Question

If the system of 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. 21
• D. 8

Answer 1

Answer: (D)

$D=\left|\begin{array}{lll}1& 1& 1\\ 1& 2& 3\\ 1& 3& \alpha \end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 0& 2& \alpha -1\end{array}\right|=(\alpha -1)-4=(\alpha -5)$
for infinite solutions $D=0\Rightarrow \alpha =5$
$D_x=0\Rightarrow \left|\begin{array}{lll}5& 1& 1\\ 9& 2& 3\\ \beta & 3& 5\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}0& 0& 1\\ -1& -1& 3\\ \beta -15& -2& 5\end{array}\right|=0$
$\Rightarrow 2+\beta -15=0\Rightarrow \beta -13=0$
on $\beta =13$ we get $D_y=D_z=0$ $\alpha =5,\beta =13$