Determinants Question 6
Question 6 - 31 January - Shift 1
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?
(1) If $\alpha=\beta=7$, then the system has no solution
(2) If $\alpha=\beta$ and $\alpha \neq 7$ then the system has a unique solution.
(3) There is a unique point $(\alpha, \beta)$ on the line $x+2 y+18=0$ for which the system has infinitely many solutions
(4) For every point $(\alpha, \beta) \neq(7,7)$ on the line $x-2 y+7=0$, the system has infinitely many solutions.
Show Answer
Answer: (4)
Solution:
Formula: System of equations with 3 variables, consistency of solutions: infinite solutions
By equation 1 and 3
$ y+2 z=8 $
$ y=8-2 z $
And $\quad x=-2+z$
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 $2 \alpha-8 \beta+3=0$
