The inverse of mat 100010001
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Answer:
Let A=[
2
5
3
7
]
We know that A=IA
∴A=[
2
5
3
7
]=[
1
0
0
1
]A
⇒[
1
5
2
3
7
]=[
2
1
0
0
1
]A(R
1
→
2
1
R
1
)
⇒[
1
0
2
3
−
2
1
]=[
2
1
−
2
5
0
1
]A(R
2
→R
2
−5R
1
)
⇒[
1
0
0
−
2
1
]=[
−7
−
2
5
3
1
]A(R
1
→R
1
−3R
2
)
⇒[
1
0
0
2
1
]=[
−7
2
5
3
−1
]A(R
2
→−R
2
)
⇒[
1
0
0
1
]=[
−7
5
3
−2
]A(R
2
→2R
2
)
∴A
−1
=[
−7
5
3
−2
]
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