Question

Express the following equations in matrix form and solve them by the method of inversion.
x + 2y + 3z = 8,
2x - y + z = 1,
3x + y - 4z = 1

Answer 1

Given equations are 
x + 2y + 3z = 8,
2x - y + z = 1,
3x + y - 4z = 1

The required matrix equation is
`[(1,2,3), (2,-1,1), (3,1,-4)] [(x), (y), (z)] = [(8), (1), (1)]`

i.e      AX = B

Now |A| = `|(1,2,3), (2,-1,1), (3,1,-4)|`
             = 1 (4 -1) -2 (-8-3) + 3( 2+ 3)
             = 3 + 22 + 15
             = 40

|A| = 40 ≠ 0          A^-1 exists 

Hence, by premultiplying the equation by A^-1 we get

`(A^-1A)X = A^-1 B`
`IX = A^-1 B`
`X = A^-1B`

As A = `[(1,2,3), (2,-1,1), (3,1,-4)]`

A₁₁ = 3,   A₁₂ = 11,  A₁₃= 5
A₂₁ = 11,  A₂₂ = -13  A₂₃ = 5
A₃₁ = 5, A₃₂ = 5, A₃₃ = -5

adj A = `[(3,11,5), (11,-13,5), (5,5,-5)]`

`A^-1 = I/|A|` (adj. A)

`A^-1 = 1/40[(3,11,5), (11,-13,5), (5,5,-5)]`

X = `1/40 [(3,11,5), (11,-13,5), (5,5,-5)] [(8), (1), (1)]`

`1/40 [(24 + 11 + 5), (88 - 13 + 5), (40 + 5 - 5)] = 1/40 [(40), (80), (40)]`

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

x = 1, y = 2, z = 1