If the system of equations + y + z = 5, x + 2y + 3z = 4, x + 3y +...

We know that, for infinitely many solutions,

$\Delta = 0 = {\Delta}_{x} = {\Delta}_{y} = {\Delta}_{z}$

So, \Delta = \left|\begin{cases}\alpha & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5\end{cases}\right| = 0

$⇒ \alpha\left(10 - 9\right) - 1\left(5 - 3\right) + 1\left(3 - 2\right) = 0$

$⇒ \alpha - 2 + 1 = 0$

$⇒ \alpha = 1$

Now checking for {\Delta}_{x} = \left|\begin{cases}5 & 1 & 1 \\ 4 & 2 & 3 \\ \beta & 3 & 5\end{cases}\right| = 0

$⇒ 5\left(10 - 9\right) - 1$\left(20 - 3\beta\right) + 1\left(12 - 2\beta\right)$</p><p>$⇒ 5 - 20 + 3\beta + 12 - 2\beta$</p><p>$⇒ - 3 + \beta = 0$</p><p>$⇒ \beta = 3$</p><p>Hence \left(\alpha, \beta\right) = \left(1, 3\right)</p>$