Let A=\left[\begin{array}{c}\\\{1\end{array}/{\sqrt{10}} , {3}/{\sqrt{10}} ; {-3}/{\sqrt{10}} , {1}/{\sqrt{10}}]} and B=\left[\begin{array}{cc}1 & -i \ 0 & 1]\end{array}, where {\i}=\sqrt{-1}. If {\M}= {\A}^{ {\T}} {\BA}, then the inverse of the matrix AM^{2023} {\A}^{ {\T}} is [25-Jan-2023 Shift 2]

\left[\begin{array}{cc}1 & -2023 i \ 0 & 1]\end{array}

\left[\begin{array}{cc}1 & 0 \ -2023 {\i\end{array} , 1]}

\left[\begin{array}{cc}1 & 0 \ 2023 {\i\end{array} , 1]}

\left[\begin{array}{cc}1 & 2023 {\i\end{array} ; 0 , 1]}

Matrices and Determinants Part 3

Section: Mathematics

Question : 72 of 100

Marks: +1, -0

Let A=[

‌

√10

‌

√10

‌

−3

√10

‌

√10

] and B=[

1	−i
0	1

], where i=√−1. If M=AT‌BA, then the inverse of the matrix AM2023AT is

[25-Jan-2023 Shift 2]

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1	−2023i
0	1

1	0
−2023i	1

1	0
2023i	1

1	2023i
0	1