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Byju's Answer
Standard XII
Mathematics
Adjoint of a Matrix
Show that i ...
Question
Show that
(
i
)
[
5
−
1
6
7
]
[
2
1
3
4
]
≠
[
2
1
3
4
]
[
5
−
1
6
7
]
(
i
i
)
⎡
⎢
⎣
1
0
0
2
1
1
3
0
0
⎤
⎥
⎦
⎡
⎢
⎣
−
1
0
2
1
−
1
3
0
1
4
⎤
⎥
⎦
≠
⎡
⎢
⎣
−
1
0
2
1
−
1
3
0
1
4
⎤
⎥
⎦
⎡
⎢
⎣
1
0
0
2
1
1
3
0
0
⎤
⎥
⎦
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Solution
(i)
[
5
−
1
6
7
]
[
2
1
3
4
]
=
[
5
(
2
)
−
1
(
3
)
5
(
1
)
−
1
(
4
)
6
(
2
)
+
7
(
3
)
6
(
1
)
+
7
(
4
)
]
=
[
10
−
3
5
−
4
12
+
21
6
+
28
]
=
[
7
1
33
34
]
[
2
1
3
4
]
[
5
−
1
6
7
]
=
[
2
(
5
)
+
1
(
6
)
2
(
−
1
)
+
1
(
7
)
3
(
5
)
+
4
(
6
)
3
(
−
1
)
+
4
(
7
)
]
=
[
10
+
6
−
2
+
7
15
+
24
−
3
+
28
]
=
[
16
5
39
25
]
∴
[
5
−
1
6
7
]
[
2
1
3
4
]
≠
[
2
1
3
4
]
[
5
−
1
6
7
]
(ii)
⎡
⎢
⎣
1
0
0
2
1
1
3
0
0
⎤
⎥
⎦
⎡
⎢
⎣
−
1
0
2
1
−
1
3
0
1
4
⎤
⎥
⎦
=
⎡
⎢
⎣
1
(
−
1
)
+
2
(
0
)
+
3
(
2
)
0
(
−
1
)
+
1
(
0
)
+
0
(
2
)
1
(
−
1
)
+
1
(
0
)
+
0
(
2
)
1
(
1
)
+
2
(
−
1
)
+
3
(
3
)
0
(
1
)
+
1
(
−
1
)
+
0
(
3
)
1
(
1
)
+
1
(
−
1
)
+
0
(
3
)
1
(
0
)
+
2
(
1
)
+
3
(
4
)
0
(
0
)
+
1
(
1
)
+
0
(
4
)
1
(
0
)
+
1
(
1
)
+
0
(
4
)
⎤
⎥
⎦
=
⎡
⎢
⎣
5
0
−
1
8
−
1
0
14
1
1
⎤
⎥
⎦
⎡
⎢
⎣
−
1
0
2
1
−
1
3
0
1
4
⎤
⎥
⎦
⎡
⎢
⎣
1
0
0
2
1
1
3
0
0
⎤
⎥
⎦
=
⎡
⎢
⎣
1
(
1
)
+
1
(
0
)
+
0
(
1
)
0
(
1
)
+
(
−
1
)
(
0
)
+
1
(
1
)
2
(
1
)
+
3
(
0
)
+
4
(
1
)
−
1
(
2
)
+
1
(
1
)
+
0
(
1
)
0
(
2
)
+
(
−
1
)
(
1
)
+
1
(
1
)
2
(
2
)
+
3
(
1
)
+
4
(
1
)
−
1
(
3
)
+
1
(
0
)
+
0
(
0
)
0
(
3
)
+
(
−
1
)
(
0
)
+
1
(
0
)
2
(
3
)
+
3
(
0
)
+
4
(
0
)
⎤
⎥
⎦
=
⎡
⎢
⎣
−
1
1
6
−
1
0
11
−
3
0
6
⎤
⎥
⎦
∴
⎡
⎢
⎣
1
0
0
2
1
1
3
0
0
⎤
⎥
⎦
⎡
⎢
⎣
−
1
0
2
1
−
1
3
0
1
4
⎤
⎥
⎦
≠
⎡
⎢
⎣
−
1
0
2
1
−
1
3
0
1
4
⎤
⎥
⎦
⎡
⎢
⎣
1
0
0
2
1
1
3
0
0
⎤
⎥
⎦
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