We wanted to solve the Machiavellian Matrix. We can began by dividing the top row by xyz to simplify it to ( 1 1 1 1 ). Then we subtract row 1 from row 3 to fin that x = -1/3. This makes x2 = 1/9. Then we can get a system of equations for y and z, that y + z = 4/3, and y2 + z2 = 8/9. If y + z = 4/3 then z = 4/3 – y, and we can substitute to get the quadratic equation. Solve to get (y – 2/3) * (y – 2/3) = 0, so y = 2/3. Our answer is thus that x = -1/3, y = 2/3, and z = 2/3.
\left(\begin{matrix} 1 & 1 & 1 & 1 \\ x & y & z & 1 \\ 4 & 1 & 1 & 0 \end{matrix}\right) \overrightarrow{R_3-R_1→R_1} \left(\begin{matrix} 3 & 0 & 0 & -1 \\ x & y & z & 1 \\ 4 & 1 & 1 & 0 \end{matrix}\right) \\[10pt] y^2+\left(\frac43-y\right)^2-\frac89=0