Question

Using matrices, solve the following system of linear equations :

x + 2y − 3z = −4
2x + 3y + 2z = 2
3x − 3y − 4z = 11

Answer 1

The system of equations can be written in the form AX = B, where

A `= [(1,2,-3),(2,3,2),(3,-3,-4)],` X`=[("x"),("y"),("z")]` and B =`[(-4),(2),(11)]`

|A| = 1 (-12+6) - 2 (-8 - 6) - 3 (-6 - 9) = 67 ≠ 0

Therefore, A is non singular and so its inverse exists.

A₁₁ = -6, A₁₂ = 14, A₁₃ = -15

A₂₁ = 17, A₂₂ = 5, A₂₃ = 9

A₃₁ = 13, A₃₂ = -8, A₃₃ = -1

 Therefore, `"A"^-1 = 1/67[(-6,17,13),(14,5,-8),(-15,9,-1)]`

So X = A^-1 B `=1/67[(-6,17,13),(14,5,-8),(-15,9,-1)][(-4),(2),(11)]`

i.e. `[("x"),("y"),("z")]=1/67[(201),(-134),(67)]=[(3),(-2),(1)]`

Hence, x = 3, y = -2 and z = 1