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)

infinitely many solutions

Answer 2

Given system of three equations
$2 x+3|y|+5[z]=0$
$x+|y|-2[z]=4$
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, $\Delta =\begin{vmatrix}2 & 3 & 5 \\\ 1 & 1 & -2 \\\ 1 & 1 & 1\end{vmatrix}$ $=2(1+2)-3(1+2)+5(1-1)=-3$
$\Delta_{1} =\begin{vmatrix}0 & 3 & 5 \\\ 4 & 1 & -2 \\\ 1 & 1 & 1\end{vmatrix}$ $=0(1+2)-3(4+2)+5(4-1)$
$=-18+15=-3$
$\Delta_{2} =\begin{vmatrix}2 & 0 & 5 \\\ 1 & 4 & -2 \\\ 1 & 1 & 1\end{vmatrix}$
$=2(4+2)-0(1+2)+5(1-4)=-3$
and $\Delta_{3}=\begin{vmatrix}2 & 3 & 0 \\\ 1 & 1 & 4 \\\ 1 & 1 & 1\end{vmatrix}$
$=2(1-4)-3(1-4)+0(1-1)=3$
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.