/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Construct a \(3 \times 3\) nonze... [FREE SOLUTION] | 91Ó°ÊÓ

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Construct a \(3 \times 3\) nonzero matrix \(A\) such that the vector \(\left[\begin{array}{l}{1} \\ {1} \\ {1}\end{array}\right]\) is a solution of \(A \mathbf{x}=\mathbf{0}\) .

Short Answer

Expert verified
Matrix \( A \) can be \( \begin{bmatrix} -1 & -1 & 2 \\ -1 & 2 & -1 \\ 2 & -1 & -1 \end{bmatrix} \).

Step by step solution

01

Understand the problem

We need to find a nonzero matrix \( A \) such that the vector \( \mathbf{x} = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \) is a solution to the homogeneous equation \( A \mathbf{x} = \mathbf{0} \). This implies that the vector \( \mathbf{x} \) must lie in the null space of \( A \).
02

Construct the equation for a matrix

Since \( A \mathbf{x} = \mathbf{0} \), this can be broken down into a system: 1. Row 1: \( a_{11} \cdot 1 + a_{12} \cdot 1 + a_{13} \cdot 1 = 0 \) 2. Row 2: \( a_{21} \cdot 1 + a_{22} \cdot 1 + a_{23} \cdot 1 = 0 \) 3. Row 3: \( a_{31} \cdot 1 + a_{32} \cdot 1 + a_{33} \cdot 1 = 0 \).
03

Create the matrix

To satisfy these equations, we need each row of matrix \( A \) to sum up to zero. One simple solution is to construct matrix \( A \) as follows:\[ A = \begin{bmatrix} -1 & -1 & 2 \ -1 & 2 & -1 \ 2 & -1 & -1 \end{bmatrix} \]In this matrix, each row sums to zero, meeting the condition that \( A \mathbf{x} = \mathbf{0} \).
04

Verify the solution

Multiply the matrix \( A \) by the vector \( \mathbf{x} \). Perform matrix-vector multiplication:\[A \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} -1 \cdot 1 + (-1) \cdot 1 + 2 \cdot 1 \ -1 \cdot 1 + 2 \cdot 1 + (-1) \cdot 1 \ 2 \cdot 1 + (-1) \cdot 1 + (-1) \cdot 1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \]All elements of the resulting vector are zeros, confirming that \( \mathbf{x} \) is indeed a solution.

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Key Concepts

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

Homogeneous Equation
In linear algebra, a homogeneous equation is a type of equation where everything is set to zero. It has the form \( A \mathbf{x} = \mathbf{0} \), where \( A \) is a matrix and \( \mathbf{x} \) is a column vector. Homogeneous equations are remarkable because the zero vector \( \mathbf{0} \) is always a solution, making them fundamentally different from non-homogeneous equations. A solution of a homogeneous equation is a member of the null space of the matrix, which we will discuss later.
This null space is the collection of all vectors that, when multiplied by the matrix, result in the zero vector. It's an essential concept when studying the properties of matrices.
  • The simplest solution to a homogeneous equation is the trivial solution, where \( \mathbf{x} = \mathbf{0} \).
  • Non-trivial solutions, if they exist, lie in the null space and tell us a lot about the structure of the matrix.
  • Homogeneous equations are particularly useful in concepts like linear dependence and independence.
Understanding these equations is crucial for deeper studies in vector spaces and linear transformations.
Matrix Construction
Constructing a matrix is a vital skill in solving many linear algebra problems. When you need a matrix with specific properties, such as a nonzero matrix whose null space includes a given vector, there are systematic methods to follow.
In the given problem, the matrix \( A \) needs to satisfy the condition that each row vector, when multiplied with \( \mathbf{x} = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \), results in zero. Thus each row should sum to zero.
  • Identify the condition required: Here, each row of matrix \( A \) should add up to zero to meet \( A \mathbf{x} = \mathbf{0} \).
  • Construct a valid matrix like \( \begin{bmatrix} -1 & -1 & 2 \ -1 & 2 & -1 \ 2 & -1 & -1 \end{bmatrix} \), where each row sums to zero.
  • Verifying that other potential matrices meet these conditions helps confirm correctness and flexibility in choice of matrix construction.
Creating matrices with desired characteristics such as those encompassing specific null space vectors can aid in analyzing and building upon solutions for various algebraic problems.
Matrix-Vector Multiplication
Matrix-vector multiplication is a fundamental operation in linear algebra. It involves multiplying a matrix \( A \) by a vector \( \mathbf{x} \), resulting in another vector. This operation is crucial in tasks such as solving systems of linear equations, transforming coordinates, and more.
In our exercise, applying matrix-vector multiplication to verify if \( \mathbf{x} \) is a solution of \( A \mathbf{x} = \mathbf{0} \) is key.
Here's how it works:
  • For each row of the matrix, multiply each element by the corresponding component of the vector.
  • Sum these products to get the resultant vector's corresponding entry.
  • If the vector resulting from this multiplication is the zero vector, \( \mathbf{x} \) lies in the null space of \( A \).
This process proves that \( \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \) is a solution by showing step-by-step that the multiplication results in zero vector \( \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \), confirming that \( \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \) lies in the null space of \( A \).

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Most popular questions from this chapter

Construct a \(3 \times 3\) matrix, not in echelon form, whose columns span \(\mathbb{R}^{3} .\) Show that the matrix you construct has the desired property.

Find the value(s) of \(h\) for which the vectors are linearly dependent. Justify each answer. \(\left[\begin{array}{r}{2} \\ {-4} \\\ {1}\end{array}\right],\left[\begin{array}{r}{-6} \\ {7} \\\ {-3}\end{array}\right],\left[\begin{array}{l}{8} \\ {h} \\\ {4}\end{array}\right]\)

Let \(A\) be an \(m \times n\) matrix and let \(\mathbf{u}\) be a vector in \(\mathbb{R}^{n}\) that satisfies the equation \(A \mathbf{x}=\mathbf{0} .\) Show that for any scalar \(c,\) the vector \(c \mathbf{u}\) also satisfies \(A \mathbf{x}=\mathbf{0} .[\text { That is, show that } A(c \mathbf{u})=\mathbf{0} .]\)

In Exercises 29–32, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first. $$ \left[\begin{array}{rrrr}{1} & {2} & {-5} & {0} \\ {0} & {1} & {-3} & {-2} \\\ {0} & {-3} & {9} & {5}\end{array}\right],\left[\begin{array}{rrrr}{1} & {2} & {-5} & {0} \\ {0} & {1} & {-3} & {-2} \\ {0} & {0} & {0} & {-1}\end{array}\right] $$

Let \(A=\left[\begin{array}{rr}{2} & {-1} \\ {-6} & {3}\end{array}\right]\) and \(\mathbf{b}=\left[\begin{array}{l}{b_{1}} \\ {b_{2}}\end{array}\right] .\) Show that the equation \(A \mathbf{x}=\mathbf{b}\) does not have a solution for all possible \(\mathbf{b},\) and describe the set of all \(\mathbf{b}\) for which \(A \mathbf{x}=\mathbf{b}\) does have a solution.

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