Question

If the system of equations αx + y + z = 5, x + 2y + 3z = 4, x + 3y +5z = 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

The Correct Option is C

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