(i)     Show that \({({A^T})^{ - 1}} = {({A^{ - 1}})^T}\).

(ii)     Show that \({(AB)^T} = {B^T}{A^T}\).

A relation \(R\) is defined on \(S\) such that \(A\) is related to \(B\) if and only if there exists an element \(X\) of \(S\) such that \(XA{X^T} = B\). Show that \(R\) is an equivalence relation.

(i)     \(A = \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)\)

\({A^T} = \left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right)\)     M1

\({({A^T})^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{array}{*{20}{c}} d&{ - c} \\ { - b}&a \end{array}} \right)\;\;\;\)(which exists because \(ad - bc \ne 0\))     A1

\({A^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{array}{*{20}{c}} d&{ - b} \\ { - c}&a \end{array}} \right)\)     M1

\({({A^{ - 1}})^T} = \frac{1}{{ad - bc}}\left( {\begin{array}{*{20}{c}} d&{ - c} \\ { - b}&a \end{array}} \right)\)     A1

hence \({({A^T})^{ - 1}} = {({A^{ - 1}})^T}\) as required AG

(ii)     \(A = \left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)\;\;\;B = \left( {\begin{array}{*{20}{c}} e&f \\ g&h \end{array}} \right)\)

\(AB = \left( {\begin{array}{*{20}{c}} {ae + bg}&{af + bh} \\ {ce + dg}&{cf + dh} \end{array}} \right)\)     M1

\({(AB)^T} = \left( {\begin{array}{*{20}{c}} {ae + bg}&{ce + dg} \\ {af + bh}&{cf + dh} \end{array}} \right)\)     A1

\({B^T} = \left( {\begin{array}{*{20}{c}} e&g \\ f&h \end{array}} \right)\;\;\;{A^T} = \left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right)\)     M1

\({B^T}{A^T} = \left( {\begin{array}{*{20}{c}} e&g \\ f&h \end{array}} \right)\left( {\begin{array}{*{20}{c}} a&c \\ b&d \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {ae + bg}&{ce + dg} \\ {af + bh}&{cf + dh} \end{array}} \right)\)     A1

hence \({(AB)^T} = {B^T}{A^T}\)     AG

\(R\) is reflexive since \(I \in S\) and \(IA{I^T} = A\)     A1

\(XA{X^T} = B \Rightarrow A = {X^{ - 1}}B{({X^T})^{ - 1}}\)     M1A1

\( \Rightarrow A = {X^{ - 1}}B{({X^{ - 1}})^T}\) from a (i)     A1

which is of the correct form, hence symmetric     AG

\(ARB \Rightarrow XA{X^T} = B\) and \(BRC = YB{Y^T} = C\)     M1

Note: Allow use of \(X\) rather than \(Y\) in this line.

\( \Rightarrow YXA{X^T}{Y^T} = YB{Y^T} = C\)     M1A1

\( \Rightarrow (YX)A{(YX)^T} = C\) from a (ii)     A1

this is of the correct form, hence transitive

hence \(R\) is an equivalence relation     AG

Part a) was successfully answered by the majority of candidates..

There were some wholly correct answers seen to part b) but a number of candidates struggled with the need to formally explain what was required.