The given equations are
`3x+4y+2z=8, " " …(i)`
`2y-3z=3, " " …(ii)`
`x-2y+6z=-2." " …(iii)`
`Let A=[{:(3 " "4 " "2),(0 " "2 " -3),(1-2 6):}],X=[(x),(y),(z):}] and B=[{:(8),(3),(-2):}]` So , the given system in matrix form AX=B Now, `|A|=|(3,4,2),(0,2,-3),(1,-2,6)|=|(0,10,-16),(0,2,-3),(1,-2,6)|[R_1 rarrR_2-3R_3]`
`=1.(-30 + 32 )=ne 0 `
So, A is invertible
Therefore , the given system has a unique solutions `X=A^(-1)B`
Now , the minors of the elements of |A| are
`M_11=6,M_12= 3 ,M_13= -2 `
`M_21 = 28 , M_22 = 16 , M_23 = -10 `
`M_31 = -16 ,M_32= -9 ,M_33 =6 `
The cofactors of the elements of |A| are
`A_11 = 6, A_12 = -3, A_13 = -2 `
`A_21 =28, A_22 = 16 , A_23 = 10 `
`A_31 = -16 , A_32 = 9 , A_33 = 6 `
`therefore (adj A) = =[(6,-3,-2),(-28,16 ,10),(-16 ,9,6)]=[(6,-28,-16),(-3,16,9),(-2,10,6)]`
`rArr A^(-1)=1/(|A|)(adj A)`
`=1/2.[(6,-28,-16),(-3, 16 ,9),(-2,10,6)]=[(3,-14, -8),(-3/2,8,9/2),(-1 ,5,3)]`
`therefore X= A^-1B rArr[(x),(y),(z)]=[(3,-14, -8),(-3/2,8,9/2),(-1 ,5,3)][(8),(3),(-2)] `
`rArr [(x),(y),(z)]=[(24-42 +16),(-12+24-9),(-8+15-6)]=[(-2),(3),(1)]`
Hence x= -2 , y = 3 and z=1
