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
abc0
D
abc+1=0

Solution

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

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