JEE MAIN - Mathematics (2021 - 26th August Morning Shift - No. 10)
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 :
$$\left( {\matrix{
{{1 \over {\sqrt 5 }}} & { - 2021} \cr
{2021} & {{1 \over {\sqrt 5 }}} \cr
} } \right)$$
$$\left( {\matrix{
1 & 0 \cr
{ - 2021i} & 1 \cr
} } \right)$$
$$\left( {\matrix{
1 & 0 \cr
{2021i} & 1 \cr
} } \right)$$
$$\left( {\matrix{
1 & { - 2021i} \cr
0 & 1 \cr
} } \right)$$
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 $${B^{2021}} = \left( {\matrix{ 1 & 0 \cr {2021i} & 1 \cr } } \right)$$
$$\therefore$$ $$A{Q^{2021}} = {A^T} = A{A^T}{B^{2021}}\,A{A^T} = I{B^{2021}}I$$
$$ \Rightarrow A{Q^{2021}}\,{A^T} = {B^{2021}} = \left( {\matrix{ 1 & 0 \cr {2021i} & 1 \cr } } \right)$$
$$\therefore$$ $${(A{Q^{2021}}\,{A^T})^{ - 1}} = {\left( {\matrix{ 1 & 0 \cr {2021i} & 1 \cr } } \right)^{ - 1}} = \left( {\matrix{ 1 & 0 \cr { - 2021i} & 1 \cr } } \right)$$
$$A{A^T} = \left( {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right) = I$$
$${Q^2} = {A^T}BA\,{A^T}BA = {A^T}BIBA$$
$$ \Rightarrow {Q^2} = {A^T}{B^2}A$$
$${Q^3} = {A^T}{B^2}A{A^T}BA \Rightarrow {Q^3} = {A^T}{B^3}A$$
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 $${B^{2021}} = \left( {\matrix{ 1 & 0 \cr {2021i} & 1 \cr } } \right)$$
$$\therefore$$ $$A{Q^{2021}} = {A^T} = A{A^T}{B^{2021}}\,A{A^T} = I{B^{2021}}I$$
$$ \Rightarrow A{Q^{2021}}\,{A^T} = {B^{2021}} = \left( {\matrix{ 1 & 0 \cr {2021i} & 1 \cr } } \right)$$
$$\therefore$$ $${(A{Q^{2021}}\,{A^T})^{ - 1}} = {\left( {\matrix{ 1 & 0 \cr {2021i} & 1 \cr } } \right)^{ - 1}} = \left( {\matrix{ 1 & 0 \cr { - 2021i} & 1 \cr } } \right)$$
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