Solve system of linear equations, using matrix method.
x−y+z=4
2x+y−3z=0
x+y+z=2
Solve system of linear equations, using matrix method.
x−y+z=4
2x+y−3z=0
x+y+z=2
Simplification of given data
Given: The system of equations is
x−y+z=4
2x+y−3z=0
x+y+z=2
Writing above equation as AX=B
⎡⎢⎣1−1121−3111⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣402⎤⎥⎦
Hence, A=⎡⎢⎣1−1121−3111⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦ and B=⎡⎢⎣402⎤⎥⎦
Calculate A−1
Calculating |A|
|A|=∣∣
∣∣1−1121−3111∣∣
∣∣
=1∣∣∣1−311∣∣∣−(−1)∣∣∣2−311∣∣∣+1∣∣∣2111∣∣∣
=1(1+3)+1(2+3)+1(2−1)
=1(4)+1(5)+1(1)
=4+5+1=10
Since |A|≠0
∴ The system of equations is consistent and has a unique solution.
A=⎡⎢⎣1−1121−3111⎤⎥⎦
Calculate adj (A)
M11=[1−311]=1+3=4
M12=[2−311]=2+3=5
M13=[2111]=2−1=1
M21=[−1111]=−1−1=−2
M22=[1111]=1−1=0
M23=[1−111]=1+1=2
M31=[−111−3]=3−1=2
A=⎡⎢⎣1−1121−3111⎤⎥⎦
M32=[112−3]=−3−2=−5
M33=[1−121]=1+2=3
Thus,
adj (A)
=⎡⎢⎣A11A21A31A12A22A32A13A23A33⎤⎥⎦T
=⎡⎢⎣M11−M21M31−M12M22−M32M13−M23M33⎤⎥⎦=⎡⎢⎣422−5051−23⎤⎥⎦
adj A=⎡⎢⎣422−5051−23⎤⎥⎦
Now,
A−1=1|A|adj A
=110⎡⎢⎣422−5051−23⎤⎥⎦
Solve for the values of x,y and z
AX=B
⇒X=A−1B
⇒⎡⎢⎣xyz⎤⎥⎦=110⎡⎢⎣422−5051−23⎤⎥⎦⎡⎢⎣402⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=110⎡⎢⎣4(4)+2(0)+2(2)−5(4)+0(0)+5(2)1(4)−2(0)+3(2)⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=110⎡⎢⎣16+0+4−20+0+104−0+6⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=110⎡⎢⎣20−1010⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣2−11⎤⎥⎦
∴x=2,y=−1 and z=1
Simplification of given data
Given: The system of equations is
x−y+z=4
2x+y−3z=0
x+y+z=2
Writing above equation as AX=B
⎡⎢⎣1−1121−3111⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣402⎤⎥⎦
Hence, A=⎡⎢⎣1−1121−3111⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦ and B=⎡⎢⎣402⎤⎥⎦
Calculate A−1
Calculating |A|
|A|=∣∣
∣∣1−1121−3111∣∣
∣∣
=1∣∣∣1−311∣∣∣−(−1)∣∣∣2−311∣∣∣+1∣∣∣2111∣∣∣
=1(1+3)+1(2+3)+1(2−1)
=1(4)+1(5)+1(1)
=4+5+1=10
Since |A|≠0
∴ The system of equations is consistent and has a unique solution.
A=⎡⎢⎣1−1121−3111⎤⎥⎦
Calculate adj (A)
M11=[1−311]=1+3=4
M12=[2−311]=2+3=5
M13=[2111]=2−1=1
M21=[−1111]=−1−1=−2
M22=[1111]=1−1=0
M23=[1−111]=1+1=2
M31=[−111−3]=3−1=2
A=⎡⎢⎣1−1121−3111⎤⎥⎦
M32=[112−3]=−3−2=−5
M33=[1−121]=1+2=3
Thus,
adj (A)
=⎡⎢⎣A11A21A31A12A22A32A13A23A33⎤⎥⎦T
=⎡⎢⎣M11−M21M31−M12M22−M32M13−M23M33⎤⎥⎦=⎡⎢⎣422−5051−23⎤⎥⎦
adj A=⎡⎢⎣422−5051−23⎤⎥⎦
Now,
A−1=1|A|adj A
=110⎡⎢⎣422−5051−23⎤⎥⎦
Solve for the values of x,y and z
AX=B
⇒X=A−1B
⇒⎡⎢⎣xyz⎤⎥⎦=110⎡⎢⎣422−5051−23⎤⎥⎦⎡⎢⎣402⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=110⎡⎢⎣4(4)+2(0)+2(2)−5(4)+0(0)+5(2)1(4)−2(0)+3(2)⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=110⎡⎢⎣16+0+4−20+0+104−0+6⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=110⎡⎢⎣20−1010⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣2−11⎤⎥⎦
∴x=2,y=−1 and z=1
