If the system of equations

$$ \begin{aligned} &x+y+z=6 \\ &2 x+5 y+\alpha z=\beta \\ &x+2 y+3 z=14 \end{aligned} $$

has infinitely many solutions, then $\alpha+\beta$ is equal to

<p>Given,</p> <p>$x + y + z = 6$ ...... (1)</p> <p>$2x + 5y + \alpha z = \beta$ ..... (2)</p> <p>$x + 2y + 3z = 14$ ...... (3)</p> <p>System of equation have infinite many solutions.</p> <p>$\therefore$ ${\Delta _x} = {\Delta _y} = {\Delta _z} = 0$ and $\Delta = 0$</p> <p>Now, $$\Delta = \left| {\matrix{ 1 & 1 & 1 \cr 2 & 5 & \alpha \cr 1 & 2 & 3 \cr } } \right| = 0$$</p> <p>${C_1} \to {C_1} - {C_3}$</p> <p>${C_2} \to {C_2} - {C_3}$</p> <p>$$ \Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr {2 - \alpha } & {5 - \alpha } & \alpha \cr { - 2} & { - 1} & 3 \cr } } \right| = 0$$</p> <p>$\Rightarrow - 2 + \alpha + 10 - 2\alpha = 0$</p> <p>$\Rightarrow 8 - \alpha = 0$</p> <p>$\Rightarrow \alpha = 8$</p> <p>Now, $x + y + z = 6$</p> <p>$2x + 5y + 8z = \beta$</p> <p>$x + 2y + 3z = 14$</p> <p>$\therefore$ $${\Delta _x} = \left| {\matrix{ 6 & 1 & 1 \cr \beta & 5 & 8 \cr {14} & 2 & 3 \cr } } \right| = 0$$</p> <p>${C_1} \to {C_1} - 6{C_3}$</p> <p>${C_2} \to {C_2} - {C_3}$</p> <p>$$ \Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr {\beta - 48} & { - 3} & 8 \cr { - 4} & { - 1} & 3 \cr } } \right| = 0$$</p> <p>$\Rightarrow - \beta + 48 - 12 = 0$</p> <p>$\Rightarrow \beta = 36$</p> <p>$\therefore$ $\alpha + \beta = 8 + 36 = 44$</p>