Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.6 Exercises - Page 55: 6

Answer

$2 . N a_{3} P O_{4}+3 . B a\left(N O_{3}\right)_{2} \rightarrow B a\left(P O_{4}\right)_{2}+6 . N a N O_{3}$

Work Step by Step

The following vectors list the numbers of atoms of sodium (Na), phosphorus (P), oxygen (O) barium (Ba), and nitrogen(N) \[ x_{1} \cdot\left[\begin{array}{c} 3 \\ 1 \\ 4 \\ 0 \\ 0 \end{array}\right]+x_{2} \cdot\left[\begin{array}{c} 0 \\ 0 \\ 6 \\ 1 \\ 2 \end{array}\right]=x_{3} \cdot\left[\begin{array}{c} 0 \\ 2 \\ 8 \\ 3 \\ 0 \end{array}\right]+x_{4} \cdot\left[\begin{array}{c} 1 \\ 0 \\ 3 \\ 0 \\ 1 \end{array}\right] \] The coefficients in the equation \[ x_{1} \cdot N a_{3} P O_{4}+x_{2} \cdot B a\left(N O_{3}\right)_{2} \rightarrow x_{3} \cdot B a\left(P O_{4}\right)_{2}+x_{4} \cdot N a N O_{3} \] Satisfy left side equation \[ \left[\begin{array}{ccccc} 3 & 0 & 0 & -1 & 0 \\ 1 & 0 & -2 & 0 & 0 \\ 4 & 6 & -8 & -3 & 0 \\ 0 & 1 & -3 & 0 & 0 \\ 0 & 2 & 0 & -1 & 0 \end{array}\right] \]Move the right terms to the left side (changing the sign of each entry in the third and fourth vectors) and row reduce the augmented matrix of the homogeneous system: 4 \[ \left[\begin{array}{ccccc} 1 & 0 & -2 & 0 & 0 \\ 3 & 0 & 0 & -1 & 0 \\ 4 & 6 & -8 & -3 & 0 \\ 0 & 1 & -3 & 0 & 0 \\ 0 & 2 & 0 & -1 & 0 \end{array}\right] \] Swapping row 1 and row 3 5 \[ \left[\begin{array}{ccccc} 1 & 0 & -2 & 0 & 0 \\ 0 & 0 & 6 & -1 & 0 \\ 0 & 6 & 0 & -3 & 0 \\ 0 & 1 & -3 & 0 & 0 \\ 0 & 2 & 0 & -1 & 0 \end{array}\right] \] 6 \[ \left[\begin{array}{ccccc} 1 & 0 & -2 & 0 & 0 \\ 0 & 1 & -3 & 0 & 0 \\ 0 & 6 & 0 & -3 & 0 \\ 0 & 0 & 6 & -1 & 0 \\ 0 & 2 & 0 & -1 & 0 \end{array}\right] \] \[ \left[\begin{array}{ccccc} 1 & 0 & -2 & 0 & 0 \\ 0 & 1 & -3 & 0 & 0 \\ 0 & 0 & 18 & -3 & 0 \\ 0 & 0 & 6 & -1 & 0 \\ 0 & 0 & 6 & -1 & 0 \end{array}\right] \] Taking 18 out as common from row 3 and \[ R_{5}=R_{5}-R_{4} \] The general solution is \[ \begin{array}{l} x_{1}=(1 / 3) x_{4} \\ x_{2}=(1 / 2) x_{4} \\ x_{3}=(1 / 6) x_{4} \end{array} \] with $x_{4}$ free. Take $x_{4}=6 .$ Then $x_{1}=2, x_{2}=3$ and $x_{3}=1 .$ The balanced equation is $2 . N a_{3} P O_{4}+3 . B a\left(N O_{3}\right)_{2} \rightarrow B a\left(P O_{4}\right)_{2}+6 . N a N O_{3}$
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