| 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.
