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 933: 39

Answer

a) The linear system can be written as $\left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 8 \\ -7 \\ 1 \\ \end{matrix} \right]$ b) the value of the inverse matrix is $ X=\left\{ \left( 2,-1,5 \right) \right\}$

Work Step by Step

(a) Consider the given system of equations: $\begin{align} & x-y+z=8 \\ & 2y-z=-7 \\ & 2x+3y=1 \end{align}$ The linear system can be written as: $ AX=B $ $\left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 8 \\ -7 \\ 1 \\ \end{matrix} \right]$ Where, $ A=\left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{matrix} \right]$ $ X=\left[ \begin{align} & x \\ & y \\ & z \\ \end{align} \right]$ $ B=\left[ \begin{align} & 8 \\ & -7 \\ & 1 \\ \end{align} \right]$ \ (b) Consider the given system of equations: $\begin{align} & x-y+z=8 \\ & 2y-z=-7 \\ & 2x+3y=1 \end{align}$ The linear system can be written as: $ AX=B $ $\left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 8 \\ -7 \\ 1 \\ \end{matrix} \right]$ Where, $ A=\left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{matrix} \right]$ $ X=\left[ \begin{align} & x \\ & y \\ & z \\ \end{align} \right]$ $ B=\left[ \begin{align} & 8 \\ & -7 \\ & 1 \\ \end{align} \right]$ Now, consider the coefficient matrix $ A=\left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{matrix} \right]$ Use the inverse of the coefficient matrix ${{\left[ A \right]}^{-1}}=\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{matrix} \right]$ Now, to find the values of the provided system: We will use the formula $ X={{A}^{-1}}B $ Where, ${{\left[ A \right]}^{-1}}=\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{matrix} \right]$ $ B=\left[ \begin{align} & 8 \\ & -7 \\ & 1 \\ \end{align} \right]$ Now, substitute the values in $ X={{A}^{-1}}B $
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