For the system of linear equations [ x+y+z=1,x+ y+z=1,x+y+ z=], which one of the following…

Question

For the system of linear equations $$\alpha x+y+z=1,x+\alpha y+z=1,x+y+\alpha z=\beta$$, which one of the following statements is NOT correct?

Accepted Answer

For infinite solution $\Delta=\Delta_x=\Delta_y=\Delta_z=0$

$$
\Delta=\left|\[\begin{array}{lll} \alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha \end{array}$$\right|=0 \Rightarrow\left(\alpha\^3-3 \alpha+2\right)=0 \Rightarrow \alpha=1,-2
\]

If $\beta=1$, then all planes are overlapping

$\therefore$ Option (A) is correct.

Option (B) :

If $\alpha=2 \Rightarrow \Delta \neq 0$

$\therefore $ Unique solution exist

$\therefore$ Option (B) is incorrect.

Option (C) :

$$
\begin{aligned}
& \alpha=2, \beta=1 \\
& 2 x+y+z=1 \\
& x+2 y+z=1 \\
& x+y+2 z=1
\end{aligned}
$$

Adding all three equations,

$$
x+y+z=\frac{3}{4}
$$

$\therefore$ option (C) is correct.

Option (D) :

If $\alpha=-2$ and $\beta=1$, then $\Delta=0, \Delta_x \neq 0$

$\therefore$ No solution.

$\therefore$ Option (D) is correct.

Solutions · 1