Question

The number of real values 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. $2$
• C. $4$
• D. $6$

Answer 1

The correct option is A $1$

The system of equations are
$x+3y+5z=\alpha x$
$5x+y+3z=\alpha y$
$3x+5y+z=\alpha z$

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

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

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

For infinite number of solutions, we must have
$\left|\begin{array}{ccc}1-\alpha & 3& 5\\ 5& 1-\alpha & 3\\ 3& 5& 1-\alpha \end{array}\right|=0$

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

$C_1\to C_2+C_2+C_3$.

Taking $(9-\alpha )$ common from the first column, we get
$\Rightarrow (9-\alpha )\left|\begin{array}{ccc}1& 3& 5\\ 1& 1-\alpha & 3\\ 1& 5& 1-\alpha \end{array}\right|=0$

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

$\Rightarrow (9-\alpha )\left|\begin{array}{cc}\alpha +2& 2\\ -2& \alpha +4\end{array}\right|=0$

$\Rightarrow (9-\alpha )({\alpha }^2+6\alpha +8+4)=0$

$(\alpha -9)({\alpha }^2+6\alpha +12)=0$

$\Rightarrow \alpha =9$ is the only real value of $\alpha$.