Question

For the system of linear equations
$ x+y+z=6 $
$\alpha x+\beta y+7 z=3$
$ x+2 y+3 z=14$
which of the following is NOT true ?

• A. If $\alpha=\beta$ and $\alpha \neq 7$, then the system has a unique solution
• B. There is a unique point $(\alpha, \beta)$ on the line $x+2 y+18=0$ for which the system has infinitely many solutions
• C. For every point $(\alpha, \beta) \neq(7,7)$ on the line $x-2 y+7=0$, the system has infinitely many solutions
• D. If $\alpha=\beta=7$, then the system has no solution

Answer 1

Answer: (C)

By equation 1 and 3
$y+2 z=8 $
$y=8-2 z $
And $x=-2+z$
Now putting in equation 2
$\alpha(z-2)+\beta(-2 z+8)+7 z=3 $
$ \Rightarrow(\alpha-2 \beta+7) z=2 \alpha-8 \beta+3$
So equations have unique solution if $\alpha-2 \beta+7 \neq 0$
And equations have no solution if $\alpha-2 \beta+7=0$ and $2 \alpha-8 \beta+3 \neq 0$
And equations have infinite solution if $\alpha-2 \beta+7=0$ and $2 \alpha-8 \beta+3=0$