Question

If the system of linear equations

2x + y – z = 7

x – 3y + 2z = 1

x + 4y + δz = k, where δ, k ∈ R has infinitely many solutions, then δ + k is equal to ______.

पर्याय

• –3

• 3

• 6

• 9

Answer 1

If the system of linear equations

2x + y – z = 7

x – 3y + 2z = 1

x + 4y + δz = k, where δ, k ∈ R has infinitely many solutions, then δ + k is equal to 3.

Explanation:

Given: A system of linear equation has infinitely many solution.

2x + y – z = 7

x – 3y + 2z = 1

x + 4y + δz = k

As we know, if a system of linear equations

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₃z = d₂

a₃x + b₂y + c₃z = d₃

has infinitely many solutions then, D = D₁ = D₂ = D₃ = 0

Where D = `|(a_1, b_1, c_1),(a_2, b_2, c_2),(a_3, b_3, c_3)|`, D₁ = `|(d_1, b_1, c_1),(d_2, b_2, c_2),(d_3, b_3, c_3)|`,

D₂ = `|(a_1, d_1, c_1),(a_2, d_2, c_2),(a_3, d_3, c_3)|` and D₃ = `|(a_1, b_1, d_1),(a_2, b_2, d_2),(a_3, b_3, d_3)|`

So, D = `|(2, 1, -1),(1, -3, 2),(1, 4, δ)|` = 0

⇒ 2(–3δ – 8) – 1(δ – 2) – 1(4 + 3) = 0

⇒ –6δ – 16 – δ + 2 – 7 = 0

⇒ 7δ = –21

⇒ δ = –3

Also, D₃ = `|(2, 1, 7),(1, -3, 1),(1, 4, k)|` = 0

⇒ 2(–3k – 4) – 1(k – 1) + 7(4 + 3) = 0

⇒ –6k – 8 – k + 1 + 49 = 0

⇒ 7k = 42

⇒ k = 6

⇒ δ + k = –3 + 6 = 3