For, suppose the system of linear equations x - y + z = 5 2 x + 2 y + z = 8 3 x - y + 4 z = has infinitely many solutions. Then and are the roots of [2023]

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Solution

Q.
For $α, β \in R,$ suppose the system of linear equations

$x - y + z = 5$

$2 x + 2 y + α z = 8$

$3 x - y + 4 z = β$

has infinitely many solutions. Then $α$ and $β$ are the roots of [2023]

1 $x^{2} - 18 x + 56 = 0$

2 $x^{2} + 14 x + 24 = 0$

3 $x^{2} - 10 x + 16 = 0$

4 $x^{2} + 18 x + 56 = 0$

Ans.
(1)

Given, $x - y + z = 5, 2 x + 2 y + α z = 8, 3 x - y + 4 z = β$

It can be written as $A X = B$

Where, $A = [\begin{matrix} 1 & - 1 & 1 \\ 2 & 2 & α \\ 3 & - 1 & 4 \end{matrix}], B = [\begin{matrix} 5 \\ 8 \\ β \end{matrix}], X = [\begin{matrix} x \\ y \\ z \end{matrix}]$

Condition for infinite solutions $| A | = 0 and (Adj A) B = 0$

$∴$ $| A | = 8 + α + 8 - 3 α - 8 = 0 \Rightarrow α = 4$

$Adj A = [\begin{matrix} 12 & 3 & - 6 \\ 4 & 1 & - 2 \\ - 8 & - 2 & 4 \end{matrix}]$

$(Adj A) B = [\begin{matrix} 12 & 3 & - 6 \\ 4 & 1 & - 2 \\ - 8 & - 2 & 4 \end{matrix}] [\begin{matrix} 5 \\ 8 \\ β \end{matrix}] = 0$

$\Rightarrow 60 + 24 - 6 β = 0 \Rightarrow β = 14$ $∴$ $α + β = 4 + 14 = 18, α β = 4 \times 14 = 56$

$∴$ $Required equation is given by x^{2} - 18 x + 56 = 0$

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