Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 27 / 5 / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 56 / 3 / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 63 / 6 / Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 1 / 6C / Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 1 / 7B / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 3 / 10H / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 22 / 4 / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 33 / 3 / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 33 / 5 / a / 2 / Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 33 / 5 / b / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 4 / 10G / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 4 / 11G / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 49 / 4 / Solution

Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 58 / 3 / Solution

= Trace
{synonym}

The trace is the sum of diagonal entries,
$$
\operatorname{tr}A=\sum_iA_{ii}.
$$
Index interchange proves $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ and hence cyclic invariance under any cyclic permutation of a product.


Matrix trace

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