Date | November 2012 | Marks available | 3 | Reference code | 12N.1.sl.TZ0.1 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 1 | Adapted from | N/A |
Question
Let \boldsymbol{A} = \left( {\begin{array}{*{20}{c}} 0&3 \\ { - 2}&4 \end{array}} \right) and \boldsymbol{B} = \left( {\begin{array}{*{20}{c}} { - 4}&0 \\ 5&1 \end{array}} \right).
Find AB .
Given that {\boldsymbol{X}} - 2{\boldsymbol{A}} = {\boldsymbol{B}}, find X.
Markscheme
evidence of multiplying (M1)
e.g. one correct element, (0 \times - 4) + (3 \times 5)
{\boldsymbol{AB}} = \left( {\begin{array}{*{20}{c}} {15}&3\\ {28}&4 \end{array}} \right) A2 N3
Note: Award A1 for three correct elements.
[3 marks]
finding 2{\boldsymbol{A}} = \left( {\begin{array}{*{20}{c}} 0&6\\ { - 4}&8 \end{array}} \right) (A1)
adding 2{\boldsymbol{A}} to both sides (may be seen first) (M1)
e.g. {\boldsymbol{X}} = {\boldsymbol{B}} +2{\boldsymbol{A}}
{\boldsymbol{X}} = \left( {\begin{array}{*{20}{c}} { - 4}&6\\ 1&9 \end{array}} \right) A1 N2
[3 marks]
Examiners report
The large majority of candidates answered this question successfully. There were only a small number of candidates who seemed to have never worked with matrices before. Occasionally a candidate would incorrectly approach part (b) by finding an inverse of matrix A.
The large majority of candidates answered this question successfully. There were only a small number of candidates who seemed to have never worked with matrices before. Occasionally a candidate would incorrectly approach part (b) by finding an inverse of matrix A.