## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Cumulative Review Exercises - Page 952: 7

#### Answer

The solution is, $x=7,y=-4,z=6$.

#### Work Step by Step

Consider the given equation system, \begin{align} & x-y+z=17 \\ & 2x+3y+z=8 \\ & -4x+y+5z=-2 \end{align} The given system of equations can be written in matrix form as $AX=b$ Where $A=\left[ \begin{array}{*{35}{r}} 1 & -1 & 1 \\ 2 & 3 & 1 \\ -4 & 1 & 5 \\ \end{array} \right];b=\left[ \begin{array}{*{35}{r}} 17 \\ 8 \\ -2 \\ \end{array} \right];X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]$ Therefore, the solution of the system of equations $X={{A}^{-1}}b$ Consider the determinant of the matrix \begin{align} & \left| A \right|=\left| \begin{array}{*{35}{r}} 1 & -1 & 1 \\ 2 & 3 & 1 \\ -4 & 1 & 5 \\ \end{array} \right| \\ & =\left( 15-1 \right)+\left( 10+4 \right)+\left( 2+12 \right) \\ & =14+14+14 \\ & =42 \end{align} Consider the adjoint of the matrix $\text{adj}\left( A \right)={{\left[ \begin{array}{*{35}{r}} +\left| \begin{array}{*{35}{r}} 3 & 1 \\ 1 & 5 \\ \end{array} \right| & -\left| \begin{array}{*{35}{r}} 2 & 1 \\ -4 & 5 \\ \end{array} \right| & +\left| \begin{array}{*{35}{r}} 2 & 3 \\ -4 & 1 \\ \end{array} \right| \\ -\left| \begin{array}{*{35}{r}} -1 & 1 \\ 1 & 5 \\ \end{array} \right| & +\left| \begin{array}{*{35}{r}} 1 & 1 \\ -4 & 5 \\ \end{array} \right| & -\left| \begin{array}{*{35}{r}} 1 & -1 \\ -4 & 1 \\ \end{array} \right| \\ +\left| \begin{array}{*{35}{r}} -1 & 1 \\ 3 & 1 \\ \end{array} \right| & -\left| \begin{array}{*{35}{r}} 1 & 1 \\ 2 & 1 \\ \end{array} \right| & +\left| \begin{array}{*{35}{r}} 1 & -1 \\ 2 & 3 \\ \end{array} \right| \\ \end{array} \right]}^{t}}$ Finally, \begin{align} & \text{adj}\left( A \right)={{\left[ \begin{array}{*{35}{r}} 14 & -14 & 14 \\ 6 & 9 & 3 \\ -4 & 1 & 5 \\ \end{array} \right]}^{t}} \\ & =\left[ \begin{array}{*{35}{r}} 14 & 6 & -4 \\ -14 & 9 & 1 \\ 14 & 3 & 5 \\ \end{array} \right] \end{align} Therefore \begin{align} & {{A}^{-1}}=\frac{1}{\left| A \right|}\text{adj}\left( A \right) \\ & =\frac{1}{42}\left[ \begin{array}{*{35}{r}} 14 & 6 & -4 \\ -14 & 9 & 1 \\ 14 & 3 & 5 \\ \end{array} \right] \end{align} Therefore, the solution of the system of equations is \begin{align} & X=\frac{1}{42}\left[ \begin{array}{*{35}{r}} 14 & 6 & -4 \\ -14 & 9 & 1 \\ 14 & 3 & 5 \\ \end{array} \right]\left[ \begin{array}{*{35}{r}} 17 \\ 8 \\ -2 \\ \end{array} \right] \\ & =\frac{1}{42}\left[ \begin{array}{*{35}{r}} 14\times 17+6\times 8+4\times 2 \\ -14\times 17+9\times 8-1\times 2 \\ 14\times 17+3\times 8-5\times 2 \\ \end{array} \right] \\ & =\frac{1}{42}\left[ \begin{array}{*{35}{r}} 294 \\ -168 \\ 252 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} 7 \\ -4 \\ 6 \\ \end{array} \right] \end{align} The solutions of the system is $x=7,y=-4,z=6$.

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