Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
Let A=[ beginarraycc(1/√10) (3/√10) (-3/√10) (1/√10) endarray] and B=[ beginarraycc1 -i 0 1 endarray], where i=√-1. If M = A T B A, then the inverse of the matrix AM 2023 A T is
Q. Let
A
=
[
10
1
10
−
3
10
3
10
1
]
and
B
=
[
1
0
−
i
1
]
, where
i
=
−
1
.
If
M
=
A
T
B
A
, then the inverse of the matrix
A
M
2023
A
T
is
685
139
JEE Main
JEE Main 2023
Determinants
Report Error
A
[
1
0
−
2023
i
1
]
B
[
1
2023
i
0
1
]
C
[
1
0
2023
i
1
]
D
[
1
−
2023
i
0
1
]
Solution:
A
A
T
=
[
10
1
10
−
3
10
3
10
1
]
[
10
1
10
3
10
−
3
10
1
]
=
[
1
0
0
1
]
B
2
=
[
1
0
−
i
1
]
[
1
0
−
i
1
]
=
[
1
0
−
2
i
1
]
B
3
=
[
1
0
−
3
i
1
]
:
:
B
2023
=
[
1
0
−
2023
i
1
]
M
=
A
T
B
A
M
2
=
M
⋅
M
=
A
T
B
A
A
T
B
A
=
A
T
B
2
A
M
3
=
M
2
⋅
M
=
A
T
B
2
A
A
T
B
A
=
A
T
B
3
A
:
:
M
2023
=
…………
..
A
T
B
2023
A
A
M
2023
A
T
=
A
A
T
B
2023
A
A
T
=
B
2023
=
[
1
0
−
2023
i
1
]
Inverse of
(
A
M
2023
A
T
)
is
[
1
0
2023
i
1
]