Question

Show that the equations x – 3y – 8z = -10; 3x + y – 4z = 0; 2x + 5y + 6z = 13 are consistent and have infinite sets of solution.

Answer 1

The last equivalent matrix is in echelon form. It has two non-zero rows. 

ρ(A) = ρ([A, B]) = 2 < number of unknowns 

The system is consistent and has infinitely many solutions 

The changed matrix equation is given by

x – 3y – 8z = -10 ……. (1) 

y + 2z = 3 ……. (2) 

(2) ⇒ y = 3 – 2z 

(1) ⇒ x = -10 + 3y + 8z 

⇒ x = -10 + 3(3 – 2z) + 8z 

⇒ x = -10 + 9 – 6z + 8z 

⇒ x = 2z – 1 

Let us take z = k, k ∈ R. We get y = 3 – 2k and x = 2k – 1. 

By giving diffemet values for k, we get different solutions.