Question:medium
Let \( \alpha, \beta \in \mathbb{R} \) be such that the system of linear equations}
\[
x + 2y + z = 5
\]
\[
2x + y + \alpha z = 5
\]
\[
8x + 4y + \beta z = 18
\]
has no solution. Then \( \frac{\beta}{\alpha} \) is equal to:
Let \( \alpha, \beta \in \mathbb{R} \) be such that the system of linear equations}
\[
x + 2y + z = 5
\]
\[
2x + y + \alpha z = 5
\]
\[
8x + 4y + \beta z = 18
\]
has no solution. Then \( \frac{\beta}{\alpha} \) is equal to:
Updated On: Apr 15, 2026
- \(-4\)
- \(4\)
- \(8\)
- \(-8\)
Show Solution
The Correct Option is C
Solution and Explanation
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