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)

For every point $(\alpha, \beta) \neq(7,7)$ on the line $x-2 y+7=0$, the system has infinitely many solutions

Answer 2

By equation 1 and 3
$y+2z=8$
And $y=8−2z$
$x=−2+z$
Now putting in equation $2$
$α(z−2)+β(−2z+8)+7z=3$
$⇒(α−2β+7)z=2α−8β+3$
So equations have unique solution if $α−2β+7\neq0$
And equations have no solution if $α−2β+7=0 \;and \;2α−8β+3\neq0$
And equations have infinite solution if $α−2β+7=0$ and $2α−8β+3=0$

The Correct Option is (C): For every point$(\alpha, \beta) \neq(7,7)$ on the line $x-2 y+7=0$, the system has infinitely many solutions