Question

Solve by matrix inversion method:

x – y + 2z = 3; 2x + z = 1; 3x + 2y + z = 4

Answer 1

The given system can be written as

`[(1,-1,2),(2,0,1),(3,2,1)][(x),(y),(z)] = [(3),(1),(4)]`

AX = B

Where A = `[(1,-1,2),(2,0,1),(3,2,1)]`, X = `[(x),(y),(z)]` and B = `[(3),(1),(4)]`

|A| = `|(1,-1,2),(2,0,1),(3,2,1)|`

= 1(0 – 2) – (-1)(2 – 3) + 2(4 – 0)

= -2 – (-1)(-1) + 2(4)

= -2 – 1 + 8

= 5

[A_ij] = `[(-2,-(-1),4),(-|(-1,2),(2,1)|,|(1,2),(3,1)|,-|(1,-1),(3,2)|),(|(-1,2),(0,1)|,-|(1,2),(2,1)|,|(1,-1),(2,0)|)]`

`= [(-2,1,4),(-(-1-4),(1-6),-(2+3)),((-1-0),-(1-4),(0+2))]`

`=> [(-2,1,4),(5,-5,-5),(-1,3,2)]`

adj A = `["A"_"ij"]^"T" = [(-2,5,-1),(1,-5,3),(4,-5,2)]`

`"A"^-1 = 1/|"A"|`(adj A)

`= 1/(5)[(-2,5,-1),(1,-5,3),(4,-5,2)]`

X = A^-1B

`= 1/(5)[(-2,5,-1),(1,-5,3),(4,-5,2)][(3),(1),(4)]`

`=> 1/(5)[(-6+5-4),(3-5+12),(12-5+8)]`

`=> 1/(5)[(-5),(10),(15)]`

`[(x),(y),(z)] = [(-1),(2),(3)]`

∴ x = -1, y = 2, z = 3.