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Question
Solve the system of equations, using matrix method2x+3y+3z=5,x−2y+z=−4,3x−y−2z=3
2x+3y+3z=5,x−2y+z=−4,3x−y−2z=3
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Solution
Given system of equations2x+3y+3z=5
x−2y+z=−4
3x−y−2z=3This can be written as AX=Bwhere A=⎡⎢⎣2331−213−1−2⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦,B=⎡⎢⎣5−43⎤⎥⎦
Here, |A|=2(4+1)−3(−2−3)+3(−1+6)⇒|A|=10+15+15=40Since, |A|≠0Hence, the system of equations is consistent and has a unique solution given by X==A−1B
A−1=adjA|A| and adjA=CT
C11=(−1)1+1∣∣∣−21−1−2∣∣∣⇒C11=4+1=5
C12=(−1)1+2∣∣∣113−2∣∣∣⇒C12=−(−2−3)=5
C13=(−1)1+3∣∣∣1−23−1∣∣∣⇒C13=−1+6=5
C21=(−1)2+1∣∣∣33−1−2∣∣∣⇒C21=−(−6+3)=3
C22=(−1)2+2∣∣∣233−2∣∣∣⇒C22=−4−9=−13
C23=(−1)2+3∣∣∣233−1∣∣∣⇒C23=−(−2−9)=11
C31=(−1)3+1∣∣∣33−21∣∣∣⇒C31=3+6=9
C32=(−1)3+2∣∣∣2311∣∣∣⇒C32=−(2−3)=1
C33=(−1)3+3∣∣∣231−2∣∣∣⇒C33=−4−3=−7
Hence, the co-factor matrix is C=⎡⎢⎣5553−131191−7⎤⎥⎦
⇒adjA=CT=⎡⎢⎣5395−131511−7⎤⎥⎦
⇒A−1=adjA|A|=140⎡⎢⎣5395−131511−7⎤⎥⎦
Solution is given by ⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣5395−131511−7⎤⎥⎦⎡⎢⎣5−43⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣25−12+2725+52+325−44−21⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣4080−40⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣12−1⎤⎥⎦
Hence, x=1,y=2,z=−1
2x+3y+3z=5
x−2y+z=−4
x−2y+z=−4
3x−y−2z=3
This can be written as
AX=B
where A=⎡⎢⎣2331−213−1−2⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦,B=⎡⎢⎣5−43⎤⎥⎦
Here, |A|=2(4+1)−3(−2−3)+3(−1+6)
⇒|A|=10+15+15=40
Since, |A|≠0
Hence, the system of equations is consistent and has a unique solution given by X==A−1B
A−1=adjA|A| and adjA=CT
C11=(−1)1+1∣∣∣−21−1−2∣∣∣
⇒C11=4+1=5
C12=(−1)1+2∣∣∣113−2∣∣∣
⇒C12=−(−2−3)=5
C13=(−1)1+3∣∣∣1−23−1∣∣∣
⇒C13=−1+6=5
C21=(−1)2+1∣∣∣33−1−2∣∣∣
⇒C21=−(−6+3)=3
C22=(−1)2+2∣∣∣233−2∣∣∣
⇒C22=−4−9=−13
C23=(−1)2+3∣∣∣233−1∣∣∣
⇒C23=−(−2−9)=11
C31=(−1)3+1∣∣∣33−21∣∣∣
⇒C31=3+6=9
C32=(−1)3+2∣∣∣2311∣∣∣
⇒C32=−(2−3)=1
C33=(−1)3+3∣∣∣231−2∣∣∣
⇒C33=−4−3=−7
Hence, the co-factor matrix is C=⎡⎢⎣5553−131191−7⎤⎥⎦
⇒adjA=CT=⎡⎢⎣5395−131511−7⎤⎥⎦
⇒A−1=adjA|A|=140⎡⎢⎣5395−131511−7⎤⎥⎦
Solution is given by
⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣5395−131511−7⎤⎥⎦⎡⎢⎣5−43⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣25−12+2725+52+325−44−21⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣4080−40⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣12−1⎤⎥⎦
Hence, x=1,y=2,z=−1
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