Using elementary transformations, find the inverse of the followng matrix.
[1−123]
Let A=[1−123],We know that A=IA
∴[1−123]=[1001]A⇒[1−105]=[10−21]A (Using R2→R2−2R1)
⇒[1−101]=[10−2515]A (Using R2→15R2)
⇒[1001]=⎡⎣3515−2515⎤⎦A (Using R1→R1+R2)
∴A−1=⎡⎣3515−2515⎤⎦(∵AA−1=I)