Question

Consider the system of linear equations :

– x + y + 2z = 0 

3x – ay + 5z = 1 

2x – 2y – az = 7

Let S₁ be the set of all a ∈ R for which the system is inconsistent and S₂ be the set of all a ∈ R for which the system has infinitely many solutions. If n(S₁) and n(S₂) denote the number of elements in S₁ and S₂ respectively, then

(1) n(S₁) = 2, n(S₂) = 2 

(2) n(S₁) = 1, n(S₂) = 0 

(3) n(S₁) = 2, n(S₂) = 0 

(4) n(S₁) = 0, n(S₂) = 2

Answer 1

Option : (3)

Δ = \(\begin{vmatrix} -1& 1 & 2 \\[0.3em] 3 & -a & 5 \\[0.3em] 2 & -2&-a \end{vmatrix}\)

= -1(a² + 10)-1(-3a-10)+2(-6+2a)

= -a²-10+3a+10-12+4a

Δ = -a²+7a-12

Δ = -[a²+7a-12]

Δ = - [(a-3)(a-4)]

Δ = \(\begin{vmatrix} 0& 1 & 2 \\[0.3em] 1 & -a & 5 \\[0.3em] 7 & -2&-a \end{vmatrix}\)

= 0 – 1 (– a – 35) + 2( – 2 + 7a)

⇒ a + 35 - 4 +14a

15a + 31 

Now,

Δ₁ = 15a + 31 

For inconsistent : Δ = 0 

∴ a = 3, a = 4 

and for a = 3 and 4 Δ₁ ≠ 0 

n(S₁) = 2 

For infinite solution : Δ = 0 

and Δ₁ = Δ₂ = Δ₃ = 0 

Not possible 

∴ n(S₂) = 0