## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 895: 56

#### Answer

The required matrices are $\left[ \begin{matrix} 1 & -3 & 2 & 5 \\ 1 & 5 & -5 & 0 \\ 3 & 0 & 4 & 7 \\ \end{matrix} \right],\left[ \begin{matrix} 1 & -4 & 2 & 3 \\ 1 & -4 & 4 & 0 \\ 2 & 0 & 7 & 4 \\ \end{matrix} \right],\left[ \begin{matrix} 1 & -3 & 2 & 0 \\ 0 & 10 & -7 & 7 \\ 2 & -2 & 1 & 3 \\ \end{matrix} \right]$

#### Work Step by Step

Consider the matrix of exercise 13: $\left[ \begin{matrix} 2 & -6 & 4 & 10 \\ 1 & 5 & -5 & 0 \\ 3 & 0 & 4 & 7 \\ \end{matrix} \right]$ The operation to be applied is ${{R}_{1}}\to \frac{1}{2}{{R}_{1}}$. Follow the steps given below to solve the provided equations: Step1: Open the Ti-83 calculator and press the “2ND” button and press “ ${{x}^{-1}}$ ”. Step2: For augmented matrix select matrix A and go to edit. Enter the order of matrix and entries of the matrix. Then again press “2ND” and “Quit”. Step3: Press “2ND” and “ ${{x}^{-1}}$ ” then go to “MATH” and select “*row (“. Step4: Enter $\frac{1}{2}$, “2ND” then press 1, again press 1, then press “Enter”. The resulting matrix is: $\left[ \begin{matrix} 1 & -3 & 2 & 5 \\ 1 & 5 & -5 & 0 \\ 3 & 0 & 4 & 7 \\ \end{matrix} \right]$ Consider the matrix of exercise 14: $\left[ \begin{matrix} 3 & -12 & 6 & 9 \\ 1 & -4 & 4 & 0 \\ 2 & 0 & 7 & 4 \\ \end{matrix} \right]$ The operation to be applied is ${{R}_{1}}\to \frac{1}{3}{{R}_{1}}$. Follow the steps given below to solve the provided equations: Step 1: Open the Ti-83 calculator and press the “2ND” button and press “ ${{x}^{-1}}$ ”. Step 2: For augmented matrix select matrix A and go to edit. Enter the order of matrix and entries of the matrix. Then again press “2ND” and “Quit”. Step 3: Press “2ND” and “ ${{x}^{-1}}$ ” then go to “MATH” and select “*row (“. Step 4: Enter $\frac{1}{3}$, “2ND” then press 1, then press 1again. The resulting matrix is: $\left[ \begin{matrix} 1 & -4 & 2 & 3 \\ 1 & -4 & 4 & 0 \\ 2 & 0 & 7 & 4 \\ \end{matrix} \right]$ Consider the matrix of exercise 15: $\left[ \begin{matrix} 1 & -3 & 2 & 0 \\ 3 & 1 & -1 & 7 \\ 0 & -2 & 1 & 3 \\ \end{matrix} \right]$ The operation to be applied is ${{R}_{2}}\to -3{{R}_{1}}+{{R}_{2}}$. Follow the steps given below to solve the provided equations: Step 1: Open the Ti-83 calculator and press the “2ND” button and press “ ${{x}^{-1}}$ ”. Step 2: For augmented matrix select matrix A and go to edit. Enter the order of matrix and entries of the matrix. Then again press “2ND” and “Quit”. Step 3: Press “2ND” and “ ${{x}^{-1}}$ ” then go to “MATH” and select “*row+ (“. Step 4: Enter $-3$, “2ND” then press 1, then again press 1, 2 and press “Enter”. The resulting matrix is: $\left[ \begin{matrix} 1 & -3 & 2 & 0 \\ 0 & 10 & -7 & 7 \\ 2 & -2 & 1 & 3 \\ \end{matrix} \right]$ The resultedmatrix are: $\left[ \begin{matrix} 1 & -3 & 2 & 5 \\ 1 & 5 & -5 & 0 \\ 3 & 0 & 4 & 7 \\ \end{matrix} \right],\left[ \begin{matrix} 1 & -4 & 2 & 3 \\ 1 & -4 & 4 & 0 \\ 2 & 0 & 7 & 4 \\ \end{matrix} \right],\left[ \begin{matrix} 1 & -3 & 2 & 0 \\ 0 & 10 & -7 & 7 \\ 2 & -2 & 1 & 3 \\ \end{matrix} \right]$

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