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Byju's Answer
Standard XII
Mathematics
Slope of a Line Connecting Two Points
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Question

Solve system of linear equations, using matrix method.
2x+y+z=1
x−2y−z=32
3y−5z=9


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Solution

Simplification of given data
Given: The system of equations is
2x+y+z=1
x−2y−z=32
3y−5z=9
Writing above equation as AX=B
⎡⎢⎣2111−2−103−5⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢ ⎢ ⎢⎣1329⎤⎥ ⎥ ⎥⎦
Hence, A=⎡⎢⎣2111−2−103−5⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦ and B=⎡⎢ ⎢ ⎢⎣1329⎤⎥ ⎥ ⎥⎦
Calculate A−1
Calculating |A|
|A|=∣∣ ∣∣2111−2−103−5∣∣ ∣∣
=2∣∣∣−2−13−5∣∣∣−1∣∣∣1−10−5∣∣∣+1∣∣∣1−203∣∣∣
=2(10+3)−1(−5−0)+1(3−0)
=26+5+3=34
Thus |A|≠0
∴ The system of equations is consistent and has a unique solution.
A=⎡⎢⎣2111−2−103−5⎤⎥⎦
Calculate adj A
M11=[−2−13−5]=10+3=13
M12=[1−10−5]=−5+0=−5
M13=[1−203]=3+0=3
M21=[113−5]=−5−3=−8
M22=[210−5]=−10−0=−10
M23=[2103]=6−0=6
M31=[11−2−1]=−1+2=1
M32=[211−1]=−2−1=−3
M33=[211−2]=−4−1=−5
Thus,
adj (A)=⎡⎢⎣A11A12A13A21A22A23A31A32A33⎤⎥⎦T
=⎡⎢⎣M11−M21M31−M12M22−M32M13−M23M33⎤⎥⎦=⎡⎢⎣13815−1033−6−5⎤⎥⎦
A−1=1|A|adj A
=134⎡⎢⎣13815−1033−6−5⎤⎥⎦
Solve for the values of x,y and z
AX=B⇒X=A−1B
⇒⎡⎢⎣xyz⎤⎥⎦=134⎡⎢⎣13815−1033−6−5⎤⎥⎦⎡⎢ ⎢ ⎢⎣1329⎤⎥ ⎥ ⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=134⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣13(1)+8(32)+1(9)5(1)+(−10)(32)+3(9)3(1)+(−6)(32)+(−5)(9)⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=134⎡⎢⎣13+12+95−15+273−9−45⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=134⎡⎢⎣3417−51⎤⎥⎦⇒⎡⎢⎣xyz⎤⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣112−32⎤⎥ ⎥ ⎥ ⎥ ⎥⎦
∴x=1,y=12 and z=−32


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