How do you solve #a-2b+20c=1#, #3a+b-4c=2# and #2a+b-8c=3# using matrices?

1 Answer
Jun 6, 2017

Answer:

The solution is #((a),(b),(c))=((2),(-7),(-3/4))#

Explanation:

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))#