Question

How do you solve `x+4y-z =3`, `2x+3y-2z =1` and `-x+2y+3z =7` using matrices?

Answer 1

The answer is `((x),(y),(z))=((1),(1),(2))`

Explanation:

We rewrite the equations with a matrix

`((1,4,-1),(2,3,-2),(-1,2,3))((x),(y),(z))=((3),(1),(7))`

Let matrix `A=((1,4,-1),(2,3,-2),(-1,2,3))`

We must calculate the inverse of `A`, `A^-1`

We calculate the determinant of `A`

`detA=|(1,4,-1),(2,3,-2),(-1,2,3)|`

`=1*|(3,-2),(2,3)|-4*|(2,-2),(-1,3)|-1*|(2,3),(-1,2)|`

`=1*(13)-4(4)-1*(7)`

`=13-16-7=-10`

`detA!=0`, so the matrix is invertible

The matrix of cofactor is

`C=((|(3,-2),(2,3)|,-|(2,-2),(-1,3)|,|(2,3),(-1,2)|),(-|(4,-1),(2,3)|,|(1,-1),(-1,3)|,-|(1,4),(-1,2)|),(|(4,-1),(3,-2)|,-|(1,-1),(2,-2)|,|(1,4),(2,3)|))`

`=((13,-4,7),(-14,2,-6),(-5,0,-5))`

We calculate the transpose of `C`

`C^T=((13,-14,-5),(-4,2,0),(7,-6,-5))`

`A^-1=C^T/detA=-1/10((13,-14,-5),(-4,2,0),(7,-6,-5))`

and

`((x),(y),(z))=-1/10((13,-14,-5),(-4,2,0),(7,-6,-5))*((3),(1),(7))`

`=-1/10((-10), (-10),(-20))=((1),(1),(2))`