The system of equations $x + y - z = 1, 3x + y-2z = 3, x-y+λz = 1$ has infinite number of solutions if is equal to

0

The correct answer is Option (1) → 0

Given system:

(1) x + y − z = 1
(2) 3x + y − 2z = 3
(3) x − y + λz = 1

Step 1: Solve equations (1) and (2) using elimination

From (1): x + y − z = 1 ⇒ x = 1 − y + z   ...(i)

Substitute (i) into (2):

3(1 − y + z) + y − 2z = 3
⇒ 3 − 3y + 3z + y − 2z = 3
⇒ 3 − 2y + z = 3
⇒ −2y + z = 0 ⇒ z = 2y   ...(ii)

Now substitute (i) and (ii) into (3):

x − y + λz = 1
From (i): x = 1 − y + z
So, 1 − y + z − y + λz = 1 ⇒ 1 − 2y + (1 + λ)z = 1

Substitute z = 2y:
⇒ 1 − 2y + (1 + λ)(2y) = 1
⇒ 1 − 2y + 2y(1 + λ) = 1 ⇒ 1 − 2y + 2y + 2λy = 1
⇒ 1 + 2λy = 1
⇒ 2λy = 0

For infinite solutions, y should be free (not fixed), so its coefficient must vanish:

2λ = 0 ⇒ λ = 0