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

The Correct Option is C

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