If the system of linear equations 2x + y - z = 3 x - y - z = α 3x +...

Given,

$2x + y - z = 3$

$x - y - z = \alpha$

$3x + 3y + \betaz = 3$

Let A = \left[\begin{cases}2 & 1 & - 1 \\ 1 & - 1 & - 1 \\ 3 & 3 & \beta\end{cases}\right] \text{ and } B = \left[$\begin{array}{ll}3 \\ \alpha \\ 3\end{array}$\right]

For Equation to have infinitely many solutions \left|A\right| = 0 \text{ and } \left(Adj . A\right) . B = 0</p><p>\left|A\right| = \left|\begin{cases}2 & 1 & - 1 \\ 1 & - 1 & - 1 \\ 3 & 3 & \beta\end{cases}\right| = 0

$⇒ - 3\beta - 3 = 0$

$⇒ \beta = - 1$

Now,

Minor of A = \left[\begin{cases}2\beta + 3 & \beta + 3 & 6 \\ \beta + 3 & 2\beta + 3 & 3 \\ - 2 & - 1 & - 3\end{cases}\right]

Cofactor of A = \left[\begin{cases}- \left(2\beta + 3\right) & - \left(\beta + 3\right) & - 6 \\ - \left(\beta + 3\right) & - \left(2\beta + 3\right) & - 3 \\ 2 & 1 & 3\end{cases}\right]

Adjoint of A = \left[\begin{cases}- \left(2\beta + 3\right) & - \left(\beta + 3\right) & 2 \\ - \left(\beta + 3\right) & - \left(2\beta + 3\right) & 1 \\ - 6 & - 3 & 3\end{cases}\right]

Now, \left(adj . A\right) . B = 0$</p><p>$⇒ \left[$\begin{cases}- \left(2\beta + 3\right) & - \left(\beta + 3\right) & 2 \\ - \left(\beta + 3\right) & - \left(2\beta + 3\right) & 1 \\ - 6 & - 3 & 3\end{cases}$\right]\left[$$\begin{array}{ll}3 \\ \alpha \\ 3\end{array}$$\right] = 0</p><p>\left[$\begin{array}{ll}- 6\beta - 9 - \alpha\beta - 3\alpha + 6 \\ - 3\beta - 9 - 2\alpha\beta - 3\alpha + 3 \\ - 18 - 3\alpha + 9\end{array}$\right] = 0

$⇒ - 18 - 3\alpha + 9 = 0$

$⇒ \alpha = - 3$

Therefore,\left|\alpha + \beta - \alpha\beta\right| = \left|- 3 - 1 - 3\right| = \left|- 7\right| = 7$