Question

If the system of equations α x + y + z = 5, x + 2 y + 3 z = 4, x + 3 y +5 z = β has infinitely many solutions, then the ordered pair (α, β) is equal to:

• A. (1, –3)
• B. (–1, 3)
• C. (1, 3)
• D. (–1, –3)

Answer 1

Answer: (C)

(1, 3)

Answer 2

The correct answer is (C) : (1, 3)
Given system of equations
αx + y + z = 5
x + 2y + 3z = 4, has infinite solution
x+3y+5z = β

∴ Δ = $\begin{vmatrix} \alpha & 1 & 1 \\\ 1 & 2 & 3 \\\ 1 & 3 & 5 \\\ \end{vmatrix}$ = 0
⇒ α(1) - 1(2) + 1(1) = 0
⇒ α = 1
and

$Δ_1$ = $\begin{vmatrix} 5 & 1 & 1 \\\ 4 & 2 & 3 \\\ \beta & 3 & 5 \\\ \end{vmatrix}$ = 0
⇒ 5(1) - 1(20 - 3β) + 1(12 - 2β) = 0
⇒ β = 3
And

$Δ_2$ = $\begin{vmatrix} 1 & 5 & 1 \\\ 1 & 4 & 3 \\\ 1 & \beta & 5 \\\ \end{vmatrix}$= 0
⇒ (20 - 3β) - 5(2) + 1(β - 4) = 0
⇒ -2β + 6 = 0
⇒ β = 3
Similarly,
⇒ $Δ_3$ = $\begin{vmatrix} 1 & 1 & 5 \\\ 1 & 2 & 4 \\\ 1 & 3 & \beta \\\ \end{vmatrix}$= 0
⇒ β = 3
∴ ( α, β ) = ( 1,3