June 2022 Paper 2 Q3
3
(a) Work out \(\qquad 3\begin{pmatrix} 4 & 2 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ -1 & 5 \end{pmatrix}\)
Give your answer as a single matrix. [3 marks]
(b) \(\begin{pmatrix} 7 & a^2 \\ b & -5 \end{pmatrix}\begin{pmatrix} 2 \\ a \end{pmatrix} = \begin{pmatrix} 78 \\ 12 \end{pmatrix}\)
Work out the values of \(a\) and \(b\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Starts by multiplying 1st matrix by 3 | ||
| \(\begin{pmatrix} 12 & 6 \\ 3 & 0 \end{pmatrix}\) | B1 | brackets may be missing but values must be in correct position in a 2 by 2 array |
| At least two values correct from evaluation of their \(\begin{pmatrix} 12 & 6 \\ 3 & 0 \end{pmatrix}\) \(\times\) \(\begin{pmatrix} 2 & 0 \\ -1 & 5 \end{pmatrix}\) | M1 | brackets may be missing but values must be in correct position in a 2 by 2 array multiplication of matrices must be in the order shown |
| \(\begin{pmatrix} 18 & 30 \\ 6 & 0 \end{pmatrix}\) | A1ft | must have brackets ft B0M1 |
| Alternative method 2 Starts by multiplying the matrices | ||
| At least two values correct in \(\begin{pmatrix} 6 & 10 \\ 2 & 0 \end{pmatrix}\) | M1 | brackets may be missing but values must be in correct position in a 2 by 2 array |
| \(\begin{pmatrix} 6 & 10 \\ 2 & 0 \end{pmatrix}\) | A1 | brackets may be missing but values must be in correct position in a 2 by 2 array |
| \(\begin{pmatrix} 18 & 30 \\ 6 & 0 \end{pmatrix}\) | B1ft | must have brackets ft \(3 \times\) their \(\begin{pmatrix} 6 & 10 \\ 2 & 0 \end{pmatrix}\) their \(\begin{pmatrix} 6 & 10 \\ 2 & 0 \end{pmatrix}\) must be a 2 by 2 array |
Additional guidance
| Alt 1 \(\begin{pmatrix} 12 & 6 \\ 3 & 0 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ -1 & 5 \end{pmatrix} = \begin{pmatrix} 18 & 42 \\ 6 & 8 \end{pmatrix}\) | B1M1A0ft |
| Alt 1 \(\begin{pmatrix} 12 & 6 \\ 3 & 0 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ -1 & 5 \end{pmatrix} = \begin{pmatrix} 24 & 35 \\ 4 & 0 \end{pmatrix}\) | B1M0A0ft |
| Alt 1 \(\begin{pmatrix} 12 & 6 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ -1 & 5 \end{pmatrix} = \begin{pmatrix} 18 & 30 \\ 2 & 0 \end{pmatrix}\) | B0M1A1ft |
| Alt 1 \(\begin{pmatrix} 7 & 5 \\ 4 & 3 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ -1 & 5 \end{pmatrix} = \begin{pmatrix} 14 & 25 \\ 5 & 20 \end{pmatrix}\) | B0M1A0ft |
| Alt 2 \(\begin{pmatrix} 6 & 10 \\ 1 & 5 \end{pmatrix}\) with answer \(\begin{pmatrix} 18 & 30 \\ 3 & 15 \end{pmatrix}\) | M1A0B1ft |
| Alt 2 \(\begin{pmatrix} 8 & 0 \\ -1 & 0 \end{pmatrix}\) with answer \(\begin{pmatrix} 24 & 0 \\ -3 & 0 \end{pmatrix}\) | M0A0B1ft |
| Alt 2 \(\begin{pmatrix} 8 & 0 \\ -1 & 0 \end{pmatrix}\) with answer \(\begin{pmatrix} 24 & 0 \\ -1 & 0 \end{pmatrix}\) | M0A0B0ft |
| For the final mark allow if there is intention to enclose the correct elements in brackets | |
| Responses that start by multiplying 2nd matrix by 3 should be marked using the principles of Alt 1 | |
| Multiplying both matrices by 3 can score a maximum of B1 \(\begin{pmatrix} 12 & 6 \\ 3 & 0 \end{pmatrix}\) or \(\begin{pmatrix} 6 & 0 \\ -3 & 15 \end{pmatrix}\) | B1M0A0ft |
| Answer | Mark | Comments |
|---|---|---|
| \(14 + a^3 = 78\) or \(2b - 5a = 12\) or \(14 + a^3\) and \(2b - 5a\) | M1 | oe eg \(a^3 = 64\) or \(2b + -5a = 12\) allow eg \(7 \times 2 + a^2 \times a\) for \(14 + a^3\) allow eg \(2 \times b - 5 \times a\) for \(2b - 5a\) |
| \(a = 4\) | A1 | |
| \(\dfrac{12 + 5 \times \text{their } a}{2}\) correctly evaluated | A1ft | accept an exact value or a value rounded to 1 dp or better |
Additional guidance
| \(\begin{pmatrix} 14 + a^3 \\ 2b - 5a \end{pmatrix}\) or \((14 + a^3, 2b - 5a)\) with or without brackets | M1 |
| \(a = 4\) (M1 is implied) | M1A1 |
| M1 for \(2b - 5a = 12\) is implied by an incorrect value for \(a\) with a correct ft answer for \(b\) eg \(a = 8 \qquad b = 26\) | M1A0A1ft |
| An incorrect but exact value for \(a\) seen in working (eg \(\dfrac{8}{3}\)) with a rounded value for \(a\) on answer line (eg 2.6) Allow ft for \(b\) from the exact or the rounded value | |
| \(a = 4\) and \(-4\) with one or both of \(b = 16\) and \(-4\) | M1A0A1ft |
| \(a = 4\) and \(-4\) (no values for \(b\) or incorrect values for \(b\)) | M1A0A0ft |