Linear Algebra and Its Applications (5th Edition)

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

Chapter 1 - Linear Equations in Linear Algebra - 1.1 Exercises: 16

Answer

The system is consistent. Infinitely many solutions will exist.

Work Step by Step

The procedure is shown with the matrix notation for simpler understanding. $x_{1} - 2x_{4} = -3$ $2x_{2} + 2x_{3} = 0$ $x_{3} + 3x_{4} = 1$ $-2x_{1} + 3x_{2} + 2x_{3} + x_{4} = 5$ This can be depicted in the augmented matrix notation as follows : $\begin{bmatrix} 1 & 0 & 0 & -2 & -3\\ 0 & 2 & 2 & 0 & 0\\ 0 & 0 & 1& 3 & 1\\ -2 & 3 & 2 & 1 & 5 \end{bmatrix}$ To eliminate the $-2x_{1}$ term in the fourth row, add $2$ times row 1 to row 4: $\begin{bmatrix} 1 & 0 & 0 & -2 & -3\\ 0 & 2 & 2 & 0 & 0\\ 0 & 0 & 1& 3 & 1\\ 0 & 3 & 2 & -3 & -1 \end{bmatrix}$ To eliminate the $3x_{2}$ term in the fourth row, add $-3/2$ times row 2 to row 4: $\begin{bmatrix} 1 & 0 & 0 & -2 & -3\\ 0 & 2 & 2 & 0 & 0\\ 0 & 0 & 1& 3 & 1\\ 0 & 0 & -1 & -3 & -1 \end{bmatrix}$ To eliminate the $-x_{3}$ term in the fourth row, add $1$ times row 3 to row 4: $\begin{bmatrix} 1 & 0 & 0 & -2 & -3\\ 0 & 2 & 2 & 0 & 0\\ 0 & 0 & 1& 3 & 1\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ $x_{1} - 2x_{4} = -3 $ $2x_{2} + 2x_{3} = 0$ $x_{3} + 3x_{4} = 1$ $0x_{4} = 0$ Now, we know that the last equation is true for infinite values of $x_{4}$. Back substituting these values into the other equations, we can find that there will be infinite subsequent values of $x_{3}$, $x_{2}$ and $x_{1}$ for each value of $x_{4}$. Hence, infinitely many solutions exist. Thus, the system is consistent.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.