If the system of linear equations x - 2 y + z = - 4 , 2 x + y + 3 z = 5 , 3 x - y + z = 3 has infinitely many solutions, then is equal to [2024]

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Solution

Q.
If the system of linear equations $x - 2 y + z = - 4,$ $2 x + α y + 3 z = 5,$ $3 x - y + β z = 3$ has infinitely many solutions, then $12 α + 13 β$ is equal to [2024]

1 54

2 64

3 58

4 60

Ans.
(3)

Given, $x - 2 y + z = - 4$

$2 x + α y + 3 z = 5, 3 x - y + β z = 3$

$Δ =$ $| \begin{matrix} 1 & - 2 & 1 \\ 2 & α & 3 \\ 3 & - 1 & β \end{matrix} |$ $= 0$

$\Rightarrow α β - 18 - 2 - 3 α + 3 + 4 β = 0 = α β - 3 α + 4 β - 17 = 0$

$Δ y =$ $| \begin{matrix} 1 & - 4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & β \end{matrix} |$ $= 0$

$Δ y = 1 (5 β - 9) + 4 (2 β - 9) + 1 (6 - 15)$

$= 5 β - 9 + 8 β - 36 + 6 - 15$

$\Rightarrow 13 β - 54 = 0 \Rightarrow 13 β = 54 \Rightarrow β = \frac{54}{13}$

$Δ z =$ $| \begin{matrix} 1 & - 2 & - 4 \\ 2 & α & 5 \\ 3 & - 1 & 3 \end{matrix} |$ $= 0$

$Δ z = 1 (3 α + 5) + 2 (6 - 15) - 4 (- 2 - 3 α)$

$\Rightarrow 3 α + 5 + 12 - 30 + 8 + 12 α$

$\Rightarrow 15 α - 5 = 0 \Rightarrow 15 α = 5 \Rightarrow α = \frac{5}{15} = \frac{1}{3}$

So, $12 α + 13 β = 12 \times \frac{1}{3} + 54 \Rightarrow 4 + 54 = 58$

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