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Question

# Find the inverse of the following matrix using elementary operations. A= ⎡⎢⎣12−2−1300−21⎤⎥⎦

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Solution

## The given matrix is A=⎡⎢⎣12−2−1300−21⎤⎥⎦ |A|=1(3−0)−2(−1−0)−2(2−0) ⇒|A|=3+2−4 ⇒|A|=1≠0 ∴A−1exists. We know that AA−1=I ⎡⎢⎣12−2−1300−21⎤⎥⎦A−1= ⎡⎢⎣100010001⎤⎥⎦ Perform R2→R2+R1, we get ⎡⎢⎣12−205−20−21⎤⎥⎦A−1= ⎡⎢⎣100110001⎤⎥⎦ Perform R2→15R2, we get ⎡⎢ ⎢ ⎢⎣12−201−250−21⎤⎥ ⎥ ⎥⎦A−1= ⎡⎢ ⎢ ⎢⎣10015150001⎤⎥ ⎥ ⎥⎦ Perform R1→R1−2R2 and R3→R3+2R2, we get ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣10−6501−250015⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A−1= ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣35−2501515025251⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦ Perform R3→5R3, we get ⎡⎢ ⎢ ⎢ ⎢ ⎢⎣10−6501−25001⎤⎥ ⎥ ⎥ ⎥ ⎥⎦A−1= ⎡⎢ ⎢ ⎢ ⎢ ⎢⎣35−25015150225⎤⎥ ⎥ ⎥ ⎥ ⎥⎦ Perform R2→R2+25R3 and R1→R1+65R3, we get ⎡⎢⎣100010001⎤⎥⎦A−1= ⎡⎢⎣326112225⎤⎥⎦ ⇒IA−1= ⎡⎢⎣326112225⎤⎥⎦ ∴A−1 = ⎡⎢⎣326112225⎤⎥⎦

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