Foundation June 2019 Paper 2 Q29
29 \(\mathbf{a} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}\qquad \mathbf{b} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\)
Work out \(\mathbf{a} - 2\mathbf{b}\) as a column vector. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\begin{pmatrix} -2 \\ 1 \end{pmatrix}\) | M1 | for \(4 - 2 \times 3\ (= -2)\) or \(5 - 2 \times 2\ (=1)\) seen as a calculation OR for \(\begin{pmatrix} 4 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \times 3 \\ 2 \times 2 \end{pmatrix}\) OR for \(\begin{pmatrix} -2 \\ b \end{pmatrix}\) where \(b \ne 1\) or \(\begin{pmatrix} a \\ 1 \end{pmatrix}\) where \(a \ne -2\) |
| A1 | cao |
Additional guidance
May be in a column vector