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Question

Find the inverse of the following matrix using elementary operations.
A= 122130021


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Solution

The given matrix is
A=122130021

|A|=1(30)2(10)2(20)
|A|=3+24
|A|=10
A1exists.


We know that AA1=I
122130021A1= 100010001

Perform R2R2+R1, we get
122052021A1= 100110001

Perform R215R2, we get
⎢ ⎢ ⎢1220125021⎥ ⎥ ⎥A1= ⎢ ⎢ ⎢10015150001⎥ ⎥ ⎥

Perform R1R12R2 and R3R3+2R2, we get
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢106501250015⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥A1= ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢352501515025251⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

Perform R35R3, we get
⎢ ⎢ ⎢ ⎢ ⎢10650125001⎥ ⎥ ⎥ ⎥ ⎥A1= ⎢ ⎢ ⎢ ⎢ ⎢3525015150225⎥ ⎥ ⎥ ⎥ ⎥

Perform R2R2+25R3 and R1R1+65R3, we get

100010001A1= 326112225

IA1= 326112225

A1 = 326112225

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