Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
We have solutions for your book!
See our solution for Question 5E from Chapter 1.6 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
Balance the chemical equations in Exercises 6–11 using the vector equation approach discussed in this section. Boron sulfide reacts violently with water to form boric acid and hydrogen sulfide gas (the smell of rotten eggs). The unbalanced equation is
Step-by-Step Solution
Given Information
We are given with an unbalanced chemical equation: \[ \mathrm { B } _ { 2 } \mathrm { S } _ { 3 } + \mathrm { H } _ { 2 } \mathrm { O } \rightarrow \mathrm { H } _ { 3 } \mathrm { BO } _ { 3 } + \mathrm { H } _ { 2 } \mathrm { S } \] We have to balance the chemical equation:
Step-1:
Let the balance equation has the form \[ x _ { 1 } \cdot \mathrm { B } _ { 2 } \mathrm { S } _ { 3 } + x _ { 2 } \cdot \mathrm { H } _ { 2 } \mathrm { O } \rightarrow x _ { 3 } \cdot \mathrm { H } _ { 3 } \mathrm { BO } _ { 3 } + x _ { 4 } \cdot \mathrm { H } _ { 2 } \mathrm { S } \] There are four types of atoms, B, S, H, O
Step-2: The matrix form is
Write the matrix form using coefficient of each atom: \[ \mathrm { B } _ { 2 } \mathrm { S } _ { 3 } \left[ \begin{array} { l } { 2 } \\ { 3 } \\ { 0 } \\ { 0 } \end{array} \right] , \mathrm { H } _ { 2 } \mathrm { O } \left[ \begin{array} { l } { 0 } \\ { 0 } \\ { 2 } \\ { 1 } \end{array} \right] , \mathrm { H } _ { 3 } \mathrm { BO } _ { 3 } \left[ \begin{array} { l } { 1 } \\ { 0 } \\ { 3 } \\ { 3 } \end{array} \right] , \mathrm { H } _ { 2 } \mathrm { O } \left[ \begin{array} { l } { 0 } \\ { 2 } \\ { 0 } \\ { 1 } \end{array} \right] , \mathrm { H } _ { 2 } \mathrm { S } \left[ \begin{array} { l } { 0 } \\ { 1 } \\ { 2 } \\ { 0 } \end{array} \right] \] So, the system of equations with coefficients is given by: \[ x _ { 1 } \left[ \begin{array} { l } { 2 } \\ { 3 } \\ { 0 } \\ { 0 } \end{array} \right] + x _ { 2 } \left[ \begin{array} { l } { 0 } \\ { 0 } \\ { 2 } \\ { 1 } \end{array} \right] = x _ { 3 } \left[ \begin{array} { l } { 1 } \\ { 0 } \\ { 3 } \\ { 3 } \end{array} \right] + x _ { 4 } \left[ \begin{array} { l } { 0 } \\ { 1 } \\ { 2 } \\ { 0 } \end{array} \right] \]
Step-3: The Row-Reduced augmented matrix form
\[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 2&0&{ - 1}&0&0\\ 3&0&0&{ - 1}&0\\ 0&2&{ - 3}&{ - 2}&0\\ 0&1&{ - 3}&0&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{r}} 1&0&{ - 1/2}&0&0\\ 3&0&0&{ - 1}&0\\ 0&2&{ - 3}&{ - 2}&0\\ 0&1&{ - 3}&0&0 \end{array}} \right]::\,\,\left\{ {{R_1} = \dfrac{{{R_1}}}{2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1/2}&0&0\\ 0&0&{3/2}&{ - 1}&0\\ 0&2&{ - 3}&{ - 2}&0\\ 0&1&{ - 3}&0&0 \end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} - 3{R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{r}} 1&0&{ - 1/2}&0&0\\ 0&2&{ - 3}&{ - 2}&0\\ 0&0&{3/2}&{ - 1}&0\\ 0&1&{ - 3}&0&0 \end{array}} \right]::\,\,\left\{ {{R_2} \Leftrightarrow {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1/2}&0&0\\ 0&1&{ - 3/2}&{ - 1}&0\\ 0&0&{3/2}&{ - 1}&0\\ 0&1&{ - 3}&0&0 \end{array}} \right]::\,\left\{ {{R_2} = \dfrac{{{R_2}}}{2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1/2}&0&0\\ 0&1&{ - 3/2}&{ - 1}&0\\ 0&0&{3/2}&{ - 1}&0\\ 0&0&{ - 3/2}&1&0 \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1/2}&0&0\\ 0&1&{ - 3/2}&{ - 1}&0\\ 0&0&1&{ - 2/3}&0\\ 0&0&{ - 3/2}&1&0 \end{array}} \right]::\left\{ {{R_2} = \dfrac{2}{3}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&{ - 1/3}&0\\ 0&1&0&{ - 2}&0\\ 0&0&1&{ - 2/3}&0\\ 0&0&0&0&0 \end{array}} \right]::\,\left\{ \begin{array}{l} {R_1} = {R_1} + \dfrac{1}{2}{R_3}\\ {R_2} = {R_2} + \dfrac{3}{2}{R_3}\\ {R_4} = {R_4} + \dfrac{3}{2}{R_3} \end{array} \right\} \end{array}\]
Step-4: System of equations
\[ \begin{array} { l } { x _ { 1 } - \dfrac { 1 } { 3 } x _ { 4 } = 0 } \\ { x _ { 2 } - 2 x _ { 4 } = 0 } \\ { x _ { 3 } - \dfrac { 2 } { 3 } x _ { 4 } = 0 } \end{array} \] On choosing, $x_4=3$: \[ \begin{array} { l } { x _ { 1 } = \dfrac { 1 } { 3 } \times 3 = 1 } \\ { x _ { 2 } = 2 ( 3 ) = 6 } \\ { x _ { 3 } = \dfrac { 2 } { 3 } ( 3 ) = 2 } \end{array} \] Therefore, the balanced equation is:
\[ \mathrm { B } _ { 2 } \mathrm { S } _ { 3 } + 6 \mathrm { H } _ { 2 } \mathrm { O } \rightarrow 2 \mathrm { H } _ { 3 } \mathrm { BO } _ { 3 } + 3 \mathrm { H } _ { 2 } \mathrm { S } \]
We are given with an unbalanced chemical equation: \[ \mathrm { B } _ { 2 } \mathrm { S } _ { 3 } + \mathrm { H } _ { 2 } \mathrm { O } \rightarrow \mathrm { H } _ { 3 } \mathrm { BO } _ { 3 } + \mathrm { H } _ { 2 } \mathrm { S } \] We have to balance the chemical equation:
Step-1:
Let the balance equation has the form \[ x _ { 1 } \cdot \mathrm { B } _ { 2 } \mathrm { S } _ { 3 } + x _ { 2 } \cdot \mathrm { H } _ { 2 } \mathrm { O } \rightarrow x _ { 3 } \cdot \mathrm { H } _ { 3 } \mathrm { BO } _ { 3 } + x _ { 4 } \cdot \mathrm { H } _ { 2 } \mathrm { S } \] There are four types of atoms, B, S, H, O
Step-2: The matrix form is
Write the matrix form using coefficient of each atom: \[ \mathrm { B } _ { 2 } \mathrm { S } _ { 3 } \left[ \begin{array} { l } { 2 } \\ { 3 } \\ { 0 } \\ { 0 } \end{array} \right] , \mathrm { H } _ { 2 } \mathrm { O } \left[ \begin{array} { l } { 0 } \\ { 0 } \\ { 2 } \\ { 1 } \end{array} \right] , \mathrm { H } _ { 3 } \mathrm { BO } _ { 3 } \left[ \begin{array} { l } { 1 } \\ { 0 } \\ { 3 } \\ { 3 } \end{array} \right] , \mathrm { H } _ { 2 } \mathrm { O } \left[ \begin{array} { l } { 0 } \\ { 2 } \\ { 0 } \\ { 1 } \end{array} \right] , \mathrm { H } _ { 2 } \mathrm { S } \left[ \begin{array} { l } { 0 } \\ { 1 } \\ { 2 } \\ { 0 } \end{array} \right] \] So, the system of equations with coefficients is given by: \[ x _ { 1 } \left[ \begin{array} { l } { 2 } \\ { 3 } \\ { 0 } \\ { 0 } \end{array} \right] + x _ { 2 } \left[ \begin{array} { l } { 0 } \\ { 0 } \\ { 2 } \\ { 1 } \end{array} \right] = x _ { 3 } \left[ \begin{array} { l } { 1 } \\ { 0 } \\ { 3 } \\ { 3 } \end{array} \right] + x _ { 4 } \left[ \begin{array} { l } { 0 } \\ { 1 } \\ { 2 } \\ { 0 } \end{array} \right] \]
Step-3: The Row-Reduced augmented matrix form
\[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 2&0&{ - 1}&0&0\\ 3&0&0&{ - 1}&0\\ 0&2&{ - 3}&{ - 2}&0\\ 0&1&{ - 3}&0&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{r}} 1&0&{ - 1/2}&0&0\\ 3&0&0&{ - 1}&0\\ 0&2&{ - 3}&{ - 2}&0\\ 0&1&{ - 3}&0&0 \end{array}} \right]::\,\,\left\{ {{R_1} = \dfrac{{{R_1}}}{2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1/2}&0&0\\ 0&0&{3/2}&{ - 1}&0\\ 0&2&{ - 3}&{ - 2}&0\\ 0&1&{ - 3}&0&0 \end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} - 3{R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{r}} 1&0&{ - 1/2}&0&0\\ 0&2&{ - 3}&{ - 2}&0\\ 0&0&{3/2}&{ - 1}&0\\ 0&1&{ - 3}&0&0 \end{array}} \right]::\,\,\left\{ {{R_2} \Leftrightarrow {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1/2}&0&0\\ 0&1&{ - 3/2}&{ - 1}&0\\ 0&0&{3/2}&{ - 1}&0\\ 0&1&{ - 3}&0&0 \end{array}} \right]::\,\left\{ {{R_2} = \dfrac{{{R_2}}}{2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1/2}&0&0\\ 0&1&{ - 3/2}&{ - 1}&0\\ 0&0&{3/2}&{ - 1}&0\\ 0&0&{ - 3/2}&1&0 \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1/2}&0&0\\ 0&1&{ - 3/2}&{ - 1}&0\\ 0&0&1&{ - 2/3}&0\\ 0&0&{ - 3/2}&1&0 \end{array}} \right]::\left\{ {{R_2} = \dfrac{2}{3}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&{ - 1/3}&0\\ 0&1&0&{ - 2}&0\\ 0&0&1&{ - 2/3}&0\\ 0&0&0&0&0 \end{array}} \right]::\,\left\{ \begin{array}{l} {R_1} = {R_1} + \dfrac{1}{2}{R_3}\\ {R_2} = {R_2} + \dfrac{3}{2}{R_3}\\ {R_4} = {R_4} + \dfrac{3}{2}{R_3} \end{array} \right\} \end{array}\]
Step-4: System of equations
\[ \begin{array} { l } { x _ { 1 } - \dfrac { 1 } { 3 } x _ { 4 } = 0 } \\ { x _ { 2 } - 2 x _ { 4 } = 0 } \\ { x _ { 3 } - \dfrac { 2 } { 3 } x _ { 4 } = 0 } \end{array} \] On choosing, $x_4=3$: \[ \begin{array} { l } { x _ { 1 } = \dfrac { 1 } { 3 } \times 3 = 1 } \\ { x _ { 2 } = 2 ( 3 ) = 6 } \\ { x _ { 3 } = \dfrac { 2 } { 3 } ( 3 ) = 2 } \end{array} \] Therefore, the balanced equation is:
\[ \mathrm { B } _ { 2 } \mathrm { S } _ { 3 } + 6 \mathrm { H } _ { 2 } \mathrm { O } \rightarrow 2 \mathrm { H } _ { 3 } \mathrm { BO } _ { 3 } + 3 \mathrm { H } _ { 2 } \mathrm { S } \]