OCR Further Pure Core AS 2024 June — Question 1 4 marks

Exam BoardOCR
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2024
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeFind inverse then solve system
DifficultyStandard +0.3 This is a straightforward application of the matrix inverse method to solve a 3×3 system. While it requires computing a 3×3 determinant and inverse (or using row reduction), it's a standard algorithmic procedure with no conceptual difficulty or problem-solving insight needed. The arithmetic is manageable and the question is typical textbook fare for Further Maths students who have learned this technique.
Spec4.03r Solve simultaneous equations: using inverse matrix

1 Use a matrix method to determine the solution of the following simultaneous equations. $$\begin{aligned} 2 x - 3 y + z & = 1 \\ x - 2 y - 4 z & = 40 \\ 5 x + 6 y - z & = 61 \end{aligned}$$

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(\begin{pmatrix} 2 & -3 & 1 \\ 1 & -2 & -4 \\ 5 & 6 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 40 \\ 61 \end{pmatrix}\)M1 Reduction of system to matrix form
\(\begin{pmatrix} 2 & -3 & 1 \\ 1 & -2 & -4 \\ 5 & 6 & -1 \end{pmatrix}^{-1} = \frac{1}{125}\begin{pmatrix} 26 & 3 & 14 \\ -19 & -7 & 9 \\ 16 & -27 & -1 \end{pmatrix}\)B1* BC. Inverse matrix correctly evaluated. Decimal form: \(\begin{pmatrix} 0.208 & 0.024 & 0.112 \\ -0.152 & -0.056 & 0.072 \\ 0.128 & -0.216 & -0.008 \end{pmatrix}\)
\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{125}\begin{pmatrix} 26 & 3 & 14 \\ -19 & -7 & 9 \\ 16 & -27 & -1 \end{pmatrix}\begin{pmatrix} 1 \\ 40 \\ 61 \end{pmatrix}\)M1 Correctly using inverse matrix: \((\mathbf{r} =)\, \mathbf{A}^{-1}\mathbf{b}\). Must be clear a matrix method is being used. Can be incorrect inverse for M1. Allow M1 for expressions of form \(\mathbf{A}^{-1}\mathbf{b}\)
\(x = 8,\ y = 2,\ z = -9\)A1*dep Could be seen in vector form but \(x\), \(y\) and \(z\) must be appropriately seen. Correct answer with no matrix forms shown (with or without other working) is 0/4. Need to have earned the B1
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix} 2 & -3 & 1 \\ 1 & -2 & -4 \\ 5 & 6 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 40 \\ 61 \end{pmatrix}$ | M1 | Reduction of system to matrix form |
| $\begin{pmatrix} 2 & -3 & 1 \\ 1 & -2 & -4 \\ 5 & 6 & -1 \end{pmatrix}^{-1} = \frac{1}{125}\begin{pmatrix} 26 & 3 & 14 \\ -19 & -7 & 9 \\ 16 & -27 & -1 \end{pmatrix}$ | B1* | BC. Inverse matrix correctly evaluated. Decimal form: $\begin{pmatrix} 0.208 & 0.024 & 0.112 \\ -0.152 & -0.056 & 0.072 \\ 0.128 & -0.216 & -0.008 \end{pmatrix}$ |
| $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{125}\begin{pmatrix} 26 & 3 & 14 \\ -19 & -7 & 9 \\ 16 & -27 & -1 \end{pmatrix}\begin{pmatrix} 1 \\ 40 \\ 61 \end{pmatrix}$ | M1 | Correctly using inverse matrix: $(\mathbf{r} =)\, \mathbf{A}^{-1}\mathbf{b}$. Must be clear a matrix method is being used. Can be incorrect inverse for M1. Allow M1 for expressions of form $\mathbf{A}^{-1}\mathbf{b}$ |
| $x = 8,\ y = 2,\ z = -9$ | A1*dep | Could be seen in vector form but $x$, $y$ and $z$ must be appropriately seen. Correct answer with no matrix forms shown (with or without other working) is 0/4. Need to have earned the B1 |

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1 Use a matrix method to determine the solution of the following simultaneous equations.

$$\begin{aligned}
2 x - 3 y + z & = 1 \\
x - 2 y - 4 z & = 40 \\
5 x + 6 y - z & = 61
\end{aligned}$$

\hfill \mbox{\textit{OCR Further Pure Core AS 2024 Q1 [4]}}