## Precalculus (6th Edition) Blitzer

The solution is, $x=-2,y=-1,z=0$.
Consider the given system of equation, \begin{align} & x-2y+z=0 \\ & y-3z=-1 \\ & 2y+5z=-2 \end{align} Therefore, the system of equations can be written in matrix form as below: $AX=b$ Where $A=\left[ \begin{array}{*{35}{r}} 1 & -2 & 1 \\ 0 & 1 & -3 \\ 0 & 2 & 5 \\ \end{array} \right];b=\left[ \begin{array}{*{35}{r}} 0 \\ -1 \\ -2 \\ \end{array} \right];X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]$ Consider the augmented matrix $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 1 & -2 & 1 & 0 \\ 0 & 1 & -3 & -1 \\ 0 & 2 & 5 & -2 \\ \end{array} \right]$ By applying elementary row operation on $A$ we will convert it to its equivalent upper triangular matrix form. Step 1: Apply the operation ${{{R}'}_{3}}={{R}_{3}}-2{{R}_{2}}$ $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 1 & -2 & 1 & 0 \\ 0 & 1 & -3 & -1 \\ 0 & 0 & 11 & 0 \\ \end{array} \right]$ Therefore, the system of equations reduces to \begin{align} & x-2y+z=0 \\ & y-3z=-1 \\ & 11z=0 \end{align} From $11z=0$ we get \begin{align} & 11z=0 \\ & z=0 \end{align} From $y-3z=-1$ we get \begin{align} & y=3z-1 \\ & =-1 \end{align} From $x-2y+z=0$ we get \begin{align} & x=2y-z \\ & =-2-0 \\ & =-2 \end{align} The solution is, $x=-2,y=-1,z=0$.