Question

The sum of real value(s) of $\alpha$ for which the system of equations
$x+3y+5z=\alpha x$
$5x+y+3z=\alpha y$
$3x+5y+z=\alpha z$
has infinite number of solutions is

• A. $1$
• B. $3$
• C. $9$
• D. $15$

Answer 1

The correct option is C $9$

$(1-\alpha )x+3y+5z=0$
$5x+(1-\alpha )y+3z=0$
$3x+5y+(1-\alpha )z=0$

For homogeneous system of linear equations to have infinitely many solutions, $\Delta =0$
$\Delta =\left|\begin{array}{ccc}1-\alpha & 3& 5\\ 5& 1-\alpha & 3\\ 3& 5& 1-\alpha \end{array}\right|=0$

$R_1\to R_1+R_2+R_3$
$\Delta =\left|\begin{array}{ccc}9-\alpha & 9-\alpha & 9-\alpha \\ 5& 1-\alpha & 3\\ 3& 5& 1-\alpha \end{array}\right|=0$

$\Rightarrow (9-\alpha )\left|\begin{array}{ccc}1& 1& 1\\ 5& 1-\alpha & 3\\ 3& 5& 1-\alpha \end{array}\right|=0$

$C_2\to C_2-C_1$ and $C_3\to C_3-C_1$
$(9-\alpha )\left|\begin{array}{ccc}1& 0& 0\\ 5& -\alpha -4& -2\\ 3& 2& -\alpha -2\end{array}\right|=0$

Expanding along $R_1,$ we get
$(9-\alpha )\left[\right(\alpha +4\left)\right(\alpha +2)+4]=0$
$\Rightarrow (9-\alpha )({\alpha }^2+6\alpha +12)=0$
$\Rightarrow \alpha =9$ is the only real root.
$[$For ${\alpha }^2+6\alpha +12,$
$D=36-48=-12<0$ and hence no real solution.$]$
$\therefore$ Sum of real values of $\alpha =9.$