Question 1 - A-Level Maths

Edexcel CP AS 2022 June — Question 1 7 marks

Exam Board	Edexcel
Module	CP AS (Core Pure AS)
Year	2022
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Matrices
Type	Matrix multiplication
Difficulty	Moderate -0.8 This is a straightforward multi-part question testing basic matrix operations (multiplication, addition, scalar multiplication) and understanding of matrix dimensions. Part (a)(i) requires routine calculation of AB and verification of an equation. Part (a)(ii) tests dimensional compatibility (BA is 2×2, not 3×3). Part (b) is standard application of matrices to solve simultaneous equations using a calculator. All parts are mechanical with no problem-solving insight required, making this easier than average.
Spec	4.03b Matrix operations: addition, multiplication, scalar 4.03o Inverse 3x3 matrix 4.03r Solve simultaneous equations: using inverse matrix

1. $$\mathbf { A } = \left( \begin{array} { r r } 4 & - 1 \\ 7 & 2 \\ - 5 & 8 \end{array} \right) \quad \mathbf { B } = \left( \begin{array} { r r r } 2 & 3 & 2 \\ - 1 & 6 & 5 \end{array} \right) \quad \mathbf { C } = \left( \begin{array} { r r r } - 5 & 2 & 1 \\ 4 & 3 & 8 \\ - 6 & 11 & 2 \end{array} \right)$$ Given that $\mathbf { I }$ is the $3 \times 3$ identity matrix,

1. show that there is an integer $k$ for which $$\mathbf { A B } - 3 \mathbf { C } + k \mathbf { I } = \mathbf { 0 }$$ stating the value of $k$
2. explain why there can be no constant $m$ such that $$\mathbf { B A } - 3 \mathbf { C } + m \mathbf { I } = \mathbf { 0 }$$
1. Show how the matrix $\mathbf { C }$ can be used to solve the simultaneous equations $$\begin{aligned} - 5 x + 2 y + z & = - 14 \\ 4 x + 3 y + 8 z & = 3 \\ - 6 x + 11 y + 2 z & = 7 \end{aligned}$$
2. Hence use your calculator to solve these equations.

Show mark scheme Show mark scheme source

Question 1:

Part (a)(i)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\mathbf{AB} = \begin{pmatrix} 4 & -1 \\ 7 & 2 \\ -5 & 8 \end{pmatrix}\begin{pmatrix} 2 & 3 & 2 \\ -1 & 6 & 5 \end{pmatrix} = \begin{pmatrix} 9 & 6 & 3 \\ 12 & 33 & 24 \\ -18 & 33 & 30 \end{pmatrix}$	M1	1.1b
$\mathbf{AB} - 3\mathbf{C} = \begin{pmatrix} 9 & 6 & 3 \\ 12 & 33 & 24 \\ -18 & 33 & 30 \end{pmatrix} - \begin{pmatrix} -15 & 6 & 3 \\ 12 & 9 & 24 \\ -18 & 33 & 6 \end{pmatrix} = \begin{pmatrix} 24 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24 \end{pmatrix}$	M1	1.1b
Hence $\mathbf{AB} - 3\mathbf{C} - 24\mathbf{I} = \mathbf{0}$ so $k = -24$	A1	1.1b

(4 marks total)

Part (a)(ii)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Need two things. One of: BA is a $2\times2$ matrix; Finds the matrix BA (must be $2\times2$). AND One of: cannot subtract a $3\times3$ matrix; finds 3C and comments different dimensions; can't subtract matrices of different sizes; 3C or C is a $3\times3$ matrix; BA needs to be a $3\times3$ matrix	B1	2.4

Part (b)(i)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{C}^{-1}\begin{pmatrix} -14 \\ 3 \\ 7 \end{pmatrix}$ or states $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{360}\begin{pmatrix} -82 & 7 & 13 \\ -56 & -4 & 44 \\ 62 & 43 & -23 \end{pmatrix}\begin{pmatrix} -14 \\ 3 \\ 7 \end{pmatrix}$	M1	1.2

Part (b)(ii)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$= \frac{1}{360}\begin{pmatrix} -82 & 7 & 13 \\ -56 & -4 & 44 \\ 62 & 43 & -23 \end{pmatrix}\begin{pmatrix} -14 \\ 3 \\ 7 \end{pmatrix} = \ldots$	M1	1.1b

Question 1 (Matrix/Simultaneous Equations):

Part (a)(i):

Answer	Marks	Guidance
Working	Mark	Guidance
Attempts to find $\mathbf{AB}$	M1	Usually done on calculator; allow at least 6 correct entries if incorrect
Uses $\mathbf{AB}$ and $3\mathbf{C}$ to find multiple $\mathbf{I}$, states value for $k$	M1	Implied by correct matrix for $\mathbf{AB} - 3\mathbf{C}$
$\mathbf{AB} - 3\mathbf{C} = \begin{pmatrix} 24 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24 \end{pmatrix}$, hence $k = -24$	A1	$k = -24$ seen explicitly (may be in equation)

Special case: If minimum working not shown and just $k = -24$ stated: M1 M0 A0

Part (a)(ii):

Answer	Marks	Guidance
Working	Mark	Guidance
Correct explanation re dimensions of $\mathbf{BA}$ and $\mathbf{C}$ (or $3\mathbf{C}$) not matching	B1	May find both matrices then comment they cannot be subtracted

Part (b):

Answer	Marks	Guidance
Working	Mark	Guidance
States/implies use of inverse matrix method	M1
Carries out multiplication using inverse; $\mathbf{C}^{-1}\begin{pmatrix}-14\\3\\7\end{pmatrix} = \ldots$	M1	Finding inverse then writing answer gains M1
$x = \dfrac{7}{2},\ y = 3,\ z = -\dfrac{5}{2}$ or $(3.5,\ 3,\ -2.5)$	A1	Must be clear; $\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}3.5\\3\\-2.5\end{pmatrix}$ acceptable

Note: Solving by simultaneous equations only: M0 M0 A0. No reference to inverse matrix with correct answers: M0 M0 A0. Two out of three correct ordinates imply M1.

# Question 1:

## Part (a)(i)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{AB} = \begin{pmatrix} 4 & -1 \\ 7 & 2 \\ -5 & 8 \end{pmatrix}\begin{pmatrix} 2 & 3 & 2 \\ -1 & 6 & 5 \end{pmatrix} = \begin{pmatrix} 9 & 6 & 3 \\ 12 & 33 & 24 \\ -18 & 33 & 30 \end{pmatrix}$ | M1 | 1.1b |
| $\mathbf{AB} - 3\mathbf{C} = \begin{pmatrix} 9 & 6 & 3 \\ 12 & 33 & 24 \\ -18 & 33 & 30 \end{pmatrix} - \begin{pmatrix} -15 & 6 & 3 \\ 12 & 9 & 24 \\ -18 & 33 & 6 \end{pmatrix} = \begin{pmatrix} 24 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24 \end{pmatrix}$ | M1 | 1.1b |
| Hence $\mathbf{AB} - 3\mathbf{C} - 24\mathbf{I} = \mathbf{0}$ so $k = -24$ | A1 | 1.1b |

**(4 marks total)**

## Part (a)(ii)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Need two things. One of: **BA** is a $2\times2$ matrix; Finds the matrix **BA** (must be $2\times2$). AND One of: cannot subtract a $3\times3$ matrix; finds **3C** and comments different dimensions; can't subtract matrices of different sizes; **3C** or **C** is a $3\times3$ matrix; **BA** needs to be a $3\times3$ matrix | B1 | 2.4 |

## Part (b)(i)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{C}^{-1}\begin{pmatrix} -14 \\ 3 \\ 7 \end{pmatrix}$ or states $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{360}\begin{pmatrix} -82 & 7 & 13 \\ -56 & -4 & 44 \\ 62 & 43 & -23 \end{pmatrix}\begin{pmatrix} -14 \\ 3 \\ 7 \end{pmatrix}$ | M1 | 1.2 |

## Part (b)(ii)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $= \frac{1}{360}\begin{pmatrix} -82 & 7 & 13 \\ -56 & -4 & 44 \\ 62 & 43 & -23 \end{pmatrix}\begin{pmatrix} -14 \\ 3 \\ 7 \end{pmatrix} = \ldots$ | M1 | 1.1b |

# Question 1 (Matrix/Simultaneous Equations):

## Part (a)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts to find $\mathbf{AB}$ | M1 | Usually done on calculator; allow at least 6 correct entries if incorrect |
| Uses $\mathbf{AB}$ and $3\mathbf{C}$ to find multiple $\mathbf{I}$, states value for $k$ | M1 | Implied by correct matrix for $\mathbf{AB} - 3\mathbf{C}$ |
| $\mathbf{AB} - 3\mathbf{C} = \begin{pmatrix} 24 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24 \end{pmatrix}$, hence $k = -24$ | A1 | $k = -24$ seen explicitly (may be in equation) |

**Special case:** If minimum working not shown and just $k = -24$ stated: M1 M0 A0

## Part (a)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Correct explanation re dimensions of $\mathbf{BA}$ and $\mathbf{C}$ (or $3\mathbf{C}$) not matching | B1 | May find both matrices then comment they cannot be subtracted |

## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| States/implies use of inverse matrix method | M1 | |
| Carries out multiplication using inverse; $\mathbf{C}^{-1}\begin{pmatrix}-14\\3\\7\end{pmatrix} = \ldots$ | M1 | Finding inverse then writing answer gains M1 |
| $x = \dfrac{7}{2},\ y = 3,\ z = -\dfrac{5}{2}$ or $(3.5,\ 3,\ -2.5)$ | A1 | Must be clear; $\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}3.5\\3\\-2.5\end{pmatrix}$ acceptable |

**Note:** Solving by simultaneous equations only: M0 M0 A0. No reference to inverse matrix with correct answers: M0 M0 A0. Two out of three correct ordinates imply M1.

---

Show LaTeX source

1.

$$\mathbf { A } = \left( \begin{array} { r r } 
4 & - 1 \\
7 & 2 \\
- 5 & 8
\end{array} \right) \quad \mathbf { B } = \left( \begin{array} { r r r } 
2 & 3 & 2 \\
- 1 & 6 & 5
\end{array} \right) \quad \mathbf { C } = \left( \begin{array} { r r r } 
- 5 & 2 & 1 \\
4 & 3 & 8 \\
- 6 & 11 & 2
\end{array} \right)$$

Given that $\mathbf { I }$ is the $3 \times 3$ identity matrix,
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item show that there is an integer $k$ for which

$$\mathbf { A B } - 3 \mathbf { C } + k \mathbf { I } = \mathbf { 0 }$$

stating the value of $k$
\item explain why there can be no constant $m$ such that

$$\mathbf { B A } - 3 \mathbf { C } + m \mathbf { I } = \mathbf { 0 }$$
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show how the matrix $\mathbf { C }$ can be used to solve the simultaneous equations

$$\begin{aligned}
- 5 x + 2 y + z & = - 14 \\
4 x + 3 y + 8 z & = 3 \\
- 6 x + 11 y + 2 z & = 7
\end{aligned}$$
\item Hence use your calculator to solve these equations.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{Edexcel CP AS 2022 Q1 [7]}}

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