Question

If the system of linear equations
$x+y+z=5$
$x+2y+3z=9$
$x+3y+\alpha z=\beta$
has infinitely many solutions, then $\beta -\alpha$ equals:

• A. $5$
• B. $18$
• C. $8$
• D. $21$

Answer 1

The correct option is C $8$

$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 3& \alpha \end{array}\right|$
$=1(2\alpha -9)-1(\alpha -3)+1(3-2)$
$=2\alpha -9-\alpha +3+1$
$=\alpha -5$
For infinitely many solutions,
$\Delta =0$
$\Rightarrow \alpha -5=0$
$\Rightarrow \alpha =5$

Now,
${\Delta }_1=\left|\begin{array}{ccc}1& 1& 5\\ 1& 2& 9\\ 1& 3& \beta \end{array}\right|$
$=1(2\beta -27)-1(\beta -9)+5(3-2)$
$=2\beta -27-\beta +9+5$
$=\beta -13$
For infinitely many solutions,
${\Delta }_1=0$
$\Rightarrow \beta -13=0$
$\Rightarrow \beta =13$

Hence, $\beta -\alpha =13-5=8$