### 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 $2 \alpha-8 \beta+3=0$