(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.