1
Question
Solve the following system of linear equation by matrix method
x−y+2z=1
2y−3z=1
3x−2y+4z=2
x−y+2z=1
2y−3z=1
3x−2y+4z=2
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Solution
A=⎡⎢⎣1−1202−33−24⎤⎥⎦;X=⎡⎢⎣xyz⎤⎥⎦;B=⎡⎢⎣112⎤⎥⎦
X=A−1B=1|A|(adjA)B
|A|=∣∣
∣∣1−1202−33−24∣∣
∣∣=1(8−6)+1(0+9)+2(0−6)=2+9−12=−1
adjA=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣∣∣∣2−3−24∣∣∣∣∣∣0−334∣∣∣∣∣∣023−2∣∣∣∣∣∣−12−24∣∣∣∣∣∣1234∣∣∣∣∣∣1−13−2∣∣∣∣∣∣−122−3∣∣∣∣∣∣120−3∣∣∣∣∣∣1−102∣∣∣⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦T=⎡⎢⎣2−9−60−2−1−132⎤⎥⎦=⎡⎢⎣20−1−9−23−6−12⎤⎥⎦
X=1|A|(adjA)B
X=−1⎡⎢⎣20−1−9−23−6−12⎤⎥⎦⎡⎢⎣112⎤⎥⎦
X=−1⎡⎢⎣2−2−5−3⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣053⎤⎥⎦
X=A−1B=1|A|(adjA)B
|A|=∣∣ ∣∣1−1202−33−24∣∣ ∣∣=1(8−6)+1(0+9)+2(0−6)=2+9−12=−1
adjA=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣∣∣∣2−3−24∣∣∣∣∣∣0−334∣∣∣∣∣∣023−2∣∣∣∣∣∣−12−24∣∣∣∣∣∣1234∣∣∣∣∣∣1−13−2∣∣∣∣∣∣−122−3∣∣∣∣∣∣120−3∣∣∣∣∣∣1−102∣∣∣⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦T=⎡⎢⎣2−9−60−2−1−132⎤⎥⎦=⎡⎢⎣20−1−9−23−6−12⎤⎥⎦
X=1|A|(adjA)B
X=−1⎡⎢⎣20−1−9−23−6−12⎤⎥⎦⎡⎢⎣112⎤⎥⎦
X=−1⎡⎢⎣2−2−5−3⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣053⎤⎥⎦
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