Question

If the equations $x+3y-4z=ax$; $x-3y+5z=ay$; $3x+y=az$ have a non-trivial solution then the values of a are

• A. 1,0
• B. 1,2
• C. -1,0
• D. -1,2

Answer 1

The correct option is C -1,0

$(1-a)x+3y-4_3=0$
$x+(-a-3)y+5_3=0$
$3x+y-a_3=0$
$\left|A\right|=0$ ,
$\left|\begin{array}{ccc}(1-a)& 3& -4\\ 1& -a-3& 5\\ 3& 1& -a\end{array}\right|=0$

$\left|\begin{array}{ccc}1-a& 3& 4\\ -1& a+3& 5\\ 3& 1& a\end{array}\right|=0$

$(a^2+3a)(1-a)-4+45-12a-36+5a+5+3a=0$
$-a^3+3a-3a^2+a^2+36-4a-36=0$
$-a^2+3-3a+a-4=0$
$a^2+2a+1=0$
$a=-1$