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Question
Using elementary transformations, find the inverse of matrix [3−1−42], if it exists.
Using elementary transformations, find the inverse of matrix [3−1−42], if it exists.
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Solution
Let A=[3−1−42]
We know that A=IA
⇒[3−1−42]=[1001]A
Applying R1→13R1
⇒⎡⎣33−13−42⎤⎦=⎡⎣130301⎤⎦A
⇒⎡⎣1−13−42⎤⎦=⎡⎣13001⎤⎦A
Applying R2→R2+4R1
⇒⎡⎢
⎢
⎢⎣1−13−4+4(1)2+4(−13)⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣1300+4(13)1+4(0)⎤⎥
⎥
⎥⎦A
⇒⎡⎢
⎢
⎢⎣1−130(23)⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣130(43)1⎤⎥
⎥
⎥⎦A
Applying R2→32R2
⇒⎡⎢
⎢
⎢⎣1−1332(0)32(23)⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣13032(43)32(1)⎤⎥
⎥
⎥⎦A
⇒⎡⎣1−1301⎤⎦=⎡⎢
⎢⎣130232⎤⎥
⎥⎦A
Applying R1→R1+13R2
⇒⎡⎣1+13(0)−13+13(1)01⎤⎦=⎡⎢
⎢
⎢⎣13+13(2)0+13(32)232⎤⎥
⎥
⎥⎦A
⇒[1001]=⎡⎢
⎢⎣3312232⎤⎥
⎥⎦A
⇒I=⎡⎢
⎢⎣112232⎤⎥
⎥⎦A
This is similar to I=A−1A
Hence, A−1=⎡⎢
⎢⎣112232⎤⎥
⎥⎦
We know that A=IA
⇒[3−1−42]=[1001]A
Applying R1→13R1
⇒⎡⎣33−13−42⎤⎦=⎡⎣130301⎤⎦A
⇒⎡⎣1−13−42⎤⎦=⎡⎣13001⎤⎦A
Applying R2→R2+4R1
⇒⎡⎢ ⎢ ⎢⎣1−13−4+4(1)2+4(−13)⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣1300+4(13)1+4(0)⎤⎥ ⎥ ⎥⎦A
⇒⎡⎢ ⎢ ⎢⎣1−130(23)⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣130(43)1⎤⎥ ⎥ ⎥⎦A
Applying R2→32R2
⇒⎡⎢ ⎢ ⎢⎣1−1332(0)32(23)⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣13032(43)32(1)⎤⎥ ⎥ ⎥⎦A
⇒⎡⎣1−1301⎤⎦=⎡⎢ ⎢⎣130232⎤⎥ ⎥⎦A
Applying R1→R1+13R2
⇒⎡⎣1+13(0)−13+13(1)01⎤⎦=⎡⎢ ⎢ ⎢⎣13+13(2)0+13(32)232⎤⎥ ⎥ ⎥⎦A
⇒[1001]=⎡⎢ ⎢⎣3312232⎤⎥ ⎥⎦A
⇒I=⎡⎢ ⎢⎣112232⎤⎥ ⎥⎦A
This is similar to I=A−1A
Hence, A−1=⎡⎢ ⎢⎣112232⎤⎥ ⎥⎦
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