Let be such that the system of linear equations x + 2y + z = 5; ; has no solution. Then is equal to : [2026]

Question

Let $α, β \in R$ be such that the system of linear equations x + 2y + z = 5; $2 x + y + α z = 5$ ; $8 x + 4 y + β z = 18$ has no solution. Then $\frac{β}{α}$ is equal to : [2026]

Smart Achievers · Accepted Answer

(2)

Given system of linear equations: x + 2y + z = 5, $2 x + y + α z = 5$ , $8 x + 4 y + β z = 18$

Since the given system of linear equations has no solution, so, $△ = 0$

$\Rightarrow △ = | \begin{matrix} 1 & 2 & 1 \\ 2 & 1 & α \\ 8 & 4 & β \end{matrix} | = 12 α - 3 β = 0$

$\Rightarrow β = 4 α$

$△_{z} = | \begin{matrix} 1 & 2 & 5 \\ 2 & 1 & 5 \\ 8 & 4 & 18 \end{matrix} | = 6 \neq 0$

$\Rightarrow z = \frac{△_{z}}{△} \Rightarrow z = \frac{6}{0}$ = not defined

Hence, $\frac{β}{α} = 4$

Solution

Q.
Let $α, β \in R$ be such that the system of linear equations x + 2y + z = 5; $2 x + y + α z = 5$ ; $8 x + 4 y + β z = 18$ has no solution. Then $\frac{β}{α}$ is equal to : [2026]

1 –4

2 4

3 8

4 –8

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Solution

Q. Let α,β∈R be such that the system of linear equations x + 2y + z = 5; 2x+y+αz=5; 8x+4y+βz=18 has no solution. Then βα is equal to : [2026]

1 –4

2 4

3 8

4 –8

Ans. (2) Given system of linear equations: x + 2y + z = 5, 2x+y+αz=5, 8x+4y+βz=18 Since the given system of linear equations has no solution, so, △=0 ⇒ △=|12121α84β|=12α–3β=0 ⇒ β=4α △z=|1252158418|=6≠0 ⇒ z=△z△ ⇒ z=60 = not defined Hence, βα=4

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Q.
Let $α, β \in R$ be such that the system of linear equations x + 2y + z = 5; $2 x + y + α z = 5$ ; $8 x + 4 y + β z = 18$ has no solution. Then $\frac{β}{α}$ is equal to : [2026]