1
Question
Find the inverse of the following matrix using elementary operations.
A= ⎡⎢⎣12−2−1300−21⎤⎥⎦
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⎤⎥⎦
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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