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Question
Using elementary transformations, find the inverse of the followng matrix.
⎡⎢⎣20−1510013⎤⎥⎦
Using elementary transformations, find the inverse of the followng matrix.
⎡⎢⎣20−1510013⎤⎥⎦
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Solution
Let A=⎡⎢⎣20−1510013⎤⎥⎦.We know that A =IA
∴⎡⎢⎣20−1510013⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A⇒⎡⎢
⎢⎣10−12510013⎤⎥
⎥⎦=⎡⎢
⎢⎣1200010001⎤⎥
⎥⎦A
(Using R1→(12)R1)
⇒⎡⎢
⎢⎣10−120152013⎤⎥
⎥⎦=⎡⎢
⎢⎣1200−5210001⎤⎥
⎥⎦A (Using R2→R2−5R1)
⇒⎡⎢
⎢
⎢⎣10−12015210−12⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣1200−521012−11⎤⎥
⎥
⎥⎦A (Using R3→R3−R2)
⇒⎡⎢
⎢⎣10−120152001⎤⎥
⎥⎦=⎡⎢
⎢⎣1200−52105−22⎤⎥
⎥⎦A (Using R3→2R3)
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣3−11−156−55−22⎤⎥⎦A
(Using R2→R2−52R3 and R1→R1+12R3)
∴A−1=⎡⎢⎣3−11−156−55−22⎤⎥⎦(∵AA−1=I)
Let A=⎡⎢⎣20−1510013⎤⎥⎦.We know that A =IA
∴⎡⎢⎣20−1510013⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A⇒⎡⎢
⎢⎣10−12510013⎤⎥
⎥⎦=⎡⎢
⎢⎣1200010001⎤⎥
⎥⎦A
(Using R1→(12)R1)
⇒⎡⎢
⎢⎣10−120152013⎤⎥
⎥⎦=⎡⎢
⎢⎣1200−5210001⎤⎥
⎥⎦A (Using R2→R2−5R1)
⇒⎡⎢
⎢
⎢⎣10−12015210−12⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣1200−521012−11⎤⎥
⎥
⎥⎦A (Using R3→R3−R2)
⇒⎡⎢
⎢⎣10−120152001⎤⎥
⎥⎦=⎡⎢
⎢⎣1200−52105−22⎤⎥
⎥⎦A (Using R3→2R3)
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣3−11−156−55−22⎤⎥⎦A
(Using R2→R2−52R3 and R1→R1+12R3)
∴A−1=⎡⎢⎣3−11−156−55−22⎤⎥⎦(∵AA−1=I)
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