Vectors as Solutions of Linear Solutions

Section 1.4 Vectors as Solutions of Linear Solutions

In Section 1.3, we saw solutions of linear systems that either had or did not have a free variable. In both of these cases, we can write the solutions in another form using vectors. In this section,

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We revisit vectors and discuss sums and scalar multiplication of vectors.
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Write a solution without a free variable as a vector.
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Write a solution with free variables(s) in vector form.
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Subsection 1.4.1 Vectors as Solutions of Linear Systems

In Subsection 1.2.2 we were introduced to matrices as a rectangular grid of numbers. Recall that if the matrix has only a single column, we called it column vector. As we will see in the first few chapters of this text, vectors are very important so we explore these a bit more here and then later in Chapter 2. A column vector \(\vec{s}\) can be written

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\begin{equation} \vec{s} = \begin{bmatrix} s_1 \\ s_2 \\ \vdots \\ s_n \end{bmatrix}\tag{1.4.1} \end{equation}

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with \(s_1, s_2, \ldots, s_n\) real numbers is a solution to the linear equation

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Definition 1.4.1.

The vector in (1.4.1) is a solution to a linear equation

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\begin{equation*} a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b \end{equation*}

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\begin{equation*} a_1 s_1 + a_2 s_2 + \cdots + a_n s_n = b. \end{equation*}

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A vector is a solution to a linear system if the points satisfy each equation in the solution. We consider the vector in (1.4.1) equivalent to the tuple \((s_1, s_2, \ldots, s_n)\text{.}\)

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Subsection 1.4.2 Addition and Scalar Multiplication of Vectors

Before examining vectors as solutions of linear system, we need to define vector sum and scalar multiplication.

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Definition 1.4.2.

The sum of two vectors \(\vec{u}\) and \(\vec{v}\) is defined as

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\begin{equation*} \vec{u} + \vec{v} = \begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{bmatrix} + \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} = \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \\ \vdots \\ u_n + v_n \end{bmatrix} \end{equation*}

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and the scalar multiplication of a vector \(\vec{u}\) by \(r \in \mathbb{R}\) is

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\begin{equation*} r \vec{u} = r \begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{bmatrix} = \begin{bmatrix} r u_1 \\ r u_2 \\ \vdots \\r u_n \end{bmatrix} \end{equation*}

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Example 1.4.3.

Let

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\begin{align*} \vec{u} \amp = \begin{bmatrix} 3 \\ -3 \\ 4 \end{bmatrix}, \amp \vec{v} \amp = \begin{bmatrix} 2 \\ 3/2 \\ 9 \end{bmatrix}. \end{align*}

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Find \(\vec{u}+\vec{v}\) and \(4 \vec{v}\text{.}\)

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Solution.

\begin{align*} \vec{u} + \vec{v} \amp = \begin{bmatrix} 3+2 \\ -3 + 3/2 \\ 4+9 \end{bmatrix} = \begin{bmatrix} 5 \\ -3/2 \\ 13 \end{bmatrix}, \amp 4 \vec{v} \amp = \begin{bmatrix} 4 \cdot 2, \\ 4 \cdot \frac{3}{2}, \\ 4 \cdot 9 \end{bmatrix} = \begin{bmatrix} 8 \\ 6 \\ 36 \end{bmatrix} \end{align*}

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Subsection 1.4.3 Writing Solutions in Vector Form

There are a couple of advantages to writing a solution to a linear system in vector form. We will see some of them later in the course, however, right now, we can write down specific points that are in the solution set quite easily. There’s little advantage to using vectors for unique and no solution linear systems, but when you have multiple (infinite number of) solutions, then vectors shine through.

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Example 1.4.4.

Consider the linear system from Subsection 1.3.4 in reduced row echelon form:

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\begin{align*} x\phantom{+2y} + 2z \amp = -6\\ y + 4z \amp = -13 \\ 0 \amp = 0 \end{align*}

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Recall that the solution in terms of \(z\) as

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\begin{equation*} \{ (-2z-6, -4z-13,z) \; | \; z \in \mathbb{R} \} \end{equation*}

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Write the solution in terms of a vector.

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Solution.

There is one parameter in this case \(z\) and thus the solution can be written as

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\begin{align*} \begin{bmatrix} -2z-6, \\ -4z-13 \\ z \end{bmatrix} \amp = \begin{bmatrix} -2z \\ -4z \\ z \end{bmatrix} + \begin{bmatrix} -6 \\ -13 \\ 0 \end{bmatrix} \\ \amp = \begin{bmatrix} -2 \\ -4 \\ 1 \end{bmatrix} z + \begin{bmatrix} -6 \\ -13 \\ 0 \end{bmatrix} \end{align*}

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And this form makes it quite easy to find individual points. If we let \(z=0\text{,}\) then \((-6,-13,0)\) is a point in the solution set, if let \(z=-3\text{,}\) then \((0,-1,-3)\) is a point.

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Note that it takes a little work to get a linear system in echelon form into a solution in vector form. We will see in the next section that there is another form that the linear system (matrix) can be put into for an easier transition to this form.

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Example 1.4.5.

Write the solution to the linear system that we saw in Example Example 1.3.13

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\begin{align*} x_1\phantom{+2x_3} + 3x_3 -9 x_4 + 11 x_5 \amp = 14, \\ 2x_3 \phantom{-9x_4} +\phantom{1} 4x_5 \amp = 10, \\ 3x_5 \amp = 27, \end{align*}

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in vector form.

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Solution.

In example Example 1.3.13, the solution set was found to be

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\begin{equation*} \{ (x_1,x_2,x_3,x_4,x_5) = (9x_4-46,x_2,-13,x_4,9) \; | \; x_2, x_4 \in \mathbb{R}\} \end{equation*}

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with the free variables \(x_2\) and \(x_4\text{.}\) This can be written as a vector as

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\begin{equation*} \begin{bmatrix} 9x_4 - 46 \\ x_2 \\ -13 \\ x_4 \\ 9 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\0 \end{bmatrix} x_2 + \begin{bmatrix} 9 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} x_4 + \begin{bmatrix} -46 \\ 0 \\ -13 \\ 0 \\ 9 \end{bmatrix} \end{equation*}

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Thus the solution can be written:

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\begin{equation*} \left\{ \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\0 \end{bmatrix} x_2 + \begin{bmatrix} 9 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} x_4 + \begin{bmatrix} -46 \\ 0 \\ -13 \\ 0 \\ 9 \end{bmatrix} \; | \; x_2, x_4 \in \mathbb{R} \right\}. \end{equation*}

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This is the most general form of the solution to the linear system in this example and as before one can write down solutions with specific values of free variables. For example, if \(x_2\) and \(x_4\) are both 0, the the point \((-46,0,-13,0,9)\) is a solution to the linear system. (Try it!) This is an example of a particular solution as we define below.

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Definition 1.4.6.

Consider a linear system. If the point \((s_1,s_2,\ldots,s_n)\) is a solution to the system and \(s_1, s_2, \ldots, s_n\) do not have any free variables, then the point \((s_1,s_2,\ldots,s_n)\) is called a particular solution.

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A few things of note before we go on:

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After the first two sections of the text, you should know how to do Gauss’s method on any linear system and be able to write down the solution set in either set builder notation or as vectors.
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Gauss’ method states that any of the 3 elementary row operations results in the same solution. Since there is no algorithm to apply to a linear system (or matrix), if we use 2 different set of row operations, do we get the same solution (or at least the same set of free variables)?
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