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Byju's Answer
Standard XII
Mathematics
Elementary Transformations
Obtain the in...
Question

Obtain the inverse of the following matrix using elementary operations A=⎡⎢⎣012123311⎤⎥⎦


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Solution

Given A=⎡⎢⎣012123311⎤⎥⎦

We know that A=IA

⇒⎡⎢⎣012123311⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
Applying R1↔R2
⇒⎡⎢⎣123012311⎤⎥⎦=⎡⎢⎣010100001⎤⎥⎦A
Applying R3→R3−3R1
⇒⎡⎢⎣1230123−3(1)1−3(2)1−3(3)⎤⎥⎦=⎡⎢⎣0101000−3(0)0−3(1)1−3(0)⎤⎥⎦A
⇒⎡⎢⎣1230120−5−8⎤⎥⎦=⎡⎢⎣0101000−31⎤⎥⎦A
Applying R1→R1−2R2
⇒⎡⎢⎣1−2(0)2−2(1)3−2(2)0120−5−8⎤⎥⎦=⎡⎢⎣0−2(1)1−2(0)0−2(0)1000−31⎤⎥⎦A
⇒⎡⎢⎣10−10120−5−8⎤⎥⎦=⎡⎢⎣−2101000−31⎤⎥⎦A
Applying R3→R3+5R2
⇒⎡⎢⎣10−10120+5(0)−5+5(1)−8+5(2)⎤⎥⎦=⎡⎢⎣−2101000+5(1)−3+5(0)1+5(0)⎤⎥⎦A
⇒⎡⎢⎣10−1012002⎤⎥⎦=⎡⎢⎣−2101005−31⎤⎥⎦A
Applying R3→12R3
⇒⎡⎢⎣10−1012001⎤⎥⎦=⎡⎢ ⎢ ⎢⎣−21010052−3212⎤⎥ ⎥ ⎥⎦A
Applying R1→R1+R3
⇒⎡⎢⎣1+00+0−1+1012001⎤⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−2+521+(−32)0+1210052−3212⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
⇒⎡⎢⎣100012001⎤⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣12−121210052−3212⎤⎥ ⎥ ⎥ ⎥ ⎥⎦A
Applying R2→R2−2R3
⇒⎡⎢⎣1000−2(0)1−2(0)2−2(1)001⎤⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣12−12121−2(52)0−2(−32)0−2(12)52−3212⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣12−1212−43−152−3212⎤⎥ ⎥ ⎥ ⎥ ⎥⎦A
This is similar to I=A−1A
Hence, A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣12−1212−43−152−3212⎤⎥ ⎥ ⎥ ⎥ ⎥⎦

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