Let A=[110310−310110] and B=[1−i01] , where i=−1 if M=ATBA then the inverse of the matrix AM2023AT

LetAAT=[110310−310110][110−310310110] =[110+910−310+310−310+310910+110]=[1001]B2=[1−i01][1−i01]=[1−2i01]Similarly ,B3=[1−3i01]B2023=[1−2023i01] AM2023=..........ATB2023A AM2023AT=AATB2023AAT=B2023=[1−2023i01]Inverse of (AM2023AT)is[12023i01]

Let A=[110310−310110] and B=[1−i01] , where i=−1 if M=ATBA then the inverse of the matrix AM2023AT

LetAAT=[110310−310110][110−310310110] =[110+910−310+310−310+310910+110]=[1001]B2=[1−i01][1−i01]=[1−2i01]Similarly ,B3=[1−3i01]B2023=[1−2023i01] AM2023=..........ATB2023A AM2023AT=AATB2023AAT=B2023=[1−2023i01]Inverse of (AM2023AT)is[12023i01]

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Let A= $[\begin{matrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{- 3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{matrix}]$ and B= $[\begin{matrix} 1 & - i \\ 0 & 1 \end{matrix}]$ , where $i = \sqrt{- 1}$ if $M = A^{T} B A$ then the inverse of the matrix $A M^{2023} A^{T}$

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$[\begin{matrix} 1 & - 2023 i \\ 0 & 1 \end{matrix}]$

$[\begin{matrix} 1 & 2023 i \\ 0 & 1 \end{matrix}]$

$[\begin{matrix} 1 & 0 \\ - 2023 i & 1 \end{matrix}]$

$[\begin{matrix} 1 & 0 \\ 2023 i & 1 \end{matrix}]$

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Detailed Solution

$L e t A A^{T} = [\begin{matrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{- 3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{matrix}] [\begin{matrix} \frac{1}{\sqrt{10}} & \frac{- 3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{matrix}]$
$= [\begin{matrix} \frac{1}{10} + \frac{9}{10} & \frac{- 3}{10} + \frac{3}{10} \\ \frac{- 3}{10} + \frac{3}{10} & \frac{9}{10} + \frac{1}{10} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$
$B^{2} = [\begin{matrix} 1 & - i \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & - i \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & - 2 i \\ 0 & 1 \end{matrix}]$

Similarly ,
$B^{3} = [\begin{matrix} 1 & - 3 i \\ 0 & 1 \end{matrix}] B^{2023} = [\begin{matrix} 1 & - 2023 i \\ 0 & 1 \end{matrix}]$
$A M^{2023} = .......... A^{T} B^{2023} A$
$A M^{2023} A^{T} = A A^{T} B^{2023} A A^{T} = B^{2023} = [\begin{matrix} 1 & - 2023 i \\ 0 & 1 \end{matrix}]$
Inverse of $(A M^{2023} A^{T}) i s [\begin{matrix} 1 & 2023 i \\ 0 & 1 \end{matrix}]$