Chapter 2: Problem 23
Solve the given system by back substitution. $$\begin{aligned}x_{1}+x_{2}-x_{3}-x_{4} &=1 \\\x_{2}+x_{3}+x_{4} &=0 \\\x_{3}-x_{4} &=0 \\\x_{4} &=1\end{aligned}$$
Short Answer
Expert verified
\(x_1 = 5\), \(x_2 = -2\), \(x_3 = 1\), \(x_4 = 1\).
Step by step solution
01
Solve for the simplest variable
Begin with the last equation, which is the simplest. We have \(x_4 = 1\).
02
Substitute into the third equation
Take the value of \(x_4\) from Step 1 and substitute it into the third equation \(x_3 - x_4 = 0\). This gives us \(x_3 - 1 = 0\), so \(x_3 = 1\).
03
Substitute into the second equation
Now substitute \(x_3 = 1\) and \(x_4 = 1\) into the second equation \(x_2 + x_3 + x_4 = 0\). This becomes \(x_2 + 1 + 1 = 0\), so \(x_2 = -2\).
04
Substitute into the first equation
Substitute \(x_2 = -2\), \(x_3 = 1\), and \(x_4 = 1\) into the first equation \(x_1 + x_2 - x_3 - x_4 = 1\). This results in \(x_1 - 2 - 1 - 1 = 1\), which simplifies to \(x_1 - 4 = 1\). Therefore, \(x_1 = 5\).
05
Solution Completed
The values satisfying the system are \(x_1 = 5\), \(x_2 = -2\), \(x_3 = 1\), and \(x_4 = 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Linear Systems
A **linear system** is essentially a collection of linear equations. Each equation represents a straight line, and together, they describe relationships between different variables. In our example, we have a system of four linear equations with four unknowns: \(x_1, x_2, x_3,\) and \(x_4\). Each equation involves sums and differences of these variables.
To solve a linear system means to find values for these variables that all equations hold true simultaneously. Solving such a system systematically often involves turning these complex relationships into a form that is easier to manage, such as by using **Gaussian Elimination** or applying the method known as **Back Substitution**.
To solve a linear system means to find values for these variables that all equations hold true simultaneously. Solving such a system systematically often involves turning these complex relationships into a form that is easier to manage, such as by using **Gaussian Elimination** or applying the method known as **Back Substitution**.
- The goal is to isolate and solve for one variable at a time.
- Start with the most simplified equation and work backwards.
- This process demonstrates how each equation can be used to find variable values step by step.
Unraveling Gaussian Elimination
**Gaussian Elimination** is a method typically used for solving linear systems and is especially useful when dealing with multiple equations and variables. The process involves three main steps: creating an augmented matrix, performing row operations, and performing back substitution. Here’s a breakdown:
- **Creating an Augmented Matrix:** Convert your linear system into a matrix form. Each row represents an equation, with coefficients of variables followed by the constant.
- **Row Operations:** These are systematic transformations you apply to simplify your system. You aim to get zeros in the lower-left part of the matrix, working towards what is known as **row-echelon form** or **reduced row-echelon form**.
- **Back Substitution:** Once the matrix is simplified, you can solve for the unknowns, starting with the simplest equation and substituting the found solutions back into the previous ones.
Solving with Matrix Equations
A **matrix equation** is another elegant way to represent a linear system. By rewriting the system of equations in matrix form, we can use more advanced mathematical techniques to find solutions. The matrix equation can be formulated as \[A\mathbf{x} = \mathbf{b}\]where \(A\) is the matrix of coefficients, \(\mathbf{x}\) is the vector of unknowns, and \(\mathbf{b}\) is the constants vector.
- For our example, the matrix \(A\) would be formed by aligning the coefficients of all the variables for each equation into rows.
- The vector \(\mathbf{x}\) would include our unknowns, \([x_1, x_2, x_3, x_4]\).
- Vector \(\mathbf{b}\) would contain the constants on the right side of each equation.