB = kI implies B = kA^(-1). Part (iv) applies the inverse to solve a linear system. All steps are standard FP1 techniques with clear scaffolding, making this slightly easier than average despite being Further Maths content.
Spec	4.03b Matrix operations: addition, multiplication, scalar 4.03o Inverse 3x3 matrix 4.03r Solve simultaneous equations: using inverse matrix

10 You are given that $\mathbf { A } = \left( \begin{array} { r r r } 3 & 4 & - 1 \\ 1 & - 1 & k \\ - 2 & 7 & - 3 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { r r c } 11 & - 5 & - 7 \\ 1 & 11 & 5 + k \\ - 5 & 29 & 7 \end{array} \right)$ and that $\mathbf { A B }$ is of the form $\mathbf { A B } = \left( \begin{array} { c c c } 42 & \alpha & 4 k - 8 \\ 10 - 5 k & - 16 + 29 k & - 12 + 6 k \\ 0 & 0 & \beta \end{array} \right)$.

Show that $\alpha = 0$ and $\beta = 28 + 7 k$.
Find $\mathbf { A B }$ when $k = 2$.
For the case when $k = 2$ write down the matrix $\mathbf { A } ^ { - 1 }$.
Use the result from part (iii) to solve the following simultaneous equations. $$\begin{aligned} 3 x + 4 y - z & = 1 \\ x - y + 2 z & = - 9 \\ - 2 x + 7 y - 3 z & = 26 \end{aligned}$$

Show mark scheme Show mark scheme source

Question 10(i):

Answer	Marks	Guidance
$\alpha=3\times-5+4\times11+-1\times29=0$	B1
$\beta=-2\times-7+7\times(5+k)+-3\times7=28+7k$	M1	Attempt at row 3 $\times$ column 3
A1 [3]

Question 10(ii):

Answer	Marks	Guidance
$\mathbf{AB}=\begin{pmatrix}42&0&0\\0&42&0\\0&0&42\end{pmatrix}$	B2 [2]	Minus 1 each error to min of 0

Question 10(iii):

Answer	Marks	Guidance
$\mathbf{A}^{-1}=\frac{1}{42}\begin{pmatrix}11&-5&-7\\1&11&7\\-5&29&7\end{pmatrix}$	M1	Use of $\mathbf{B}$
B1	$\frac{1}{42}$
A1 [3]	Correct inverse, allow decimals to 3 sf

Question 10(iv):

Answer	Marks	Guidance
$\frac{1}{42}\begin{pmatrix}11&-5&-7\\1&11&7\\-5&29&7\end{pmatrix}\begin{pmatrix}1\\-9\\26\end{pmatrix}=\begin{pmatrix}x\\y\\z\end{pmatrix}$	M1	Attempt to pre-multiply by $\mathbf{A}^{-1}$
$=\frac{1}{42}\begin{pmatrix}-126\\84\\-84\end{pmatrix}=\begin{pmatrix}-3\\2\\-2\end{pmatrix}$		SC B2 for Gaussian elimination with 3 correct solutions, $-1$ each error to min of 0
$x=-3,\ y=2,\ z=-2$	A3 [4]	Minus 1 each error

# Question 10(i):

$\alpha=3\times-5+4\times11+-1\times29=0$ | B1 |
$\beta=-2\times-7+7\times(5+k)+-3\times7=28+7k$ | M1 | Attempt at row 3 $\times$ column 3
 | A1 [3] |

# Question 10(ii):

$\mathbf{AB}=\begin{pmatrix}42&0&0\\0&42&0\\0&0&42\end{pmatrix}$ | B2 [2] | Minus 1 each error to min of 0

# Question 10(iii):

$\mathbf{A}^{-1}=\frac{1}{42}\begin{pmatrix}11&-5&-7\\1&11&7\\-5&29&7\end{pmatrix}$ | M1 | Use of $\mathbf{B}$
 | B1 | $\frac{1}{42}$
 | A1 [3] | Correct inverse, allow decimals to 3 sf

# Question 10(iv):

$\frac{1}{42}\begin{pmatrix}11&-5&-7\\1&11&7\\-5&29&7\end{pmatrix}\begin{pmatrix}1\\-9\\26\end{pmatrix}=\begin{pmatrix}x\\y\\z\end{pmatrix}$ | M1 | Attempt to pre-multiply by $\mathbf{A}^{-1}$
$=\frac{1}{42}\begin{pmatrix}-126\\84\\-84\end{pmatrix}=\begin{pmatrix}-3\\2\\-2\end{pmatrix}$ | | SC B2 for Gaussian elimination with 3 correct solutions, $-1$ each error to min of 0
$x=-3,\ y=2,\ z=-2$ | A3 [4] | Minus 1 each error

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10 You are given that $\mathbf { A } = \left( \begin{array} { r r r } 3 & 4 & - 1 \\ 1 & - 1 & k \\ - 2 & 7 & - 3 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { r r c } 11 & - 5 & - 7 \\ 1 & 11 & 5 + k \\ - 5 & 29 & 7 \end{array} \right)$ and that $\mathbf { A B }$ is of the form $\mathbf { A B } = \left( \begin{array} { c c c } 42 & \alpha & 4 k - 8 \\ 10 - 5 k & - 16 + 29 k & - 12 + 6 k \\ 0 & 0 & \beta \end{array} \right)$.\\
(i) Show that $\alpha = 0$ and $\beta = 28 + 7 k$.\\
(ii) Find $\mathbf { A B }$ when $k = 2$.\\
(iii) For the case when $k = 2$ write down the matrix $\mathbf { A } ^ { - 1 }$.\\
(iv) Use the result from part (iii) to solve the following simultaneous equations.

$$\begin{aligned}
3 x + 4 y - z & = 1 \\
x - y + 2 z & = - 9 \\
- 2 x + 7 y - 3 z & = 26
\end{aligned}$$

\hfill \mbox{\textit{OCR MEI FP1 2009 Q10 [12]}}

This paper (10 questions)

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Q1 5 Q2 4 Q3 5 Q4 3 Q5 6 Q6 6 Q7 7 Q8 12 Q9 12 Q10 12

Answer	Marks	Guidance
\(\alpha=3\times-5+4\times11+-1\times29=0\)	B1
\(\beta=-2\times-7+7\times(5+k)+-3\times7=28+7k\)	M1	Attempt at row 3 \(\times\) column 3
A1 [3]

Answer	Marks	Guidance
\(\mathbf{A}^{-1}=\frac{1}{42}\begin{pmatrix}11&-5&-7\\1&11&7\\-5&29&7\end{pmatrix}\)	M1	Use of \(\mathbf{B}\)
B1	\(\frac{1}{42}\)
A1 [3]	Correct inverse, allow decimals to 3 sf

Answer	Marks	Guidance
\(\frac{1}{42}\begin{pmatrix}11&-5&-7\\1&11&7\\-5&29&7\end{pmatrix}\begin{pmatrix}1\\-9\\26\end{pmatrix}=\begin{pmatrix}x\\y\\z\end{pmatrix}\)	M1	Attempt to pre-multiply by \(\mathbf{A}^{-1}\)
\(=\frac{1}{42}\begin{pmatrix}-126\\84\\-84\end{pmatrix}=\begin{pmatrix}-3\\2\\-2\end{pmatrix}\)		SC B2 for Gaussian elimination with 3 correct solutions, \(-1\) each error to min of 0
\(x=-3,\ y=2,\ z=-2\)	A3 [4]	Minus 1 each error