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Question

# Solve the following system of linear equations, using matrix method 2x+3y+3z=5,x−2y+z=−4,3x−y−2z=3

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Solution

## The given system can be written as AX=B, where A=⎡⎢⎣2331−213−1−2⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦ and B=⎡⎢⎣5−43⎤⎥⎦ Here, |A|=⎡⎢⎣2331−213−1−2⎤⎥⎦=2(4+1)−3(−2−3))+3(−1+6) Thus, A is non-singular, Therefore, its inverse exists. Therefore, the given system is consistent and has a unique solution given by X=A−1B. Cofactors of A are A11=4+1=5,A12=−(−2−3)=−5,A13=(−1+6)=5A21=−(−6+3)=3,A22(−4−9)=−13,A23=−(−2−9)=11A31=3+6=9,A32=−(2−3)=1,A33=−4−3=−7 adj(A)=⎡⎢⎣5553−131191−7⎤⎥⎦T=⎡⎢⎣5395−131511−7⎤⎥⎦ ∴A−1=1|A|(adj A)=140⎡⎢⎣5395−131511−7⎤⎥⎦∴ A−1=1|A|(adj A)=140⎡⎢⎣5395−131511−7⎤⎥⎦Now,X=A−1B ⇒⎡⎢⎣xyz⎤⎥⎦=140⎡⎢⎣5395−131511−7⎤⎥⎦⎡⎢⎣5−43⎤⎥⎦=140⎡⎢⎣25−12+2725+52+325−44−21⎤⎥⎦=140⎡⎢⎣4080−40⎤⎥⎦=⎡⎢⎣12−1⎤⎥⎦ Hence, x=1, y=2 and z=-1.

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