Question

If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x+3|y|+5[z]=0,x+|y|-2[z]=4,x+|y|+|z|=1$ has

• A. a unique solution
• B. finitely many solutions
• C. infinitely many solutions
• D. no solution

Answer 1

Answer: (C)

Given system of three equations
$2x+3|y|+5[z]=0$
$x+|y|-2[z]=4$
and $x+|y|+[z]=1$
According to Cramer's rule,
$x=\frac{{\Delta }_1}{\Delta },|y|=\frac{{\Delta }_2}{\Delta }$
and $[z]=\frac{{\Delta }_3}{\Delta }$
where, $\Delta =\left|\begin{array}{ccc}2& 3& 5\\ 1& 1& -2\\ 1& 1& 1\end{array}\right|$
$=2(1+2)-3(1+2)+5(1-1)=-3$
${\Delta }_1=\left|\begin{array}{ccc}0& 3& 5\\ 4& 1& -2\\ 1& 1& 1\end{array}\right|$
$=0(1+2)-3(4+2)+5(4-1)$
$=-18+15=-3$
${\Delta }_2=\left|\begin{array}{ccc}2& 0& 5\\ 1& 4& -2\\ 1& 1& 1\end{array}\right|$
$=2(4+2)-0(1+2)+5(1-4)=-3$
and ${\Delta }_3=\left|\begin{array}{ccc}2& 3& 0\\ 1& 1& 4\\ 1& 1& 1\end{array}\right|$
$=2(1-4)-3(1-4)+0(1-1)=3$
Now, $x=\frac{-3}{-3}=1,|y|=\frac{-3}{-3}=1$
and $[z]=\frac{-3}{3}=-1$
$\therefore x=1,|y|=1$
$\Rightarrow y=\pm 1$ and $[z]=-1$
$\Rightarrow z\in [-1,0)$
So, the given system of three equations has infinitely many solution.