Question

If the system of equations $\alpha x+y+z=5,x+2y+3z=4,x+3y+5z=\beta$ has infinitely many solutions, then the ordered pair $(\alpha ,\beta )$ is equal to:

• A. $(1,-3)$
• B. $(-1,3)$
• C. $(1,3)$
• D. $(-1,-3)$

Answer 1

The correct option is C $(1,3)$

Sol. Given system of equations

$\alpha x+y+z=5$

$x+2y+3z=4,\text{has infinite solution}$

$x+3y+5z=\beta$

$\therefore \Delta =\left|\begin{array}{ccc}\alpha & 1& 1\\ 1& 2& 3\\ 1& 3& 5\end{array}\right|=0\Rightarrow \alpha \left(1\right)-1\left(2\right)+1\left(1\right)=0$

$\Rightarrow \alpha =1$

and ${\Delta }_1=\left|\begin{array}{ccc}5& 1& 1\\ 4& 2& 3\\ \beta & 3& 5\end{array}\right|=0$

$\Rightarrow 5\left(1\right)-1(20-3\beta )+1(12-2\beta )=0$

$\Rightarrow \beta =3$

And ${\Delta }_2=\left|\begin{array}{ccc}1& 5& 1\\ 1& 4& 3\\ 1& \beta & 5\end{array}\right|=0\Rightarrow (20-3\beta )-5\left(2\right)+1(\beta -4)=0$

$\Rightarrow -2\beta +6=0$

$\Rightarrow \beta =3$

Similarly,

$\Rightarrow {\Delta }_3=\left|\begin{array}{ccc}1& 1& 5\\ 1& 2& 4\\ 1& 3& \beta \end{array}\right|=0\Rightarrow \beta =3$

$(\alpha ,\beta )=(1,3)$