2
Question
Using elementary transformations, find the inverse of matrix [2−61−2], if it exists.
Using elementary transformations, find the inverse of matrix [2−61−2], if it exists.
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Solution
Let A=[2−61−2]
We know that A=IA
⇒[2−61−2]=[1001]A
Applying R1→R1−R2
⇒[2−1−6−(−2)1−2]=[1−00−101]A
⇒[1−41−2]=[1−101]A
Applying R2→R2−R1
⇒[1−41−1−2−(−4)]=[1−10−11−(−1)]A
⇒[1−402]=[1−1−12]A
Applying R2→12R2
⇒[1−401]=⎡⎣1−1−121⎤⎦A
Applying R1→R1+4R2
⇒[1+4(0)−4+4(1)01]=⎡⎢
⎢
⎢⎣1+4(−12)−1+4(1)−121⎤⎥
⎥
⎥⎦A
⇒[1001]=⎡⎣−13−121⎤⎦A
⇒I=⎡⎣−13−121⎤⎦A
This is similar to I=A−1A
Thus, A−1=⎡⎣−13−121⎤⎦
We know that A=IA
⇒[2−61−2]=[1001]A
Applying R1→R1−R2
⇒[2−1−6−(−2)1−2]=[1−00−101]A
⇒[1−41−2]=[1−101]A
Applying R2→R2−R1
⇒[1−41−1−2−(−4)]=[1−10−11−(−1)]A
⇒[1−402]=[1−1−12]A
Applying R2→12R2
⇒[1−401]=⎡⎣1−1−121⎤⎦A
Applying R1→R1+4R2
⇒[1+4(0)−4+4(1)01]=⎡⎢ ⎢ ⎢⎣1+4(−12)−1+4(1)−121⎤⎥ ⎥ ⎥⎦A
⇒[1001]=⎡⎣−13−121⎤⎦A
⇒I=⎡⎣−13−121⎤⎦A
This is similar to I=A−1A
Thus, A−1=⎡⎣−13−121⎤⎦
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