AP EAMCET · Maths · Matrices
If the values \(x=\alpha, y=\beta, z=\gamma\) satisfy all the 3 equations \(x+2 y+3 z=4\), \(3 x+y+z=3\) and \(x+3 y+3 z=2\), then \(3 \alpha+\gamma=\)
- A \(\beta\)
- B \(2 \beta\)
- C \(1-2 \beta\)
- D \(2 \beta+1\)
Answer & Solution
Correct Answer
(C) \(1-2 \beta\)
Step-by-step Solution
Detailed explanation
\(y = (x+3y+3z) - (x+2y+3z) = 2 - 4 = -2\) \(\beta = -2\) \(x+2(-2)+3z=4 \Rightarrow x+3z=8\) \(3x+(-2)+z=3 \Rightarrow 3x+z=5\) \(3(3x+z)-(x+3z) = 3(5)-8 \Rightarrow 9x+3z-x-3z=15-8 \Rightarrow 8x=7 \Rightarrow x=\frac{7}{8}\) \(\alpha = \frac{7}{8}\)…
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