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Chapter 1: Problem 7

In Exercises \(7-12\) , describe all solutions of \(A \mathbf{x}=0\) in parametric vector form, where \(A\) is row equivalent to the given matrix. $$ \left[\begin{array}{cccc}{1} & {3} & {-3} & {7} \\ {0} & {1} & {-4} & {5}\end{array}\right] $$

Short Answer

Expert verified
The solution is \(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = t\begin{bmatrix} -9 \\ 4 \\ 1 \\ 0 \end{bmatrix} + s\begin{bmatrix} 8 \\ -5 \\ 0 \\ 1 \end{bmatrix}\), where \(t, s\) are parameters.

Step by step solution

01

Understand the Matrix and System

We are given that matrix \(A\) is row equivalent to another matrix. The task is to solve the homogeneous system \(A\mathbf{x} = \mathbf{0}\) for the row equivalent form: \[ \left[\begin{array}{cccc}{1} & {3} & {-3} & {7} \ {0} & {1} & {-4} & {5}\end{array}\right] \]This means we want to find all vectors \(\mathbf{x}\) that make this matrix equation true.
02

Write the System of Equations

The matrix \( \left[\begin{array}{cccc}{1} & {3} & {-3} & {7} \ {0} & {1} & {-4} & {5}\end{array}\right] \) represents the following system of equations:\[1x_1 + 3x_2 - 3x_3 + 7x_4 = 0 \1x_2 - 4x_3 + 5x_4 = 0\]
03

Express Variables in Terms of Free Variables

Here, notice that \(x_3\) and \(x_4\) can be considered the free variables because they do not have leading coefficients in each row. Solve the equations in terms of these variables:- From the second equation: \(x_2 = 4x_3 - 5x_4\)- Substitute \(x_2\) into the first equation: \[ x_1 + 3(4x_3 - 5x_4) - 3x_3 + 7x_4 = 0 \] \[ x_1 + 12x_3 - 15x_4 - 3x_3 + 7x_4 = 0 \] \[ x_1 + 9x_3 - 8x_4 = 0 \] Thus, \(x_1 = -9x_3 + 8x_4\).
04

Write the Solution in Parametric Vector Form

The solution to the system can be expressed with the free variables \(x_3\) and \(x_4\). Let:\[ x_3 = t \quad \text{and} \quad x_4 = s \]Where \(t\) and \(s\) are parameters that can take any real number.Thus, \[\begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 \end{bmatrix} =\begin{bmatrix} -9t + 8s \ 4t - 5s \ t \ s \end{bmatrix} = t\begin{bmatrix} -9 \ 4 \ 1 \ 0 \end{bmatrix} + s\begin{bmatrix} 8 \ -5 \ 0 \ 1 \end{bmatrix}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Vector Form
In a homogeneous system like the one given, finding the parametric vector form allows us to express all potential solutions. After identifying the free variables, we assign a parameter to each (e.g., let \( x_3 = t \) and \( x_4 = s \)). This transforms the general solution into a linear combination of vectors.
  • The solution vector \( \mathbf{x} \) is expressed as a sum of products of the parameters and vectors.
  • In this example, we constructed \(\mathbf{x} = t\begin{bmatrix} -9 \ 4 \ 1 \ 0 \\end{bmatrix} + s\begin{bmatrix} 8 \ -5 \ 0 \ 1 \\end{bmatrix}\).
This form clearly shows how any solution is a linear combination of two vector expressions, modulated by the free variables. Each vector in the combination is multiplied by its corresponding parameter, \( t \) or \( s \). This representation is invaluable as it provides a complete picture of all solutions in the system.
Free Variables
Free variables are the crux in finding all solutions of a homogeneous system. They exist in a system whenever a column in a row-reduced matrix lacks a leading 1.
  • In our example, \( x_3 \) and \( x_4 \) are free variables.
  • These variables offer flexibility, enabling infinite solutions.
To find the parametric solution, express the leading variables \( (x_1, x_2) \) in terms of the free ones \( (x_3, x_4) \).By assigning parameters (such as \( t \) or \( s \)) to free variables, we interpret any potential solution in vector form. This approach simplifies solving the system by reducing it to terms of independent variables, providing a scalable method to handle larger systems.
Row Reduction
Row reduction is a key technique for simplifying matrix systems to understand the nature of their solutions, often used to convert a matrix to row-echelon or reduced row-echelon form.
  • In this process, elementary row operations are applied to a matrix.
  • The goal is to isolate pivot positions, ideally leading one in each row.
For our given matrix, row reduction helps us quickly identify free variables and organize the system's equations. By rearranging and simplifying, row reduction breaks down complex systems into easier, more understandable terms. This step is preparatory, laying the groundwork for expressing the final solution in parametric vector form.
System of Linear Equations
A system of linear equations consists of multiple linear equations involving the same set of variables. In this exercise, we are working with a homogeneous system which means all equations are set to zero.
  • This type of system always includes the trivial solution where all variables equal zero.
  • However, our goal is to determine if there are non-trivial solutions.
The given matrix simplifies to this system: \( \begin{aligned}x_1 + 3x_2 - 3x_3 + 7x_4 &= 0 \x_2 - 4x_3 + 5x_4 &= 0\end{aligned} \)Each equation represents a plane in multi-dimensional space. Analyzing solutions involves understanding how these planes intersect, leading possibly to infinite solutions influenced by free variables. This allows us to express all solutions in a structured form using parameters.

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Most popular questions from this chapter

In a certain region, about 6% of a city鈥檚 population moves to the surrounding suburbs each year, and about 4% of the suburban population moves into the city. In 2015, there were 10,000,000 residents in the city and 800,000 in the suburbs. Set up a difference equation that describes this situation, where \(\mathbf{x}_{0}\) is the initial population in 2015 . Then estimate the populations in the city and in the suburbs two years later, in 2017 .

Exercises \(17-20\) refer to the matrices \(A\) and \(B\) below. Make appropriate calculations that justify your answers and mention an appropriate theorem. $$ A=\left[\begin{array}{rrrr}{1} & {3} & {0} & {3} \\ {-1} & {-1} & {-1} & {1} \\\ {0} & {-4} & {2} & {-8} \\ {2} & {0} & {3} & {-1}\end{array}\right] \quad B=\left[\begin{array}{rrrr}{1} & {3} & {-2} & {2} \\ {0} & {1} & {1} & {-5} \\\ {1} & {2} & {-3} & {7} \\ {-2} & {-8} & {2} & {-1}\end{array}\right] $$ Can every vector in \(\mathbb{R}^{4}\) be written as a linear combination of the columns of the matrix \(B\) above? Do the columns of \(B\) span \(\mathbb{R}^{3} ?\)

In a certain region, about 7\(\%\) of a city's population moves to the surrounding suburbs each year, and about 5\(\%\) of the suburban population moves into the city. In 2015 , there were \(800,000\) residents in the city and \(500,000\) in the suburbs. Set up a difference equation that describes this situation, where \(\mathbf{x}_{0}\) is the initial population in 2015 . Then estimate the populations in the city and in the suburbs two years later, in 2017 . (Ignore other factors that might influence the population sizes.)

Suppose an \(m \times n\) matrix \(A\) has \(n\) pivot columns. Explain why for each \(\mathbf{b}\) in \(\mathbb{R}^{m}\) the equation \(A \mathbf{x}=\mathbf{b}\) has at most one solution. [Hint: Explain why \(A \mathbf{x}=\mathbf{b}\) cannot have infinitely many solutions.]

Let \(T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) be a linear transformation such that \(T\left(x_{1}, x_{2}\right)=\left(x_{1}+x_{2}, 4 x_{1}+5 x_{2}\right) .\) Find \(\mathbf{x}\) such that \(T(\mathbf{x})=\) \((3,8) .\)
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