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Question

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


A
abc=1
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B
abc+2=4
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C
abc0
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D
abc+1=0
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Solution

The correct option is C $$abc\neq 0$$
$$bx + ay =c$$
$$cx + az = b$$
$$cy + bz = a$$
for unique solution $$\begin{vmatrix} b&a&0 \\c&0 &a\\0&c&b  \end{vmatrix} \neq 0$$
$$\Rightarrow b(o - ac)-a(bc - 0) + 0 (c^2 - 0) \neq 0$$
$$\Rightarrow -2abc \neq 0$$
$$\Rightarrow abc \neq 0$$

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