Question

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

Answer 1

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))`