6
Question
Using elementary transformations, find the inverse of matrix [2174], if it exists.
Using elementary transformations, find the inverse of matrix [2174], if it exists.
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Solution
Let A=[2174]
We know that A=IA
⇒[2174]=[1001]A
Applying R1→R1−17R2
⇒⎡⎣2−17(7)1−17(4)74⎤⎦=⎡⎣1−17(0)0−17(1)01⎤⎦A
⇒⎡⎣2−11−4774⎤⎦=⎡⎣1−0−1701⎤⎦A
Applying R2→R2−7R1
⇒⎡⎢
⎢
⎢⎣1377−7(1)4−7(37)⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣1−170−7(1)1−7(−17)⎤⎥
⎥
⎥⎦A
⇒⎡⎣1377−74−3⎤⎦=⎡⎣1−170−71+1⎤⎦A
⇒⎡⎣13701⎤⎦=⎡⎣1−17−72⎤⎦A
Applying R1→R1−37R2
⇒⎡⎣1−37(0)37−37(1)01⎤⎦=⎡⎣1−37(−7)−17−37(2)−72⎤⎦A
⇒⎡⎣1−037−3701⎤⎦=⎡⎣1+3−17−67−72⎤⎦A
⇒[1001]=[4−1−72]A
⇒I=[4−1−72]A
This is similar to I=A−1A
Thus, A−1=[4−1−72]
We know that A=IA
⇒[2174]=[1001]A
Applying R1→R1−17R2
⇒⎡⎣2−17(7)1−17(4)74⎤⎦=⎡⎣1−17(0)0−17(1)01⎤⎦A
⇒⎡⎣2−11−4774⎤⎦=⎡⎣1−0−1701⎤⎦A
Applying R2→R2−7R1
⇒⎡⎢ ⎢ ⎢⎣1377−7(1)4−7(37)⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣1−170−7(1)1−7(−17)⎤⎥ ⎥ ⎥⎦A
⇒⎡⎣1377−74−3⎤⎦=⎡⎣1−170−71+1⎤⎦A
⇒⎡⎣13701⎤⎦=⎡⎣1−17−72⎤⎦A
Applying R1→R1−37R2
⇒⎡⎣1−37(0)37−37(1)01⎤⎦=⎡⎣1−37(−7)−17−37(2)−72⎤⎦A
⇒⎡⎣1−037−3701⎤⎦=⎡⎣1+3−17−67−72⎤⎦A
⇒[1001]=[4−1−72]A
⇒I=[4−1−72]A
This is similar to I=A−1A
Thus, A−1=[4−1−72]
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