4
Question
Examine the consistency of the system of equations3x−y−2z=2,2y−z=−1,3x−5y=3
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Solution
Given system of equations3x−y−2z=2
2y−z=−1
3x−5y=3This can be written as AX=Bwhere A=⎡⎢⎣3−1−202−13−50⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦,B=⎡⎢⎣2−13⎤⎥⎦
Here, |A|=3(0−5)+1(0+3)−2(0−6)⇒|A|=0Since, |A|=0Hence, the system of equations has either infinitely many solutions (consistent) or no solution (inconsistent).
We need to find (adjA)BC11=(−1)1+1∣∣∣2−1−50∣∣∣⇒C11=0−5=−5
C12=(−1)1+2∣∣∣0−130∣∣∣⇒C12=−(0+3)=−3
C13=(−1)1+3∣∣∣023−5∣∣∣⇒C13=0−6=−6
C21=(−1)2+1∣∣∣−1−2−50∣∣∣⇒C21=−(0−10)=10
C22=(−1)2+2∣∣∣3−230∣∣∣⇒C22=0+6=6
C23=(−1)2+3∣∣∣3−13−5∣∣∣⇒C23=−(−15+3)=12
C31=(−1)3+1∣∣∣−1−22−1∣∣∣⇒C31=1+4=5
C32=(−1)3+2∣∣∣3−20−1∣∣∣⇒C32=−(−3−0)=3
C33=(−1)3+3∣∣∣3−102∣∣∣⇒C33=6−0=6
Hence, the co-factor matrix is C=⎡⎢⎣−5−3−610612536⎤⎥⎦
⇒adjA=CT=⎡⎢⎣−5105−363−6126⎤⎥⎦
Now, (adjA)B=⎡⎢⎣−5105−363−6126⎤⎥⎦⎡⎢⎣2−13⎤⎥⎦
=⎡⎢⎣−10−10+15−6−6+9−12−12+18⎤⎥⎦
⇒(adjA)B=⎡⎢⎣−5−3−6⎤⎥⎦
Since, (adjA)B≠OHence, the system of equations is inconsistent.
3x−y−2z=2
2y−z=−1
2y−z=−1
3x−5y=3
This can be written as
AX=B
where A=⎡⎢⎣3−1−202−13−50⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦,B=⎡⎢⎣2−13⎤⎥⎦
Here, |A|=3(0−5)+1(0+3)−2(0−6)
⇒|A|=0
Since, |A|=0
Hence, the system of equations has either infinitely many solutions (consistent) or no solution (inconsistent).
We need to find (adjA)B
C11=(−1)1+1∣∣∣2−1−50∣∣∣
⇒C11=0−5=−5
C12=(−1)1+2∣∣∣0−130∣∣∣
⇒C12=−(0+3)=−3
C13=(−1)1+3∣∣∣023−5∣∣∣
⇒C13=0−6=−6
C21=(−1)2+1∣∣∣−1−2−50∣∣∣
⇒C21=−(0−10)=10
C22=(−1)2+2∣∣∣3−230∣∣∣
⇒C22=0+6=6
C23=(−1)2+3∣∣∣3−13−5∣∣∣
⇒C23=−(−15+3)=12
C31=(−1)3+1∣∣∣−1−22−1∣∣∣
⇒C31=1+4=5
C32=(−1)3+2∣∣∣3−20−1∣∣∣
⇒C32=−(−3−0)=3
C33=(−1)3+3∣∣∣3−102∣∣∣
⇒C33=6−0=6
Hence, the co-factor matrix is C=⎡⎢⎣−5−3−610612536⎤⎥⎦
⇒adjA=CT=⎡⎢⎣−5105−363−6126⎤⎥⎦
Now, (adjA)B=⎡⎢⎣−5105−363−6126⎤⎥⎦⎡⎢⎣2−13⎤⎥⎦
=⎡⎢⎣−10−10+15−6−6+9−12−12+18⎤⎥⎦
⇒(adjA)B=⎡⎢⎣−5−3−6⎤⎥⎦
Since, (adjA)B≠O
Hence, the system of equations is inconsistent.
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