$$\left| {\matrix{ {[x + 1]} & {[x + 2]} & {[x + 3]} \cr {[x]} & {[x + 3]} & {[x + 3]} \cr {[x]} & {[x + 2]} & {[x + 4]} \cr } } \right| = 192$$

R₁ $$\to$$ R₁ $$-$$ R₃ & R₂ $$\to$$ R₂ $$-$$ R₃

$$\left[ {\matrix{ 1 & 0 & { - 1} \cr 0 & 1 & { - 1} \cr {[x]} & {[x] + 2} & {[x] + 4} \cr } } \right] = 192$$

$$2[x] + 6 + [x] = 192 \Rightarrow [x] = 62$$

Given matrix $$A = \left[ {\matrix{ 0 & 2 \cr k & { - 1} \cr } } \right]$$

$${A^4} + 3IA = 2I$$

$$ \Rightarrow {A^4} = 2I - 3A$$

Also characteristic equation of A is $$|A - \lambda I|\, = 0$$

$$ \Rightarrow \left| {\matrix{ {0 - \lambda } & 2 \cr k & { - 1 - \lambda } \cr } } \right| = 0$$

$$ \Rightarrow \lambda + {\lambda ^2} - 2k = 0$$

$$ \Rightarrow A + {A^2} = 2K.I$$

$$ \Rightarrow {A^2} = 2KI - A$$

$$ \Rightarrow {A^4} = 4{K^2}I + {A^2} - 4AK$$

Put $${A^2} = 2KI - A$$

and $${A^4} = 2I - 3A$$

$$2I - 3A = 4{K^2}I + 2KI - A - 4AK$$

$$ \Rightarrow I(2 - 2K - 4{K^2}) = A(2 - 4K)$$

$$ \Rightarrow - 2I(2{K^2} + K - 1) = 2A(1 - 2K)$$

$$ \Rightarrow - 2I(2K - 1)(K + 1) = 2A(1 - 2K)$$

$$ \Rightarrow (2K - 1)(2A) - 2I(2K - 1)(K + 1) = 0$$

$$ \Rightarrow (2K - 1)[2A - 2I(K + 1)] = 0$$

$$ \Rightarrow K = {1 \over 2}$$

$$A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right] \Rightarrow {A^2} = \left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right]$$

$${A^3} = \left[ {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right] \Rightarrow {A^4} = \left[ {\matrix{ 1 & 0 & 0 \cr 3 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right]$$

$${A^n} = \left[ {\matrix{ 1 & 0 & 0 \cr {n - 1} & 1 & 1 \cr 1 & 0 & 0 \cr } } \right]$$

$${A^{2025}} - {A^{2020}} = \left[ {\matrix{ 0 & 0 & 0 \cr 5 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]$$

$${A^6} - A = \left[ {\matrix{ 0 & 0 & 0 \cr 5 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]$$

$$A{A^T} = \left( {\matrix{ {{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \cr {{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr } } \right)\left( {\matrix{ {{1 \over {\sqrt 5 }}} & {{{ - 2} \over {\sqrt 5 }}} \cr {{2 \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr } } \right)$$

$$A{A^T} = \left( {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right) = I$$

$${Q^2} = {A^T}BA\,{A^T}BA = {A^T}BIBA$$

$$ \Rightarrow {Q^2} = {A^T}{B^2}A$$

$${Q^3} = {A^T}{B^2}A{A^T}BA \Rightarrow {Q^3} = {A^T}{B^3}A$$

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 $${B^{2021}} = \left( {\matrix{ 1 & 0 \cr {2021i} & 1 \cr } } \right)$$

$$\therefore$$ $$A{Q^{2021}} = {A^T} = A{A^T}{B^{2021}}\,A{A^T} = I{B^{2021}}I$$

$$ \Rightarrow A{Q^{2021}}\,{A^T} = {B^{2021}} = \left( {\matrix{ 1 & 0 \cr {2021i} & 1 \cr } } \right)$$

$$\therefore$$ $${(A{Q^{2021}}\,{A^T})^{ - 1}} = {\left( {\matrix{ 1 & 0 \cr {2021i} & 1 \cr } } \right)^{ - 1}} = \left( {\matrix{ 1 & 0 \cr { - 2021i} & 1 \cr } } \right)$$