If $$A = \left( {\matrix{
{{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \cr
{{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr
} } \right)$$, $$B = \left( {\matrix{
1 & 0 \cr
i & 1 \cr
} } \right)$$, $$i = \sqrt { - 1} $$, and Q = AT BA, then the inverse of the matrix A Q2021 AT is equal to :
A
$$\left( {\matrix{
{{1 \over {\sqrt 5 }}} & { - 2021} \cr
{2021} & {{1 \over {\sqrt 5 }}} \cr
} } \right)$$
B
$$\left( {\matrix{
1 & 0 \cr
{ - 2021i} & 1 \cr
} } \right)$$
C
$$\left( {\matrix{
1 & 0 \cr
{2021i} & 1 \cr
} } \right)$$
D
$$\left( {\matrix{
1 & { - 2021i} \cr
0 & 1 \cr
} } \right)$$
CHECK ANSWER
Explanation $$A{A^T} = \left( {\matrix{
{{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \cr
{{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr
} } \right)\left( {\matrix{
{{1 \over {\sqrt 5 }}} & {{{ - 2} \over {\sqrt 5 }}} \cr
{{2 \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr
} } \right)$$ $$A{A^T} = \left( {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right) = I$$ $${Q^2} = {A^T}BA\,{A^T}BA = {A^T}BIBA$$ $$ \Rightarrow {Q^2} = {A^T}{B^2}A$$ $${Q^3} = {A^T}{B^2}A{A^T}BA \Rightarrow {Q^3} = {A^T}{B^3}A$$ Similarly : $${Q^{2021}} = {A^T}{B^{2021}}A$$ Now, $${B^2} = \left( {\matrix{
1 & 0 \cr
i & 1 \cr
} } \right)\left( {\matrix{
1 & 0 \cr
i & 1 \cr
} } \right) = \left( {\matrix{
1 & 0 \cr
{2i} & 1 \cr
} } \right)$$ $${B^3} = \left( {\matrix{
1 & 0 \cr
{2i} & 1 \cr
} } \right)\left( {\matrix{
1 & 0 \cr
i & 1 \cr
} } \right) \Rightarrow {B^3} = \left( {\matrix{
1 & 0 \cr
{3i} & 1 \cr
} } \right)$$ Similarly $${B^{2021}} = \left( {\matrix{
1 & 0 \cr
{2021i} & 1 \cr
} } \right)$$ $$\therefore$$ $$A{Q^{2021}} = {A^T} = A{A^T}{B^{2021}}\,A{A^T} = I{B^{2021}}I$$ $$ \Rightarrow A{Q^{2021}}\,{A^T} = {B^{2021}} = \left( {\matrix{
1 & 0 \cr
{2021i} & 1 \cr
} } \right)$$ $$\therefore$$ $${(A{Q^{2021}}\,{A^T})^{ - 1}} = {\left( {\matrix{
1 & 0 \cr
{2021i} & 1 \cr
} } \right)^{ - 1}} = \left( {\matrix{
1 & 0 \cr
{ - 2021i} & 1 \cr
} } \right)$$