Chapter 6: Problem 25
Each augmented matrix is in row echelon form and represents a linear system. Use back-substitution to solve the system if possible. $$\left[\begin{array}{rrr|r} 1 & 1 & -1 & 4 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 1 & 1 \end{array}\right]$$
Short Answer
Expert verified
The solution is \(x = 2\), \(y = 3\), \(z = 1\).
Step by step solution
01
Identify the equations
First, write down the linear equations that correspond to the given augmented matrix. The matrix\[\begin{bmatrix}1 & 1 & -1 & | & 4 \0 & 1 & -1 & | & 2 \0 & 0 & 1 & | & 1\end{bmatrix}\]represents the following system of equations:1. \(x + y - z = 4\)2. \(y - z = 2\)3. \(z = 1\).
02
Substitute and solve for the third equation
Since the third equation is \(z = 1\), we can directly substitute \(z = 1\) in the second and first equations. This step provides us the value of \(z\) right away.
03
Substitute for the second equation
Using the value of \(z=1\) in the second equation \(y - z = 2\), we substitute and solve for \(y\):\[y - 1 = 2\]Solving for \(y\), we get:\[y = 3\].
04
Substitute for the first equation
Now, use the values \(y = 3\) and \(z = 1\) in the first equation \(x + y - z = 4\):\[x + 3 - 1 = 4\]Simplify and solve for \(x\):\[x + 2 = 4\]\[x = 2\].
05
Write the solution
Having found all the variables, the solution to the system of equations is \(x = 2\), \(y = 3\), and \(z = 1\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Augmented Matrix
An augmented matrix is a powerful tool used to represent a system of linear equations in a compact form. This matrix combines the coefficients of the variables and the constants from the equations into one coherent structure. If you look at a typical linear equation system like:
- Equation 1: \(a_1x + b_1y + c_1z = d_1\)
- Equation 2: \(a_2x + b_2y + c_2z = d_2\)
- Equation 3: \(a_3x + b_3y + c_3z = d_3\)
Row Echelon Form
The Row Echelon Form, often abbreviated as REF, is a simplified kind of matrix that makes solving systems of linear equations easier using back-substitution. In this form, a matrix has:
- Each leading entry (the first non-zero number in a row from the left) in a row is to the right of the leading entry in the row above.
- The leading entry in any non-zero row is 1.
- Any row of all zeroes (if present) is at the bottom of the matrix.
- The first row begins with a 1, and the second row starts with a zero, then a 1 in the second column, following these rules.
- The third row begins with two zeros and then has a 1 in the third column.
Linear System
A linear system consists of multiple linear equations that involve the same set of variables. The primary objective of solving a linear system is to find the values of these variables that satisfy all equations simultaneously. Consider the system:
- Your first equation might be of the form \(x + y - z = 4\).
- Your second equation is \(y - z = 2\).
- Your third equation is \(z = 1\).