If A = 1 5 2 5 - 2 5 1 5 , B = 1 0 i 1 , i = - 1 , and Q = A T B A , then the inverse of the matrix A Q 2021 A T is equal to:

If A = 1 5 2 5 - 2 5 1 5 , B = 1 0 i 1 , i = - 1 , and Q = A T B A , then the in | TestHub

If $A = [\begin{array}{ll} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{- 2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{array}], B = [\begin{array}{l} 1 & 0 \\ i & 1 \end{array}], i = \sqrt{- 1}$ , and $Q = A^{T} B A$ , then the inverse of the matrix $A Q^{2021} A^{T}$ is equal to:

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Solution:

${AA}^{⊤} = [\begin{array}{l} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{- 2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{array}] [\begin{array}{l} \frac{1}{\sqrt{5}} & \frac{- 2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{array}] = [\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}] = I$

$Q = A^{⊤} B A$

$\Rightarrow Q^{2} = (A^{⊤} B A) (A^{⊤} B A)$

$= A^{⊤} B (A A^{⊤}) B A$

$= A^{⊤} B (I) B A$

$= A^{⊤} B^{2} A$

$\Rightarrow Q^{3} = A^{⊤} B^{3} A$

$\Rightarrow Q^{2021} = A^{T} B^{2021} A$

Now let $P = {AQ}^{2021} A^{T}$

$P = A (A^{⊤} B^{2021} A) A^{⊤}$

given $A A^{⊤} = I$

$P = B^{2021}$

$B^{2} = [\begin{array}{l} 1 & 0 \\ i & 1 \end{array}] [\begin{array}{l} 1 & 0 \\ i & 1 \end{array}] = [\begin{matrix} 1 & 0 \\ 2 i & 1 \end{matrix}]$

$B^{3} = [\begin{array}{l} 1 & 0 \\ 2 & 1 \end{array}] [\begin{array}{l} 1 & 0 \\ i & 1 \end{array}] = [\begin{array}{l} 1 & 0 \\ 3 i & 1 \end{array}]$

$B^{2021} = [\begin{matrix} 1 & 0 \\ 2021 i & 1 \end{matrix}]$

inverse of $P = (P^{- 1}) = {(B^{2021})}^{- 1} = [\begin{matrix} 1 & 0 \\ - 2021 i & 1 \end{matrix}]$

Mathematics - Matrices Question with Solution | TestHub

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