Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 7E from Chapter 1.6 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 7E
Chapter:
Problem:
Balance the chemical equations in Exercises 6–11 using the vector equation approach discussed in this section. Alka-Seltzer contains sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7). When a tablet is dissolved in water, the following reaction produces sodium citrate, water, and carbon dioxide (gas):
Step-by-Step Solution
Given Information
We are given with an unbalanced chemical equation: \[ \mathrm { NaHCO } _ { 3 } + \mathrm { H } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } \rightarrow \mathrm { Na } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } + \mathrm { H } _ { 2 } \mathrm { O } + \mathrm { CO } _ { 2 } \] We have to balance the chemical equation:
Step-1:
Let the balance equation has the form \[ x _ { 1 } \cdot \mathrm { NaHCO } _ { 3 } + x _ { 2 } \cdot \mathrm { H } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } \rightarrow x _ { 3 } \cdot \mathrm { Na } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } + x _ { 4 } \cdot \mathrm { H } _ { 2 } \mathrm { O } + x _ { 5 } \mathrm { CO } _ { 2 } \] There are four types of atoms, Na, H, C, O
Step-2: The matrix form is
Write the matrix form using coefficient of each atom: \[ \mathrm { NaHCO } _ { 3 } \left[ \begin{array} { l } { 1 } \\ { 1 } \\ { 1 } \\ { 3 } \end{array} \right] , \mathrm { H } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } \left[ \begin{array} { l } { 0 } \\ { 8 } \\ { 6 } \\ { 7 } \end{array} \right] , \mathrm { Na } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } \left[ \begin{array} { l } { 3 } \\ { 5 } \\ { 6 } \\ { 7 } \end{array} \right] , \mathrm { H } _ { 2 } \mathrm { O } \left[ \begin{array} { l } { 0 } \\ { 2 } \\ { 0 } \\ { 1 } \end{array} \right] , \mathrm { CO } _ { 2 } \left[ \begin{array} { l } { 0 } \\ { 0 } \\ { 1 } \\ { 2 } \end{array} \right] \] So, the system of equations with coefficients is given by: \[ x _ { 1 } \left[ \begin{array} { l } { 1 } \\ { 1 } \\ { 1 } \\ { 3 } \end{array} \right] + x _ { 2 } \left[ \begin{array} { l } { 0 } \\ { 8 } \\ { 6 } \\ { 7 } \end{array} \right] = x _ { 3 } \left[ \begin{array} { l } { 3 } \\ { 5 } \\ { 6 } \\ { 7 } \end{array} \right] + x _ { 4 } \left[ \begin{array} { l } { 0 } \\ { 2 } \\ { 0 } \\ { 1 } \end{array} \right] + x _ { 5 } \left[ \begin{array} { l } { 0 } \\ { 0 } \\ { 1 } \\ { 2 } \end{array} \right] \]
Step-3: The Row-Reduced augmented matrix form
\[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 3}&0&0&0\\ 1&8&{ - 5}&{ - 2}&0&0\\ 1&6&{ - 6}&0&{ - 1}&0\\ 3&7&{ - 7}&{ - 1}&{ - 2}&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 3}&0&0&0\\ 0&8&{ - 2}&{ - 2}&0&0\\ 0&6&{ - 3}&0&{ - 1}&0\\ 0&7&2&{ - 1}&{ - 2}&0 \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_2} = {R_2} - {R_1}\\ {R_3} = {R_3} - {R_1}\\ {R_4} = {R_4} - 3{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 3}&0&0&0\\ 0&1&{ - 1/4}&{ - 1/4}&0&0\\ 0&6&{ - 3}&0&{ - 1}&0\\ 0&7&2&{ - 1}&{ - 2}&0 \end{array}} \right]::\,\,\left\{ {{R_2} = \dfrac{{{R_2}}}{8}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 3}&0&0&0\\ 0&1&{ - 1/4}&{ - 1/4}&0&0\\ 0&0&{ - 3/2}&{3/2}&{ - 1}&0\\ 0&0&{15/4}&{3/4}&{ - 2}&0 \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_3} = {R_3} - 6{R_2}\\ {R_4} = {R_4} - 7{R_2} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 3}&0&0&0\\ 0&1&{ - 1/4}&{ - 1/4}&0&0\\ 0&0&1&{ - 1}&{2/3}&0\\ 0&0&{15/4}&{3/4}&{ - 2}&0 \end{array}} \right]::\,\left\{ {{R_2} = \dfrac{2}{3}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 3}&0&0&0\\ 0&1&{ - 1/4}&{ - 1/4}&0&0\\ 0&0&1&{ - 1}&{2/3}&0\\ 0&0&0&{9/2}&{ - 9/2}&0 \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - \dfrac{{15}}{4}{R_3}} \right\} \end{array}\]
Step-4: System of equations
\[ \begin{array} { r } { x _ { 1 } - 3 x _ { 3 } = 0 } \\ { x _ { 2 } - \dfrac { 1 } { 4 } x _ { 3 } - \dfrac { 1 } { 4 } x _ { 4 } = 0 } \\ { x _ { 3 } - x _ { 4 } + \dfrac { 2 } { 3 } x _ { 5 } = 0 } \\ { \dfrac { 9 } { 2 } x _ { 4 } - \dfrac { 9 } { 2 } x _ { 5 } = 0 } \end{array} \] On choosing, $x_5=3$: \[ \begin{array} { l } { x _ { 1 } = x _ { 5 } = 3 } \\ { x _ { 2 } = \dfrac { 1 } { 3 } x _ { 5 } = \dfrac { 1 } { 3 } \times 3 = 1 } \\ { x _ { 3 } = \dfrac { 1 } { 3 } x _ { 5 } = \dfrac { 1 } { 3 } \times 3 = 1 } \\ { x _ { 4 } = x _ { 5 } = 3 } \end{array} \] Therefore, the balanced equation is:
\[ 3 \mathrm { NaHCO } _ { 3 } + \mathrm { H } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } \rightarrow \mathrm { Na } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } + 3 \mathrm { H } _ { 2 } \mathrm { O } + 3 \mathrm { CO } _ { 2 } \]
We are given with an unbalanced chemical equation: \[ \mathrm { NaHCO } _ { 3 } + \mathrm { H } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } \rightarrow \mathrm { Na } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } + \mathrm { H } _ { 2 } \mathrm { O } + \mathrm { CO } _ { 2 } \] We have to balance the chemical equation:
Step-1:
Let the balance equation has the form \[ x _ { 1 } \cdot \mathrm { NaHCO } _ { 3 } + x _ { 2 } \cdot \mathrm { H } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } \rightarrow x _ { 3 } \cdot \mathrm { Na } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } + x _ { 4 } \cdot \mathrm { H } _ { 2 } \mathrm { O } + x _ { 5 } \mathrm { CO } _ { 2 } \] There are four types of atoms, Na, H, C, O
Step-2: The matrix form is
Write the matrix form using coefficient of each atom: \[ \mathrm { NaHCO } _ { 3 } \left[ \begin{array} { l } { 1 } \\ { 1 } \\ { 1 } \\ { 3 } \end{array} \right] , \mathrm { H } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } \left[ \begin{array} { l } { 0 } \\ { 8 } \\ { 6 } \\ { 7 } \end{array} \right] , \mathrm { Na } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } \left[ \begin{array} { l } { 3 } \\ { 5 } \\ { 6 } \\ { 7 } \end{array} \right] , \mathrm { H } _ { 2 } \mathrm { O } \left[ \begin{array} { l } { 0 } \\ { 2 } \\ { 0 } \\ { 1 } \end{array} \right] , \mathrm { CO } _ { 2 } \left[ \begin{array} { l } { 0 } \\ { 0 } \\ { 1 } \\ { 2 } \end{array} \right] \] So, the system of equations with coefficients is given by: \[ x _ { 1 } \left[ \begin{array} { l } { 1 } \\ { 1 } \\ { 1 } \\ { 3 } \end{array} \right] + x _ { 2 } \left[ \begin{array} { l } { 0 } \\ { 8 } \\ { 6 } \\ { 7 } \end{array} \right] = x _ { 3 } \left[ \begin{array} { l } { 3 } \\ { 5 } \\ { 6 } \\ { 7 } \end{array} \right] + x _ { 4 } \left[ \begin{array} { l } { 0 } \\ { 2 } \\ { 0 } \\ { 1 } \end{array} \right] + x _ { 5 } \left[ \begin{array} { l } { 0 } \\ { 0 } \\ { 1 } \\ { 2 } \end{array} \right] \]
Step-3: The Row-Reduced augmented matrix form
\[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 3}&0&0&0\\ 1&8&{ - 5}&{ - 2}&0&0\\ 1&6&{ - 6}&0&{ - 1}&0\\ 3&7&{ - 7}&{ - 1}&{ - 2}&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 3}&0&0&0\\ 0&8&{ - 2}&{ - 2}&0&0\\ 0&6&{ - 3}&0&{ - 1}&0\\ 0&7&2&{ - 1}&{ - 2}&0 \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_2} = {R_2} - {R_1}\\ {R_3} = {R_3} - {R_1}\\ {R_4} = {R_4} - 3{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 3}&0&0&0\\ 0&1&{ - 1/4}&{ - 1/4}&0&0\\ 0&6&{ - 3}&0&{ - 1}&0\\ 0&7&2&{ - 1}&{ - 2}&0 \end{array}} \right]::\,\,\left\{ {{R_2} = \dfrac{{{R_2}}}{8}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 3}&0&0&0\\ 0&1&{ - 1/4}&{ - 1/4}&0&0\\ 0&0&{ - 3/2}&{3/2}&{ - 1}&0\\ 0&0&{15/4}&{3/4}&{ - 2}&0 \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_3} = {R_3} - 6{R_2}\\ {R_4} = {R_4} - 7{R_2} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 3}&0&0&0\\ 0&1&{ - 1/4}&{ - 1/4}&0&0\\ 0&0&1&{ - 1}&{2/3}&0\\ 0&0&{15/4}&{3/4}&{ - 2}&0 \end{array}} \right]::\,\left\{ {{R_2} = \dfrac{2}{3}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 3}&0&0&0\\ 0&1&{ - 1/4}&{ - 1/4}&0&0\\ 0&0&1&{ - 1}&{2/3}&0\\ 0&0&0&{9/2}&{ - 9/2}&0 \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - \dfrac{{15}}{4}{R_3}} \right\} \end{array}\]
Step-4: System of equations
\[ \begin{array} { r } { x _ { 1 } - 3 x _ { 3 } = 0 } \\ { x _ { 2 } - \dfrac { 1 } { 4 } x _ { 3 } - \dfrac { 1 } { 4 } x _ { 4 } = 0 } \\ { x _ { 3 } - x _ { 4 } + \dfrac { 2 } { 3 } x _ { 5 } = 0 } \\ { \dfrac { 9 } { 2 } x _ { 4 } - \dfrac { 9 } { 2 } x _ { 5 } = 0 } \end{array} \] On choosing, $x_5=3$: \[ \begin{array} { l } { x _ { 1 } = x _ { 5 } = 3 } \\ { x _ { 2 } = \dfrac { 1 } { 3 } x _ { 5 } = \dfrac { 1 } { 3 } \times 3 = 1 } \\ { x _ { 3 } = \dfrac { 1 } { 3 } x _ { 5 } = \dfrac { 1 } { 3 } \times 3 = 1 } \\ { x _ { 4 } = x _ { 5 } = 3 } \end{array} \] Therefore, the balanced equation is:
\[ 3 \mathrm { NaHCO } _ { 3 } + \mathrm { H } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } \rightarrow \mathrm { Na } _ { 3 } \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { O } _ { 7 } + 3 \mathrm { H } _ { 2 } \mathrm { O } + 3 \mathrm { CO } _ { 2 } \]