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Solve the system of equations `x+2y+3z=1,2x+3y+2z=2 and 3x+3y+4z=1` with the help of matrix inversion. |
Answer» we have `x+2y+3z=1,2x+3y+2z=2 and 3x+3y+4z=1` the given system of equation in the matrix form are written as below. `[(1,2,3),(2,3,2),(3,3,4)]=[(,1),(,2),(,1)]` ` AX=B` `rArr X=A^(-1)B` where` " " A=[(1,2,3),(2,3,2),(3,3,4)],X=[(,x),(,y),(,z)]and B= [(,1),(,2),(,1)]` `|A|=1(12-6)-2(8-6)+3(6-9)` ` =6-4-9=-7!=0` ` therefore A^(-1)` exixts and has unique solution. Let C be the matrix of confactor of elements in `|A|.` Now, confactor along `R_(1)=6,-2,-3` confactor along `R_(2)=1,-5,3` and confacators along `R_(3)=-5,4,-1` `therefore " " C=[(6,-2,-3),(1,-5,3),(-5,4,-1)]` `therefore" " A=C^(T)` `rArr " " adjA=[(6,-2,-3),(1,-5,3),(-5,4,-1)]=[(6,1,-5),(-2,-5,4),(-3,3,-1)]` `rArr" " A^(-1)=(adjA)/(|A|)=-(1)/(7)[(6,1,-5),(-2,-5,4),(-3,3,-1)]` `=[(-(6)/(7),-(1)/(7),(5)/(7)),((2)/(7),(5)/(7),-(4)/(7)),((3)/(7),-(3)/(7),(1)/(7))]` from Eq. (i),`X=A^(-1)B` `rArr " " [(,x),(,y),(,z)]=[(-(6)/(7),-(1)/(7),(5)/(7)),((2)/(7),(5)/(7),-(4)/(7)),((3)/(7),-(3)/(7),(1)/(7))][(,1),(,2),(,1)]=[(-(3)/(7)),((8)/(7)),(-(2)/(7))]` Hence `x=-(3)/(7),y=(8)/(7) and z=-(2)/(7)` is the required solution. |
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