Consider the following system of linear equations $$\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} \alpha \\ 5 \\ 7 \end{array} \right)$$ Notice that the second and the third columns of the coefficient matrix are linearly dependent. For how many values of $\alpha$, does this system of equations have infinitely many solutions?

$\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} \alpha \\ 5 \\ 7 \end{array} \right)$ We can write for this linear equation 2x + y - 4z = a             4x + 3y - 12z = 5 x + 2y - 8z = 7 For infinitely solutions, D = 0 $\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) = 0$ Because, 2^nd and 3^rd columns are linearly dependent For x D = $\left( \begin{array}{ccc} \alpha & 1 & -4 \\ 5 & 3 & -12 \\ 7 & 2 & -8 \end{array} \right) = 0$ Because, 2^nd and 3^rd columns are linearly dependent For y $\left( \begin{array}{ccc} 2 & \alpha & -4 \\ 4 & 5 & -12 \\ 1 & 7 & -8 \end{array} \right) = 0$ $\Rightarrow$2(- 40 + 84) -$\alpha$(- 32 + 12) - 4 ( 28 -5) = 0 $\Rightarrow$ 88 + 20$\alpha$-92 = 0 $\Rightarrow$$\alpha = \dfrac{4}{20} = \dfrac{1}{5}$ --- (i) For z D = $\left( \begin{array}{ccc} 2 & 1 & \alpha \\ 4 & 3 & 5 \\ 1 & 2 & 7 \end{array} \right) = 0$

$\Rightarrow$2 (21-10) - 1 ( 28 - 5) +$\alpha$ (8 - 3) = 0 $\Rightarrow$ 22 - 23 + 5 $\alpha$ = 0 $\Rightarrow$ $\alpha = \dfrac{1}{5}$--(ii) Hence from equation, we find that, $\alpha$ have only one value for infinitely solution.