2
Question
Using elementary transformations, find the inverse of the followng matrix.
⎡⎢⎣2−332233−22⎤⎥⎦
Using elementary transformations, find the inverse of the followng matrix.
⎡⎢⎣2−332233−22⎤⎥⎦
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Solution
Let A=⎡⎢⎣2−332233−22⎤⎥⎦.We know that A=IA
∴⎡⎢⎣2−332233−22⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A⇒⎡⎢⎣1142233−22⎤⎥⎦=⎡⎢⎣11−1010001⎤⎥⎦A
(Using R1)→R1+R2−R3
⇒⎡⎢⎣11400−50−5−10⎤⎥⎦=⎡⎢⎣11−1−2−12−3−34⎤⎥⎦A
(Using R2→2R1 ~and~ R3→R3−3R1)
⇒⎡⎢⎣1140−5−1000−5⎤⎥⎦=⎡⎢⎣11−1−3−34−2−12⎤⎥⎦A (Using R2↔R3)
⇒⎡⎢⎣114012001⎤⎥⎦=⎡⎢
⎢
⎢⎣11−13535−452515−25⎤⎥
⎥
⎥⎦A (Using R2→−15R2andR3→−15R3)
⇒⎡⎢⎣110010001⎤⎥⎦=⎡⎢
⎢
⎢⎣−351535−151502515−25⎤⎥
⎥
⎥⎦A (Using R2→R2−2R3 and R1→4R3)
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢
⎢
⎢⎣−25035−151502515−25⎤⎥
⎥
⎥⎦A(UsingR1→R1−R2)⇒I3=−15⎡⎢⎣20−31−10−20−3⎤⎥⎦A∴A−1=−15⎡⎢⎣20−31−10−2−12⎤⎥⎦(∵AA−1=I3)
Let A=⎡⎢⎣2−332233−22⎤⎥⎦.We know that A=IA
∴⎡⎢⎣2−332233−22⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A⇒⎡⎢⎣1142233−22⎤⎥⎦=⎡⎢⎣11−1010001⎤⎥⎦A
(Using R1)→R1+R2−R3
⇒⎡⎢⎣11400−50−5−10⎤⎥⎦=⎡⎢⎣11−1−2−12−3−34⎤⎥⎦A
(Using R2→2R1 ~and~ R3→R3−3R1)
⇒⎡⎢⎣1140−5−1000−5⎤⎥⎦=⎡⎢⎣11−1−3−34−2−12⎤⎥⎦A (Using R2↔R3)
⇒⎡⎢⎣114012001⎤⎥⎦=⎡⎢
⎢
⎢⎣11−13535−452515−25⎤⎥
⎥
⎥⎦A (Using R2→−15R2andR3→−15R3)
⇒⎡⎢⎣110010001⎤⎥⎦=⎡⎢
⎢
⎢⎣−351535−151502515−25⎤⎥
⎥
⎥⎦A (Using R2→R2−2R3 and R1→4R3)
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢ ⎢ ⎢⎣−25035−151502515−25⎤⎥ ⎥ ⎥⎦A(UsingR1→R1−R2)⇒I3=−15⎡⎢⎣20−31−10−20−3⎤⎥⎦A∴A−1=−15⎡⎢⎣20−31−10−2−12⎤⎥⎦(∵AA−1=I3)
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