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Byju's Answer
Standard XII
Mathematics
Elementary Transformations
Using element...
Question

Using elementary transformations, find the inverse of matrix ⎡⎢⎣20−1510013⎤⎥⎦, if it exists.


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Solution

Let A=⎡⎢⎣20−1510013⎤⎥⎦
We know that A=IA
⇒⎡⎢⎣20−1510013⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
Applying R1→12R1
⇒⎡⎢ ⎢ ⎢⎣10−12510013⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣1200010001⎤⎥ ⎥ ⎥⎦A
Applying R2→R2−5R1
⇒⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣10−125−5(1)1−5(0)0−5(−12)013⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣12000−5(12)1−5(0)0−5(0)001⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
⇒⎡⎢ ⎢ ⎢ ⎢ ⎢⎣10−120152013⎤⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1200−(52)10001⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
Applying R3→R3−R2
⇒⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣10−12015201−13−52⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1200−(52)100+520−11−0⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
⇒⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣10−1201520012⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1200−(52)1052−11⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
Applying R3→2R3
⇒⎡⎢ ⎢ ⎢ ⎢ ⎢⎣10−120152001⎤⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1200−(52)105−22⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
Applying R1→R1+12R3
⇒⎡⎢ ⎢ ⎢ ⎢ ⎢⎣1+12(0)0+12(0)−12+12(1)0152001⎤⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣12+12(5)0+12(−2)0+12(2)−52105−22⎤⎥ ⎥ ⎥ ⎥ ⎥⎦A
⇒⎡⎢ ⎢ ⎢⎣1000152001⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣3−11−52105−22⎤⎥ ⎥ ⎥⎦A
Applying R2→R2−52R3
⇒⎡⎢ ⎢ ⎢⎣1000−52(0)1−52(0)52−52(1)001⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣3−11−52−52(5)1−52(−2)0−52(2)5−22⎤⎥ ⎥ ⎥⎦A
⇒⎡⎢⎣100010000⎤⎥⎦=⎡⎢ ⎢ ⎢⎣3−11−3026−55−22⎤⎥ ⎥ ⎥⎦A
⇒I=⎡⎢⎣3−11−156−55−22⎤⎥⎦A
This is similar to I=A−1A
Thus, A−1=⎡⎢⎣3−11−156−55−22⎤⎥⎦

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