Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 1 - Systems of Linear Equations - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises: 73


There are infinitely many solutions to this system and they are given by $$x=\frac{2}{5-t};\quad y=\frac{1}{-1+4t};\quad z=\frac{1}{t}$$ where $t\in R/\{5,\frac{1}{4},0\}.$

Work Step by Step

As we are instructed in the problem we will put $A=1/x$, $B=1/y$ and $C=1/z$. The we get a system of LINEAR equations in $A$, $B$ and $C$: \begin{align*} 2A+B-3C=&4\\ 4A\qquad+2C=&10\\ -2A+3B-13C=&-8 \end{align*} Now let's solve this system by transforming it to Row-Echelon form: Step 1: Substitute the second and the third equation and rewrite the term on order of appearing $B$. then $A$ and then $C$: \begin{align*} B+2A-3C=&4\\ 3B-2A-13C=&-8\\ 4A+2C=&10. \end{align*} Step 2: To eliminate $B$ from the second equation add to it term by term the first one multiplied by $3$: $$3B-3B-2A-3\times2A-13C-3\times(-3C) = -8-3\times4$$ Which becomes $$-8A-4C=-20.$$ This can be simplified by dividing with $-4$: $$2A+C=5.$$ We can also simplify the last equation by dividing wiht $2$. Now our system reads: \begin{align*} B+2A-3C=&4\\ 2A+C=&5\\ 2A+C=&5. \end{align*} Step 3: Note that the last two equations are identical so we can just drop the third one. Now we end up with the system of $2$ equations with three unknowns and we will have infinitely many solutions. To find their parametrized form equate one of the variables, let's say $C$ with $t$: \begin{align*} B+2A-3C=&4\\ 2A+C=&5\\ C=t. \end{align*} Step 4: Put $C=t$ into the second equation and express $A$ in terms of $t$: $$2A+t=5\Rightarrow 2A=5-t$$ and that gives $$A=\frac{5}{2}-\frac{1}{2}t.$$ Step 5: Now put $C=t$ and $A=frac{5}{2}-\frac{1}{2}t$ into the first equation and express $B$ in terms of $t$: $$B+2\times\left(\frac{5}{2}-\frac{1}{2}t\right)-3t=4$$ which becomes $$B+5-t-3t=4\Rightarrow B+5-4t=4$$ and this gives $$B=-1+4t.$$ Now we have to solve for $x$, $y$ and $z$: Since $$A=\frac{1}{x}$$ then $$x=\frac{1}{A} =\frac{1}{\frac{5}{2}-\frac{1}{2}t} = \frac{2}{5-t}.$$ Since $$B=\frac{1}{y}$$ then $$y=\frac{1}{B} = \frac{1}{-1+4t}.$$ Since $$C=\frac{1}{z}$$ then $$z=\frac{1}{C} = \frac{1}{t}.$$ IMPORTANT NOTE: Parameter $t$ cannot take all real values. This is because the expressions for $x$, $y$ and $z$ in terms of $t$ have to be valid expressions so their denominator must not be zero! This means that $$5-t\neq 0,\quad -1+4t\neq 0, \quad t\neq0$$ This means that $$t\neq 5\text{ and } t\neq \frac{1}{4} \text{ and } t\neq 0$$ so $t$ can take any real value but not the three listed above.
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