If the system of linear equations [ aligned & 3 x+y+ z=3 & 2 x+ y-z=-3 & x+2 y+z=4 aligned…

Question

If the system of linear equations

$$ \begin{aligned} & 3 x+y+\beta z=3 \ & 2 x+\alpha y-z=-3 \ & x+2 y+z=4 \end{aligned} $$

has infinitely many solutions, then the value of $22 \beta-9 \alpha$ is :

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& 3 x+y+\beta z=3 \
& 2 x+\alpha y-z=-3 \
& x+2 y+z=4
\end{aligned}\
\&\text { has infinite solution }\
\&\begin{aligned}
& \Rightarrow \Delta=0, \Delta_1=\Delta_2=\Delta_3 \
& \Delta=0 \Rightarrow\left|\[\begin{array}{ccc} 3 & 1 & \beta \ 2 & \alpha & -1 \ 1 & 2 & 1 \end{array}\end{aligned}\end{aligned}$$\right|=0 \
\& \Delta\_2=0 \Rightarrow\left|$$\begin{array}{ccc} 3 & 3 & \beta \ 2 & -3 & -1 \ 1 & 4 & 1 \end{array}$$\right|=0 \
\& \Rightarrow 3(-3+4)-3(2+1)+\beta(8+3)=0 \
\& \Rightarrow 3-9+11 \beta=0 \
\& \Rightarrow \quad \beta=\frac{6}{11} \
\& \Delta\_3=0 \Rightarrow\left|$$\begin{array}{ccc} 3 & 1 & 3 \ 2 & \alpha & -3 \ 1 & 2 & 4 \end{array}$$\right|=0 \
\& \Rightarrow \quad 3(4 \alpha+6)-1(8+3)+3(4-\alpha)=0 \
\& \quad 12 \alpha+18-11+12-3 \alpha=0 \
\& \quad 9 \alpha=-19 \
\& \quad \alpha=\frac{-19}{9}

\]

$\therefore \quad 22 \beta-9 \alpha=31$

Solutions · 1