1 problem found
Consider the system of equations \begin{alignat*}{1} 2yz+zx-5xy & =2\\ yz-zx+2xy & =1\\ yz-2zx+6xy & =3 \end{alignat*} Show that \[xyz=\pm 6\] and find the possible values of \(x\), \(y\) and \(z\).
Solution: Consider the linear \(3\times 3\) system in \(yz, zx, xy\), then \begin{align*} \left(\begin{array}{ccc|c} 2 & 1 & -5 & 2 \\ 1 & -1 & 2 & 1 \\ 1 & -2 & 6 & 3 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 1 & -2 & 6 & 3 \\ 2 & 1 & -5 & 2 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 0 & -1 & 4 & 2 \\ 0 & 3 & -9 & 0 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 0 & -1 & 4 & 2 \\ 0 & 0 & 3 & 6 \\ \end{array}\right) \\ \end{align*} Therefore \(yz = 2, zx = 6, xy = 3 \Rightarrow (xyz)^2 = 36 \Rightarrow xyz = \pm 6\). If \(xyz = 6, x = 3, y = 1, z = 2\), if \(xyz = -6, x = -3, y = -1, z = -2\)