Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 932: 37

Answer

a) The linear system can be written as $\left[ \begin{matrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 8 \\ 10 \\ 9 \\ \end{matrix} \right]$ b) $ X=\left\{ \left( 1,2,-1 \right) \right\}$

Work Step by Step

(a) Consider the given system of equations: $\begin{align} & 2x+6y+6z=8 \\ & 2x+7y+6z=10 \\ & 2x+7y+7z=9 \end{align}$ The linear system can be written as: $ AX=B $ $\left[ \begin{matrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 8 \\ 10 \\ 9 \\ \end{matrix} \right]$ Where, $ A=\left[ \begin{matrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \\ \end{matrix} \right]$, $ X=\left[ \begin{align} & x \\ & y \\ & z \\ \end{align} \right]$, $ B=\left[ \begin{align} & 8 \\ & 10 \\ & 9 \\ \end{align} \right]$ (b) Consider the given system of equations: $\begin{align} & 2x+6y+6z=8 \\ & 2x+7y+6z=10 \\ & 2x+7y+7z=9 \end{align}$ The linear system can be written as: $ AX=B $ $\left[ \begin{matrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 8 \\ 10 \\ 9 \\ \end{matrix} \right]$ Where, $ A=\left[ \begin{matrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \\ \end{matrix} \right]$ $ X=\left[ \begin{align} & x \\ & y \\ & z \\ \end{align} \right]$ $ B=\left[ \begin{align} & 8 \\ & 10 \\ & 9 \\ \end{align} \right]$ Now, consider the coefficient matrix $ A=\left[ \begin{matrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \\ \end{matrix} \right]$ Use the inverse of the coefficient matrix. ${{\left[ A \right]}^{-1}}=\left[ \begin{matrix} \frac{7}{2} & 0 & -3 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \\ \end{matrix} \right]$ Now, to find the values of the provided system: Using formula $ X={{A}^{-1}}B $ we get, Where, ${{A}^{-1}}=\left[ \begin{matrix} \frac{7}{2} & 0 & -3 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \\ \end{matrix} \right]$ $ B=\left[ \begin{align} & 8 \\ & 10 \\ & 9 \\ \end{align} \right]$ Now, substitute the values in $ X={{A}^{-1}}B $ to get,
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