5
Question
Using elementary transformations, find the inverse of matrix [4534], if it exists.
Using elementary transformations, find the inverse of matrix [4534], if it exists.
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Solution
Let A=[4534]
We know that A=IA
⇒[4534]=[1001]A
Applying R1→R1−R2
⇒[4−35−434]=[1−00−101]A
⇒[1134]=[1−101]A
Applying R2→R2−3R1
⇒[113−3(1)4−3(1)]=[1−10−3(1)1−3(−1)]A
⇒[1101]=[1−10−31+3]A
⇒[1101]=[1−1−34]A
Applying R1→R1−R2
⇒[1−01−101]=[1−(−3)−1−4−34]A
⇒[1001]=[4−5−34]A
⇒I=[4−5−34]A
This is similar to I=A−1A
Thus, A−1=[4−5−34]
We know that A=IA
⇒[4534]=[1001]A
Applying R1→R1−R2
⇒[4−35−434]=[1−00−101]A
⇒[1134]=[1−101]A
Applying R2→R2−3R1
⇒[113−3(1)4−3(1)]=[1−10−3(1)1−3(−1)]A
⇒[1101]=[1−10−31+3]A
⇒[1101]=[1−1−34]A
Applying R1→R1−R2
⇒[1−01−101]=[1−(−3)−1−4−34]A
⇒[1001]=[4−5−34]A
⇒I=[4−5−34]A
This is similar to I=A−1A
Thus, A−1=[4−5−34]
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