A=[1 2 2 2 1 2 2 2 1] find A^-1 by elementary row transformation
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Step-by-step explanation:
Solution :- Given A=[
1
2
2
−1
]
Now we find A
−1
by elementary row transformation
so, we can write
A=I.A
[
1
2
2
−1
]=[
1
0
0
1
].A
R
2
→R
2
−2R
1
[
1
0
2
−5
]=[
1
−2
0
1
].A
R
1
→R
1
+
5
2
R
2
[
1
0
0
−5
]=[
5
1
−2
5
2
1
].A
R
2
→
−5
R
2
[
1
0
0
1
]=
⎣
⎢
⎢
⎡
5
1
5
2
5
2
5
−1
⎦
⎥
⎥
⎤
.A
this form is I=A
−1
.A So, A
−1
=
⎣
⎢
⎢
⎡
5
1
5
2
5
2
5
−1
⎦
⎥
⎥
⎤
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