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 - Page 11: 51


This system has infinitely many solutions and all of them are given by: $$x_1 = \frac{1}{2}(5-t),\quad x_2=4t-1;\quad x_3 = t$$ where $t$ is any real number.

Work Step by Step

We will transform this system into the row-echelon form performing the following steps: Step 1: Substitute the last two equations and rewrite all of the equations reordering the terms by putting $x_1$ in the second place and $x_2$ in the first place: \begin{align*} x_2 + 2x_1-3x_3=&4\\ 3x_2-2x_1-13x_3 = &-8\\ 4x_1+2x_3=&10 \end{align*} Step 2: Add to the second equation the first one multiplied by $-3$ to eliminate $x_2$ from the second equation: The second equation will be $$3x_2 - 3x_2 - 2x_1-6x_1-13x_3+9x_3 = -8-3\times 4$$ which is rewritten as $$-8x_1-4x_3=-20$$ we can simplify this by dividing it by $-4$: $$2x_1+x_3=5$$ Rewriting all of the equations we are left at: \begin{align} x_2+2x_1-3x_3 =& 4\\ 2x_1+x_3 =&5\\ 4x_1+2x_3=&10 \end{align} Step 3: Divide the last equation by $2$: \begin{align} x_2+2x_1-3x_3 =& 4\\ 2x_1+x_3 =&5\\ 2x_1+x_3=&5 \end{align} Step 4: We see that the last two equations are exactly identical. This means that we actually have $2$ conditions with $3$ unknowns. This means that we can equate one of them with parameter $t$, lets say $x_3=t$ \begin{align} x_2+2x_1-3x_3 =& 4\\ 2x_1+x_3 =&5\\ x_3= &t \end{align} Step 5: We can express both $x_1$ and $x_2$ in terms of $t$. From example substitute $x_3$ for $t$ in the second equation to obtain $$x_1 = \frac{1}{2}(5-t).$$ Now substituting both $x_1$ and $x_3$ in the second equation we express $x_2$ in terms of $t$: $$x_2 = 4t-1$$ Step 6: Just summarizing all we did we have that this system has infinitely many solutions given by $$x_1 = \frac{1}{2}(5-t),\quad x_2=4t-1;\quad x_3 = t$$ where $t$ is any real number.
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