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))
