## Precalculus (6th Edition) Blitzer

The solution set of the provided system of equations is $x=e,y={{e}^{-3}},z={{e}^{-2}}\text{ and }w={{e}^{-1}}$.
Consider, $\ln w=A,\ln x=B,\ln y=C,\ln z=D$ The resulting system of equations is: \begin{align} & 2A+B+3C-2D=-6 \\ & 4A+3B+C-D=-2 \\ & A+B+C+D=-5 \\ & A+B-C-D=-5 \end{align} First write the augmented matrix for the given system of equations: Augmented matrix: $\left[ \left. \begin{matrix} 2 & 1 & 3 & -2 \\ 4 & 3 & 1 & -1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ \end{matrix} \right|\begin{matrix} -6 \\ -2 \\ -5 \\ 5 \\ \end{matrix} \right]$ Now, use row operation to reduce the matrix to row echelon form. First apply ${{R}_{1}}\leftrightarrow {{R}_{3}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 4 & 3 & 1 & -1 \\ 2 & 1 & 3 & -2 \\ 1 & 1 & -1 & -1 \\ \end{matrix} \right|\begin{matrix} -5 \\ -2 \\ -6 \\ 5 \\ \end{matrix} \right]$ Now apply ${{R}_{2}}\to {{R}_{2}}-4{{R}_{1}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & -1 & -3 & -5 \\ 2 & 1 & 3 & -2 \\ 1 & 1 & -1 & -1 \\ \end{matrix} \right|\begin{matrix} -5 \\ 18 \\ -6 \\ 5 \\ \end{matrix} \right]$ Apply ${{R}_{2}}\to \left( -1 \right){{R}_{2}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 5 \\ 2 & 1 & 3 & -2 \\ 1 & 1 & -1 & -1 \\ \end{matrix} \right|\begin{matrix} -5 \\ -18 \\ -6 \\ 5 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to {{R}_{3}}-2{{R}_{1}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 5 \\ 0 & -1 & 1 & -4 \\ 1 & 1 & -1 & -1 \\ \end{matrix} \right|\begin{matrix} -5 \\ -18 \\ 4 \\ 5 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to {{R}_{3}}+{{R}_{2}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 5 \\ 0 & 0 & 4 & 1 \\ 1 & 1 & -1 & -1 \\ \end{matrix} \right|\begin{matrix} -5 \\ -18 \\ -14 \\ 5 \\ \end{matrix} \right]\,$ Apply ${{R}_{4}}\to {{R}_{4}}-{{R}_{1}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 5 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & -2 & -2 \\ \end{matrix} \right|\begin{matrix} -5 \\ -18 \\ -14 \\ 10 \\ \end{matrix} \right]$ Apply ${{R}_{4}}\to 2{{R}_{4}}+{{R}_{3}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 5 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & -3 \\ \end{matrix} \right|\begin{matrix} -5 \\ -18 \\ -14 \\ 6 \\ \end{matrix} \right]$ Write the system of equations corresponding to the reduced matrix $A+B+C+D=-5$ …… (I) $B+3C+5D=-18$ …… (II) $4C+D=-14$ …… (III) $-3D=6$ …… (IV) From equation (IV), $D=\frac{-6}{3}=-2$ To find $C$, substitute the value of $D$ in equation (III), \begin{align} & 4C-2=-14 \\ & 4C=-14+2 \\ & 4C=-12 \\ & C=\frac{-12}{4} \end{align} Simplify further to get, $C=-3$ To find $B$, substitute the value of $C,D$ in equation (II), \begin{align} & B+3\times -3+5\times -2=-18 \\ & B-9-10=-18 \\ & B-19=-18 \\ & B=-18+19 \end{align} After further simplification, we get, So, $B=1$ To find $A$, substitute the value of $B,C,D$ in equation (I), \begin{align} & A+1-3-2=-5 \\ & A+1-5=-5 \\ & A-4=-5 \\ & A=-5+4 \end{align} Thus, $A=-1$ Now, \begin{align} & \ln w=A \\ & w={{e}^{A}} \\ & w={{e}^{-1}} \end{align} And \begin{align} & \ln x=B \\ & x={{e}^{B}} \\ & x=e \end{align} And \begin{align} & \ln y=C \\ & y={{e}^{C}} \\ & y={{e}^{-3}} \end{align} and, \begin{align} & \ln z=D \\ & z={{e}^{D}} \\ & z={{e}^{-2}} \end{align} Hence, the solution set of the provided system of equations is $\left( {{e}^{-1}},e,{{e}^{-3}},{{e}^{-2}} \right)$.