If the system of equations

$\alpha$x + y + z = 5, x + 2y + 3z = 4, x + 3y + 5z = $\beta$

has infinitely many solutions, then the ordered pair ($\alpha$, $\beta$) is equal to :

<p>Given system of equations</p> <p>$\alpha x + y + z = 5$</p> <p>$x + 2y + 3z = 4$, has infinite solution</p> <p>$x + 3y + 5z = \beta$</p> <p>$\therefore$ $$\Delta = \left| {\matrix{ \alpha & 1 & 1 \cr 1 & 2 & 3 \cr 1 & 3 & 5 \cr } } \right| = 0 \Rightarrow \alpha (1) - 1(2) + 1(1) = 0$$</p> <p>$\Rightarrow \alpha = 1$</p> <p>and $${\Delta _1} = \left| {\matrix{ 5 & 1 & 1 \cr 4 & 2 & 3 \cr \beta & 3 & 5 \cr } } \right| = 0$$</p> <p>$\Rightarrow 5(1) - 1(20 - 3\beta ) + 1(12 - 2\beta ) = 0$</p> <p>$\Rightarrow \beta = 3$</p> <p>and $${\Delta _2} = \left| {\matrix{ 1 & 5 & 1 \cr 1 & 4 & 3 \cr 1 & \beta & 5 \cr } } \right| = 0 \Rightarrow (20 - 3\beta ) - 5(2) + 1(\beta - 4) = 0$$</p> <p>$\Rightarrow - 2\beta + 6 = 0$</p> <p>$\Rightarrow \beta = 3$</p> <p>Similarly,</p> <p>$$ \Rightarrow {\Delta _3} = \left| {\matrix{ 1 & 1 & 5 \cr 1 & 2 & 4 \cr 1 & 3 & \beta \cr } } \right| = 0 \Rightarrow \beta = 3$$</p> <p>$\therefore$ ($\alpha$, $\beta$) = (1, 3)</p>