Solve system of linear equations, using matrix method.
2x+3y+3z=5
x−2y+z=−4
3x−y−2z=3
Solve system of linear equations, using matrix method.
2x+3y+3z=5
x−2y+z=−4
3x−y−2z=3
Simplification of given data
Given: The system of equations is
2x+3y+3z=5
x−2y+z=−4
3x−y−2z=3
Writing above equation as AX=B
⎡⎢⎣2331−213−1−2⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣5−43⎤⎥⎦
Hence, A=⎡⎢⎣2331−213−1−2⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦ & B=⎡⎢⎣5−43⎤⎥⎦
Calculate A−1
Calculating |A|
|A|=∣∣
∣∣2331−213−1−2∣∣
∣∣
=2∣∣∣−21−1−2∣∣∣−3∣∣∣113−2∣∣∣+3∣∣∣1−23−1∣∣∣
=2(4+1)−3(−2−3)+3(−1+6)
=2(5)−3(−5)+3(5)
=10+15+15=40
Since |A|≠0
∴ The system of equations is consistent & has a unique solution.
A=⎡⎢⎣2331−213−1−2⎤⎥⎦
Calculate adj (A)
M11=[−21−1−2]=4+1=5
M12=[113−2]=−2−3=−5
M13=[1−23−1]=−1+6=5
M21=[33−1−2]=−6+3=−3
M22=[233−2]=−4−9=−13
M23=[233−1]=−2−9=−11
M31=[33−21]=3+6=9
M32=[2311]=2−3=−1
M33=[231−2]=−4−3=−7
Thus,
adj (A)=⎡⎢⎣A11A12A13A21A22A23A31A32A33⎤⎥⎦′
=⎡⎢⎣A11A21A31A12A22A32A13A23A33⎤⎥⎦′
=⎡⎢⎣M11−M21M31−M12M22−M32M13−M23M33⎤⎥⎦=⎡⎢⎣5395−131511−7⎤⎥⎦
adj A=⎡⎢⎣5395−131511−7⎤⎥⎦
Now,
A−1=1|A|adj A
=140⎡⎢⎣5395−131511−7⎤⎥⎦
Solve for the values of X,Y & Z
AX=B
⇒X=A−1B
⇒⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣5395−131511−7⎤⎥⎦⎡⎢⎣5−43⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣5(5)+3(−4)+9(3)5(5)+(−13)(−4)+1(3)5(5)+11(−4)+(−7)(3)⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣25−12+2725+52+325−44−21⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣4080−40⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣12−1⎤⎥⎦
∴x=1 & y=2 & z=−1
Simplification of given data
Given: The system of equations is
2x+3y+3z=5
x−2y+z=−4
3x−y−2z=3
Writing above equation as AX=B
⎡⎢⎣2331−213−1−2⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣5−43⎤⎥⎦
Hence, A=⎡⎢⎣2331−213−1−2⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦ & B=⎡⎢⎣5−43⎤⎥⎦
Calculate A−1
Calculating |A|
|A|=∣∣
∣∣2331−213−1−2∣∣
∣∣
=2∣∣∣−21−1−2∣∣∣−3∣∣∣113−2∣∣∣+3∣∣∣1−23−1∣∣∣
=2(4+1)−3(−2−3)+3(−1+6)
=2(5)−3(−5)+3(5)
=10+15+15=40
Since |A|≠0
∴ The system of equations is consistent & has a unique solution.
A=⎡⎢⎣2331−213−1−2⎤⎥⎦
Calculate adj (A)
M11=[−21−1−2]=4+1=5
M12=[113−2]=−2−3=−5
M13=[1−23−1]=−1+6=5
M21=[33−1−2]=−6+3=−3
M22=[233−2]=−4−9=−13
M23=[233−1]=−2−9=−11
M31=[33−21]=3+6=9
M32=[2311]=2−3=−1
M33=[231−2]=−4−3=−7
Thus,
adj (A)=⎡⎢⎣A11A12A13A21A22A23A31A32A33⎤⎥⎦′
=⎡⎢⎣A11A21A31A12A22A32A13A23A33⎤⎥⎦′
=⎡⎢⎣M11−M21M31−M12M22−M32M13−M23M33⎤⎥⎦=⎡⎢⎣5395−131511−7⎤⎥⎦
adj A=⎡⎢⎣5395−131511−7⎤⎥⎦
Now,
A−1=1|A|adj A
=140⎡⎢⎣5395−131511−7⎤⎥⎦
Solve for the values of X,Y & Z
AX=B
⇒X=A−1B
⇒⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣5395−131511−7⎤⎥⎦⎡⎢⎣5−43⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣5(5)+3(−4)+9(3)5(5)+(−13)(−4)+1(3)5(5)+11(−4)+(−7)(3)⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣25−12+2725+52+325−44−21⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣4080−40⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣12−1⎤⎥⎦
∴x=1 & y=2 & z=−1
