Question

A matrix \(A\) and vectors \(\vec{b}, \bar{u}\) and \(\vec{v}\) are given. Verify that \(A \vec{u}=\overrightarrow{0}, A \vec{v}=\vec{b}\) and \(A(\vec{u}+\vec{v})=\vec{b}\).$$ A=\left[\begin{array}{ccc} 1 & -1 & 3 \\ 3 & -3 & -3 \\ -1 & 1 & 1 \end{array}\right] $$ $$ \vec{b}=\left[\begin{array}{c} -1 \\ -3 \\ 1 \end{array}\right], \vec{u}=\left[\begin{array}{l} 2 \\ 2 \\ 0 \end{array}\right], \vec{v}=\left[\begin{array}{l} 2 \\ 3 \\ 0 \end{array}\right] $$

Short answer

All three conditions are verified successfully.

Step-by-step solution

Verify A\(\vec{u}=\overrightarrow{0}\)

To verify that \(A\vec{u}=\overrightarrow{0}\), calculate the product of matrix \(A\) and vector \(\vec{u}\):\[A\vec{u} = \begin{bmatrix} 1 & -1 & 3 \ 3 & -3 & -3 \ -1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 2 \ 2 \ 0 \end{bmatrix}\]Perform the multiplication:1. First row: \((1 \times 2) + (-1 \times 2) + (3 \times 0) = 2 - 2 + 0 = 0\)2. Second row: \((3 \times 2) + (-3 \times 2) + (-3 \times 0) = 6 - 6 + 0 = 0\)3. Third row: \((-1 \times 2) + (1 \times 2) + (1 \times 0) = -2 + 2 + 0 = 0\)Thus, \(A\vec{u} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}=\overrightarrow{0}\).

Verify A\(\vec{v}=\vec{b}\)

To verify that \(A\vec{v}=\vec{b}\), calculate the product of matrix \(A\) and vector \(\vec{v}\):\[A\vec{v} = \begin{bmatrix} 1 & -1 & 3 \ 3 & -3 & -3 \ -1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 2 \ 3 \ 0 \end{bmatrix}\]Perform the multiplication:1. First row: \((1 \times 2) + (-1 \times 3) + (3 \times 0) = 2 - 3 + 0 = -1\)2. Second row: \((3 \times 2) + (-3 \times 3) + (-3 \times 0) = 6 - 9 + 0 = -3\)3. Third row: \((-1 \times 2) + (1 \times 3) + (1 \times 0) = -2 + 3 + 0 = 1\)Thus, \(A\vec{v} = \begin{bmatrix} -1 \ -3 \ 1 \end{bmatrix}=\vec{b}\).

Verify A(\(\vec{u}+\vec{v}\))=\(\vec{b}\)

First, calculate \(\vec{u}+\vec{v}\):\[\vec{u}+\vec{v} = \begin{bmatrix} 2 \ 2 \ 0 \end{bmatrix} + \begin{bmatrix} 2 \ 3 \ 0 \end{bmatrix} = \begin{bmatrix} 4 \ 5 \ 0 \end{bmatrix}\]Next, calculate \(A(\vec{u}+\vec{v})\):\[A\begin{bmatrix} 4 \ 5 \ 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 3 \ 3 & -3 & -3 \ -1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 4 \ 5 \ 0 \end{bmatrix}\]Perform the multiplication:1. First row: \((1 \times 4) + (-1 \times 5) + (3 \times 0) = 4 - 5 + 0 = -1\)2. Second row: \((3 \times 4) + (-3 \times 5) + (-3 \times 0) = 12 - 15 + 0 = -3\)3. Third row: \((-1 \times 4) + (1 \times 5) + (1 \times 0) = -4 + 5 + 0 = 1\)Thus, \(A(\vec{u}+\vec{v}) = \begin{bmatrix} -1 \ -3 \ 1 \end{bmatrix}=\vec{b}\).

Key concepts

Linear Combinations

In linear algebra, a linear combination involves creating a new vector by adding together scaled versions of other vectors. For instance, consider vectors \(\vec{u}\) and \(\vec{v}\). A linear combination of these vectors can be expressed as \(c_1 \vec{u} + c_2 \vec{v}\), where \(c_1\) and \(c_2\) are scalars.
Linear combinations are key in understanding how vectors interact and are used in processes like vector spaces, basis, and span. By modifying the scalars, you can explore different combinations and contexts in which these vectors are applicable.
This concept is particularly useful in the exercise because it illustrates how the original vectors \(\vec{u}\) and \(\vec{v}\) relate to matrix \(A\) and vector \(\vec{b}\) through their combinations.

Matrix Vector Product

Matrix vector product, also known as matrix multiplication with a vector, is an essential operation in linear algebra. It involves multiplying a matrix \(A\) by a vector \(\vec{x}\), which results in a new vector. The product \(A\vec{x}\) is calculated by performing the dot product of each row of the matrix with the vector.

• The resulting vector has its components determined by the sum of products from corresponding entries.
• This operation is pivotal in transformations and solving systems of linear equations.

In the exercise, multiplying the matrix \(A\) with \(\vec{u}, \vec{v}, \) and \(\vec{u} + \vec{v}\) demonstrates verifying conditions such as \(A\vec{u} = \overrightarrow{0}\) and \(A\vec{v} = \vec{b}\), showing how the matrix influences these vectors.

Homogeneous Systems

Homogeneous systems refer to systems of linear equations where all the constants on the right side are zero. In such a system, the equation can be written as \(A\vec{x} = \overrightarrow{0}\). This implies that the solution involves finding vectors that, when multiplied by a given matrix, yield the zero vector.
Solutions to homogeneous systems often include a trivial solution where the vector \(\vec{x}\) is the zero vector itself. However, there may be infinite non-trivial solutions depending on the matrix and vector dimensions.
In the exercise, the condition \(A\vec{u} = \overrightarrow{0}\) indicates a homogeneous equation where \(\vec{u}\) is a non-trivial solution, demonstrating a scenario where certain changes to a vector yield no change when transformed by the matrix \(A\).

Vector Addition

Vector addition is a straightforward operation where two vectors are combined to produce a third vector, often used to understand their sum of components. The sum \(\vec{u} + \vec{v}\) involves adding corresponding entries of each vector to form a new vector.

• It is an associative operation, meaning the order of addition does not affect the result.
• This operation bridges the gap between individual vector properties and linear transformations when interacting with a matrix.

In the given exercise, vector addition provides insight into how \(A(\vec{u} + \vec{v}) = \vec{b}\), demonstrating how the combination of two vectors results in a specific transformation and eventual alignment with another vector upon multiplication by \(A\).