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 $D_{y}=D_{z}=0$ $\alpha=5, \beta=13$