Question

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

• A. If $\alpha =\beta$ and $\alpha \ne 7$, then the system has a unique solution
• B. There is a unique point $(\alpha ,\beta )$ on the line $x+2y+18=0$ for which the system has infinitely many solutions
• C. For every point $(\alpha ,\beta )\ne (7,7)$ on the line $x-2y+7=0$, 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
$y+2z=8$
$y=8-2z$
And $x=-2+z$
Now putting in equation 2
$\alpha (z-2)+\beta (-2z+8)+7z=3$
$\Rightarrow (\alpha -2\beta +7)z=2\alpha -8\beta +3$
So equations have unique solution if $\alpha -2\beta +7\ne 0$
And equations have no solution if $\alpha -2\beta +7=0$ and $2\alpha -8\beta +3\ne 0$
And equations have infinite solution if $\alpha -2\beta +7=0$ and $2\alpha -8\beta +3=0$