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In Exercises \(57-62,\) determine whether \(\mathbf{b}\) is in the column space of \(A\). If it is, write \(b\) as a linear combination of the column vectors of \(A\). $$A=\left[\begin{array}{rrr}-1 & -1 & 1 \\\1 & 0 & 1 \\\\-3 & -2 & 1\end{array}\right], \quad \mathbf{b}=\left[\begin{array}{r}0 \\\3 \\\\-3\end{array}\right]$$

Short Answer

Expert verified
The vector \(\mathbf{b}\) is in the column space of matrix \(A\) and can be expressed as the linear combination of the columns of \(A\), \(2\mathbf{a}_1 - 1\mathbf{a}_2 + 1\mathbf{a}_3\).

Step by step solution

01

Identify Column Vectors of A

The matrix \(A\) has three column vectors: \(\mathbf{a}_1 = [-1, 1, -3]^T\), \(\mathbf{a}_2 = [-1, 0, -2]^T\), and \(\mathbf{a}_3 = [1, 1, 1]^T\).
02

Set up System of Linear Equations

We write \(\mathbf{b}\) as a linear combination of the columns of \(A\), \(\mathbf{b} = x\mathbf{a}_1 + y\mathbf{a}_2 + z\mathbf{a}_3\), which equates to a system of linear equations. The system is: \[ \begin{align*} -x - y + z & = 0 \ x + z & = 3 \ -3x - 2y + z &= -3 \end{align*} \]
03

Solve the System of Equations

Solving the system of equations, we find that the values of \(x\), \(y\), and \(z\) are 2, -1, and 1 respectively. This indicates that the vector \(\mathbf{b}\) can be written as a linear combination of the columns of \(A\) which is \(2\mathbf{a}_1 - 1\mathbf{a}_2 + 1\mathbf{a}_3\). So, \(\mathbf{b}\) is in the column space of \(A\).

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Key Concepts

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

Understanding Linear Combination
A linear combination involves creating a new vector by multiplying each vector in a set by a scalar and then adding the results. It's a foundational idea used to explore sets of vectors and their capabilities. In the context of matrix operations, this concept is essential for understanding how the column vectors of a matrix relate to each other and to other vectors.

Take matrix A from our exercise. When we say vector b can be written as a linear combination of the column vectors of A, we mean that there's a way to multiply each column of A by some number (these numbers are called scalars) and, when these products are added together, they give us vector b. If such scalars exist, b is part of the column space of A—a space filled with all possible vectors you can get through linear combinations of A's columns.

In our solution, the scalars applied to the columns of A were found to be 2, -1, and 1. This discovery confirms that b is indeed a resident of the column space of A because the equation 2a1 - a2 + a3 = b has a solution.
Solving a System of Linear Equations
A system of linear equations is a collection of one or more linear equations involving the same set of variables. Think of it as a puzzle where you're trying to find the values for the variables that make all the equations true at once.

In our exercise, we're looking for scalars x, y, and z that satisfy the equations created by the linear combination of the columns of matrix A. We set each component of the resulting vector equal to the corresponding component of the vector b. This gives us our system. Solving it can be done using various methods, such as substitution, elimination, or matrix operations like row reduction.

The successful calculation of x=2, y=-1, and z=1 as the solutions to our system means that these values, when used as scalars in the linear combination of A's columns, will yield vector b. It's a clear demonstration of the interconnectedness of these concepts—linear combinations, vectors, and systems of equations—are all pieces of a larger mathematical picture.
Vector Representation in Column Space
Working with vector representation is a crucial part of understanding linear algebra, particularly when dealing with column space. It enables us to visualize and manipulate vectors in relation to the matrices they come from.

In the context of a matrix, each vector can be thought of as a direction or point in space. When we use vectors to form a matrix, such as the column vectors of matrix A from our exercise, we can start examining the 'space' those vectors span. This space is what's known as the column space of the matrix. All vectors in the column space can be represented as linear combinations of the original column vectors.

For vector b, its representation in the column space of A was confirmed when we solved the system of linear equations and found scalars that, when multiplied with A's columns and added together, equaled b. That means we can effectively place b in that space defined by A's columns, solidifying its representation within that matrix's spanned area.

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Most popular questions from this chapter

(a) find the transition matrix from \(B\) to \(B^{\prime},(b)\) find the transition matrix from \(B^{\prime}\) to \(B\) (c) verify that the two transition matrices are inverses of each other, and (d) find the coordinate matrix \([\mathrm{x}]_{n},\) given the coordinate matrix \([\mathbf{x}]_{B^{* *}}\) $$\begin{aligned}&B=\\{(1,1,1),(1,-1,1),(0,0,1)\\}\\\&B^{\prime}=\\{(2,2,0),(0,1,1),(1,0,1)\\}\\\&[\mathbf{x}]_{B^{r}}=\left[\begin{array}{l}2 \\\3 \\\1\end{array}\right]\end{aligned}$$

determine whether the set, together with the standard operations, is a vector space. If it is not, identify at least one of the ten vector space axioms that fails. The set of all quadratic functions whose graphs pass through the origin

Determine whether each vector is a scalar multiple of \(\mathbf{z}=\left(\frac{1}{2},-\frac{2}{3}, \frac{3}{4}\right)\) (a) \(\mathbf{u}=(6,-4,9)\) (b) \(\mathbf{v}=\left(-1, \frac{4}{3},-\frac{3}{2}\right)\)

Finding a Basis and Dimension In Exercises \(43-48\), find (a) a basis for and (b) the dimension of the solution space of the homogeneous system of linear equations. $$ \begin{array}{r} x-2 y+3 z=0 \\ -3 x+6 y-9 z=0 \end{array} $$

Determine whether each statement is true or false. If a statement is true, give a reason or cite an appropriate statement from the text. If a statement is false, provide an example that shows the statement is not true in all cases or cite an appropriate statement from the text. (a) If \(P\) is the transition matrix from a basis \(B^{\prime}\) to \(B\) then \(P^{-1}\) is the transition matrix from \(B\) to \(B^{\prime}\). (b) To perform the change of basis from a nonstandard basis \(B^{\prime}\) to the standard basis \(B\), the transition matrix \(P^{-1}\) is simply \(B^{\prime}\). (c) The coordinate matrix of \(p=-3+x+5 x^{2}\) relative to the standard basis for \(P_{2}\) is \([p]_{s}=\left[\begin{array}{lll}5 & 1 & -3\end{array}\right]^{T}\).

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