1
Question
Using elementary transformations, find the inverse of matrix [2357], if it exists.
Using elementary transformations, find the inverse of matrix [2357], if it exists.
Open in App
Solution
Let A=[2357]
We know that A=IA
⇒[2357]=[1001]A
Applying R1→12R1
⇒⎡⎣223257⎤⎦=⎡⎣120201⎤⎦A
⇒⎡⎣13257⎤⎦=⎡⎣12001⎤⎦A
Applying R2→R2−5R1
⇒⎡⎢
⎢
⎢⎣1325−57−5(32)⎤⎥
⎥
⎥⎦=⎡⎢
⎢⎣1200−521−5(0)⎤⎥
⎥⎦A
⇒⎡⎢
⎢⎣1320−12⎤⎥
⎥⎦=⎡⎢
⎢⎣120−521⎤⎥
⎥⎦A
Applying R2→−2R2
⇒⎡⎢
⎢
⎢⎣132−2(0)−2(−12)⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣120−2(−52)−2(1)⎤⎥
⎥
⎥⎦A
⇒⎡⎣13201⎤⎦=⎡⎣1205−2⎤⎦A
Applying R1→R1−32R2
⇒⎡⎣1−32(0)32−32(1)01⎤⎦=⎡⎣12−32(5)0−32(−2)5−2⎤⎦A
⇒[1001]=[−735−2]A
This is similar to I=A−1A
Thus, A−1=[−735−2]
We know that A=IA
⇒[2357]=[1001]A
Applying R1→12R1
⇒⎡⎣223257⎤⎦=⎡⎣120201⎤⎦A
⇒⎡⎣13257⎤⎦=⎡⎣12001⎤⎦A
Applying R2→R2−5R1
⇒⎡⎢ ⎢ ⎢⎣1325−57−5(32)⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢⎣1200−521−5(0)⎤⎥ ⎥⎦A
⇒⎡⎢ ⎢⎣1320−12⎤⎥ ⎥⎦=⎡⎢ ⎢⎣120−521⎤⎥ ⎥⎦A
Applying R2→−2R2
⇒⎡⎢ ⎢ ⎢⎣132−2(0)−2(−12)⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣120−2(−52)−2(1)⎤⎥ ⎥ ⎥⎦A
⇒⎡⎣13201⎤⎦=⎡⎣1205−2⎤⎦A
Applying R1→R1−32R2
⇒⎡⎣1−32(0)32−32(1)01⎤⎦=⎡⎣12−32(5)0−32(−2)5−2⎤⎦A
⇒[1001]=[−735−2]A
This is similar to I=A−1A
Thus, A−1=[−735−2]
Suggest Corrections
3
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
