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 = ATBA, then the inverse of the matrix A Q2021 AT is equal to :
Solution
$$A{A^T} = \left( {\matrix{
{{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \cr
{{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr
} } \right)\left( {\matrix{
{{1 \over {\sqrt 5 }}} & {{{ - 2} \over {\sqrt 5 }}} \cr
{{2 \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr
} } \right)$$<br><br>$$A{A^T} = \left( {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right) = I$$<br><br>${Q^2} = {A^T}BA\,{A^T}BA = {A^T}BIBA$<br><br>$\Rightarrow {Q^2} = {A^T}{B^2}A$<br><br>${Q^3} = {A^T}{B^2}A{A^T}BA \Rightarrow {Q^3} = {A^T}{B^3}A$<br><br>Similarly : ${Q^{2021}} = {A^T}{B^{2021}}A$<br><br>Now, $${B^2} = \left( {\matrix{
1 & 0 \cr
i & 1 \cr
} } \right)\left( {\matrix{
1 & 0 \cr
i & 1 \cr
} } \right) = \left( {\matrix{
1 & 0 \cr
{2i} & 1 \cr
} } \right)$$<br><br>$${B^3} = \left( {\matrix{
1 & 0 \cr
{2i} & 1 \cr
} } \right)\left( {\matrix{
1 & 0 \cr
i & 1 \cr
} } \right) \Rightarrow {B^3} = \left( {\matrix{
1 & 0 \cr
{3i} & 1 \cr
} } \right)$$<br><br>Similarly $${B^{2021}} = \left( {\matrix{
1 & 0 \cr
{2021i} & 1 \cr
} } \right)$$<br><br>$\therefore$ $A{Q^{2021}} = {A^T} = A{A^T}{B^{2021}}\,A{A^T} = I{B^{2021}}I$<br><br>$$ \Rightarrow A{Q^{2021}}\,{A^T} = {B^{2021}} = \left( {\matrix{
1 & 0 \cr
{2021i} & 1 \cr
} } \right)$$<br><br>$\therefore$ $${(A{Q^{2021}}\,{A^T})^{ - 1}} = {\left( {\matrix{
1 & 0 \cr
{2021i} & 1 \cr
} } \right)^{ - 1}} = \left( {\matrix{
1 & 0 \cr
{ - 2021i} & 1 \cr
} } \right)$$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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