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Determine whether the given set of vectors is linearly independent. If linearly dependent, find a linear relation among them. The vectors are written as row vectors to save space, but may be considered as column vectors; that is, the transposes of the given vectors may be used instead of the vectors themselves. $$ \begin{array}{l}{\mathbf{x}^{(1)}=(1,2,2,3), \quad \mathbf{x}^{(2)}=(-1,0,3,1), \quad \mathbf{x}^{(3)}=(-2,-1,1,0)} \\\ {\mathbf{x}^{(4)}=(-3,0,-1,3)}\end{array} $$

Short Answer

Expert verified
If they are linearly dependent, find a linear relation among them. Answer: The given vectors are linearly independent.

Step by step solution

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01

Form the augmented matrix

Form the augmented matrix A by writing the given vectors as rows of a matrix: $$ A = \begin{bmatrix} 1 & 2 & 2 & 3 \\ -1 & 0 & 3 & 1 \\ -2 & -1 & 1 & 0 \\ -3 & 0 & -1 & 3 \end{bmatrix} $$
02

Row reduce the matrix using Gaussian elimination

Perform Gaussian elimination on the matrix A to row reduce it into row-echelon form: $$ \begin{bmatrix} 1 & 2 & 2 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & 2/3 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$
03

Determine linear independence

The row-reduced augmented matrix has a pivot element in each row, which means the given vectors are linearly independent. So, the given set of vectors \(\mathbf{x}^{(1)}, \mathbf{x}^{(2)}, \mathbf{x}^{(3)}, \text{and}\ \mathbf{x}^{(4)}\) is linearly independent.

Key Concepts

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

Gaussian elimination
Imagine you come across a maze of numbers in a matrix and your challenge is to clear a path to simplicity. This is where Gaussian elimination shines—it's a method we use in linear algebra to simplify matrices, a step towards understanding systems of linear equations or examining the properties of vectors like linear independence.

Think of it like tidying up a room. We start with a cluttered space and systematically put things in order until we can easily see what's what. In the world of matrices, this means performing row operations like swapping two rows, multiplying a row by a non-zero number, and adding multiples of one row to another to reduce the matrix to a simpler form. In the context of our exercise, Gaussian elimination was the tool that helped us clean up the matrix formed by vectors. It transformed a daunting array of numbers into a tidy, row-reduced echelon form, revealing whether our vectors were stepping on each other's toes (linearly dependent) or marching to their own beat (linearly independent).
Pivot elements
In a grand dance of numbers, pivot elements are the leading performers on the matrix stage. These are the first non-zero numbers in each row of a matrix that has been primped and preened through Gaussian elimination. They are like beacons that guide us through the sea of zeros to determine the matrix's rank and the solutions to a system of equations.

When looking for linear independence among vectors, finding a pivot in every row during Gaussian elimination is a lot like finding a key for every lock—each vector brings something unique to the table, and no vector can be formed by a combination of the others. In our exercise, the appearance of pivot elements in each row of the row-reduced matrix was the decisive moment, as it confirmed the vectors' independence.
Row-reduced echelon form
The row-reduced echelon form of a matrix is like its fingerprint—unique and revealing. It's the end result of diligent Gaussian elimination where each row starts with a pivot, and all the pivot columns are cleared out above and below these leading 1s. It's a streamlined version of the original matrix from which we can read off solutions to equations or identify linearly independent vectors.

Creating the row-reduced echelon form is akin to solving a mystery; it strips away the unnecessary and highlights the structure that was hidden in plain sight. In the exercise, our detective work through Gaussian elimination led us to this form, proving without a shadow of a doubt that our vectors were independent, as if each had its own unique DNA strand.
Vectors in linear algebra
In the realm of linear algebra, vectors are not just arrows in space—they are essential characters in the story of multidimensional geometry. Vectors can represent points, movements, forces, and various other quantities in mathematics and science.

When we speak of vectors relative to linear independence, we're essentially looking at whether a specific set of vectors can be used to reach any point in their span without redundancy. Linear independence means that each vector in a set has its role, and none can be booted out without changing what can be reached or described by that group. It's like a recipe where each ingredient is crucial—omit one, and the dish falls apart. Our given vectors' ability to stand alone demonstrated that they indeed each add a distinct flavor to the mathematical concoction.

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Most popular questions from this chapter

Find the solution of the given initial value problem. Draw the trajectory of the solution in the \(x_{1} x_{2}-\) plane and also the graph of \(x_{1}\) versus \(t .\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{cc}{-\frac{5}{2}} & {\frac{3}{2}} \\\ {-\frac{3}{2}} & {\frac{1}{2}}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{c}{3} \\ {-1}\end{array}\right) $$

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{2} & {-1} \\ {3} & {-2}\end{array}\right) \mathbf{x} $$

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{cc}{1} & {\sqrt{3}} \\ {\sqrt{3}} & {-1}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{e^{\prime}} \\\ {\sqrt{3} e^{-t}}\end{array}\right) $$

In each of Problems 13 through 20 the coefficient matrix contains a parameter \(\alpha\). In each of these problems: (a) Determine the eigervalues in terms of \(\alpha\). (b) Find the critical value or values of \(\alpha\) where the qualitative nature of the phase portrait for the system changes. (c) Draw a phase portrait for a value of \(\alpha\) slightly below, and for another value slightly above, each crititical value. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\alpha} & {1} \\ {-1} & {\alpha}\end{array}\right) \mathbf{x} $$

Find the solution of the given initial value problem. Draw the trajectory of the solution in the \(x_{1} x_{2}-\) plane and also the graph of \(x_{1}\) versus \(t .\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{3} & {9} \\ {-1} & {-3}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{l}{2} \\ {4}\end{array}\right) $$

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