Question

If the system of equations $bx+ay=c$, $cx+az=b,cy+bz=a$ has unique solution, then

• A. $abc=1$
• B. $abc+2=-4$
• C. $abc\ne 0$
• D. $abc+1=0$

Answer 1

The correct option is C $abc\ne 0$

$bx+ay=c$

$cx+az=b$

$cy+bz=a$

for unique solution $\left|\begin{array}{ccc}b& a& 0\\ c& 0& a\\ 0& c& b\end{array}\right|\ne 0$

$\Rightarrow b(o-ac)-a(bc-0)+0(c^2-0)\ne 0$

$\Rightarrow -2abc\ne 0$

$\Rightarrow abc\ne 0$