Question

Solve by matrix inversion method:

2x – z = 0; 5x + y = 4; y + 3z = 5

Answer 1

The given system can be written as

`[(2,0,-1),(5,1,0),(0,1,3)][(x),(y),(z)] = [(0),(4),(5)]`

AX = B

Where A = `[(2,0,-1),(5,1,0),(0,1,3)]`, X = `[(x),(y),(z)]` and B = `[(0),(4),(5)]`

|A| = `|(2,0,-1),(5,1,0),(0,1,3)|`

= 2(3 – 0) – 0(15 – 0) – 1(5 – 0)

= 2(3) – 0(15) – 1(5)

= 6 – 0 – 5

= 1

[A_ij] = `[(3,-15,5),(-|(0,-1),(1,3)|,|(2,-1),(0,3)|,-|(2,0),(0,1)|),(|(0,-1),(1,0)|,-|(2,-1),(5,0)|,|(2,0),(5,1)|)]`

`= [(3,-15,5),(-1,6,-2),(1,-5,2)]`

adj A = `["A"_"ij"]^"T" = [(3,-1,1),(-15,6,-5),(5,-2,2)]`

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

`= 1/1[(3,-1,1),(-15,6,-5),(5,-2,2)] => [(3,-1,1),(-15,6,-5),(5,-2,2)]`

X = A^-1B

`[(x),(y),(z)] = [(3,-1,1),(-15,6,-5),(5,-2,2)][(0),(4),(5)] => [(0-4+5),(0+24-25),(0-8+10)]`

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

∴ x = 1, y = -1, z = 2.