We write the equations in matrix form
#((1,-2,20),(3,1,-4),(2,1,-8))*((a),(b),(c))=((1),(2),(3))#
or,
#A*x=b#
We must calculate #A^-1# the inverse of #A#
To see if the matrix is invertible, we calculate the determinant
#detA=|(1,-2,20),(3,1,-4),(2,1,-8)|#
#=1*|(1,-4),(1,-8)|+2|(3,-4),(2,-8)|+20*|(3,1),(2,1)|#
#=1(-4)+2(-16)+20(1)#
#=-4-32+20=-16#
As,
#detA!=0#, the matrix is invertible
We calculate the matrix of co-factors
#C=((|(1,-4),(1,-8)|,-|(3,-4),(2,-8)|,|(3,1),(2,1)|),(-|(-2,20),(1,-8)|,|(1,20),(2,-8)|,-|(1,-2),(2,1)|),(|(-2,20),(1,-4)|,-|(1,20),(3,-4)|,|(1,-2),(3,1)|))#
#=((-4,16,1),(4,-48,-5),(-12,64,7))#
The transpose of #C# is
#C_T=((-4,4,-12),(16,-48,64),(1,-5,7))#
The inverse is
#A^-1=(C_T)/(detA)#
#=-1/16((-4,4,-12),(16,-48,64),(1,-5,7))#
#=((1/4,-1/4,3/4),(-1,3,-4),(-1/16,5/16,-7/16))#
Therefore,
#((a),(b),(c))=((1/4,-1/4,3/4),(-1,3,-4),(-1/16,5/16,-7/16))*((1),(2),(3))#
#=((1/4-2/4+9/4),(-1+6-12),(-1/16+10/16-21/16))#
#=((2),(-7),(-3/4))#
