If the system of linear equations [3x+y+ z=3, 2x+ y-z=-3, x+2y+z=4] has infinitely many so…

Question

If the system of linear equations
$$3x+y+\beta z=3,\qquad 2x+\alpha y-z=-3,\qquad x+2y+z=4$$
has infinitely many solutions, then the value of $22\beta-9\alpha$ is:

Accepted Answer

For infinitely many solutions, $\Delta=\Delta_1=\Delta_2=\Delta_3=0$.

$\Delta=\begin{vmatrix}3&1&\beta\2&\alpha&-1\1&2&1\end{vmatrix}=0$: expanding gives $3(\alpha+2)-1(2+1)+\beta(4-\alpha)=0$.

From the $\Delta_2=0$ (constants) condition: $3(-3+4)-3(2+1)+\beta(8+3)=0\Rightarrow 3-9+11\beta=0\Rightarrow\beta=\dfrac6{11}$.

From the $\Delta_3=0$ condition: $3(4\alpha+6)-1(8+3)+3(4-\alpha)=0\Rightarrow 9\alpha=-19\Rightarrow \alpha=-\dfrac{19}9$.

$$22\beta-9\alpha=22\left(\frac6{11}\right)-9\left(-\frac{19}9\right)=12+19=31.$$

Solutions · 1