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Question
Using elementary transformations, find the inverse of matrix [31027], if it exists.
Using elementary transformations, find the inverse of matrix [31027], if it exists.
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Solution
Let A=[31027]
We know that A=IA
⇒[31027]=[1001]A
Applying R1→R1−R2
⇒[3−210−727]=[1−00−101]A
⇒[1327]=[1−101]A
Applying R2→R2−2R1
⇒[132−2(1)7−2(3)]=[1−10−2(1)1−2(−1)]A
⇒[1301]=[1−1−23]A
Applying R1→R1−3R2
⇒[1−3(0)3−3(1)01]=[1−3(−2)−1−3(3)−23]A
⇒[1001]=[7−10−23]A
⇒I=[7−10−23]A
This is similar to I=A−1A
Thus, A−1=[7−10−23]
We know that A=IA
⇒[31027]=[1001]A
Applying R1→R1−R2
⇒[3−210−727]=[1−00−101]A
⇒[1327]=[1−101]A
Applying R2→R2−2R1
⇒[132−2(1)7−2(3)]=[1−10−2(1)1−2(−1)]A
⇒[1301]=[1−1−23]A
Applying R1→R1−3R2
⇒[1−3(0)3−3(1)01]=[1−3(−2)−1−3(3)−23]A
⇒[1001]=[7−10−23]A
⇒I=[7−10−23]A
This is similar to I=A−1A
Thus, A−1=[7−10−23]
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