Question

A matrix \(A\) and vectors \(\vec{b}, \vec{u}\) and \(\vec{v}\) are given. Verify that \(\vec{u}\) and \(\vec{v}\) are both solutions to the equation \(A \vec{x}=\vec{b} ;\) that is, show that \(A \vec{u}=A \vec{v}=\vec{b}\). $$ \begin{array}{l} A=\left[\begin{array}{cc} 1 & -2 \\ -3 & 6 \end{array}\right] \\ \vec{b}=\left[\begin{array}{l} 0 \\ 0 \end{array}\right], \vec{u}=\left[\begin{array}{l} 2 \\ 1 \end{array}\right], \vec{v}=\left[\begin{array}{c} -10 \\ -5 \end{array}\right] \end{array} $$

Short answer

Both \(\vec{u}\) and \(\vec{v}\) are solutions to the equation \(A\vec{x}=\vec{b}\).

Step-by-step solution

Set-Up Matrix Multiplication for \(A \vec{u}\)

We need to multiply matrix \(A\) by vector \(\vec{u}\) to check if it equals \(\vec{b}\). The equation is\[A = \begin{bmatrix} 1 & -2 \ -3 & 6 \end{bmatrix}, \quad \vec{u} = \begin{bmatrix} 2 \ 1 \end{bmatrix}\]

Perform Matrix Multiplication for \(A \vec{u}\)

Calculate the product \(A \vec{u}\): \[\begin{bmatrix} 1 & -2 \ -3 & 6 \end{bmatrix} \begin{bmatrix} 2 \ 1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 2 + (-2) \cdot 1 \ (-3) \cdot 2 + 6 \cdot 1 \end{bmatrix} = \begin{bmatrix} 2 - 2 \ -6 + 6 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]

Verify the Result of \(A \vec{u}\)

The output of \(A \vec{u}\) is \(\begin{bmatrix} 0 \ 0 \end{bmatrix}\), which matches \(\vec{b}\). Therefore, \(\vec{u}\) is a solution to the equation \(A \vec{x} = \vec{b}\).

Set-Up Matrix Multiplication for \(A \vec{v}\)

We need to multiply matrix \(A\) by vector \(\vec{v}\) to check if it equals \(\vec{b}\). The equation is\[A = \begin{bmatrix} 1 & -2 \ -3 & 6 \end{bmatrix}, \quad \vec{v} = \begin{bmatrix} -10 \ -5 \end{bmatrix}\]

Perform Matrix Multiplication for \(A \vec{v}\)

Calculate the product \(A \vec{v}\): \[\begin{bmatrix} 1 & -2 \ -3 & 6 \end{bmatrix} \begin{bmatrix} -10 \ -5 \end{bmatrix} = \begin{bmatrix} 1 \cdot (-10) + (-2) \cdot (-5) \ (-3) \cdot (-10) + 6 \cdot (-5) \end{bmatrix} = \begin{bmatrix} -10 + 10 \ 30 - 30 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]

Verify the Result of \(A \vec{v}\)

The output of \(A \vec{v}\) is \(\begin{bmatrix} 0 \ 0 \end{bmatrix}\), which matches \(\vec{b}\). Therefore, \(\vec{v}\) is also a solution to the equation \(A \vec{x} = \vec{b}\).

Key concepts

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra. It involves calculating the product of a matrix and a vector, or two matrices. The process is systematic, requiring each element of a row in the first matrix to be multiplied by the corresponding element of a column in the second matrix. Then, you sum these products for each row-column pair.

To multiply a matrix by a vector, such as in our problem with matrix \( A \) and vectors \( \vec{u} \) and \( \vec{v} \), follow these steps:

• Take each row of the matrix \( A \).
• Multiply it by the vector \( \vec{x} \), which could be either \( \vec{u} \) or \( \vec{v} \).
• Sum the results to get a single number for each entry of the resulting product vector.

In our example, the product \( A\vec{u} \) is calculated by multiplying each element of the rows of \( A \) by the corresponding element in \( \vec{u} \), and similarly for \( A\vec{v} \).

This operation allows us to transform vectors and solve matrix equations efficiently.

Solution Verification

Solution verification is crucial to ensure that the vectors \( \vec{u} \) and \( \vec{v} \) are indeed solutions to the matrix equation \( A\vec{x} = \vec{b} \). In practice, this involves computing the result of the multiplication \( A\vec{x} \) and comparing it with the vector \( \vec{b} \).

In this problem, we verified whether both \( \vec{u} \) and \( \vec{v} \) are solutions by:

• Calculating \( A\vec{u} \) and checking if it equals the zero vector \( \vec{b} \).
• Doing the same for \( A\vec{v} \).

Upon finding that both results equal \( \vec{b} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \), we confirmed that both \( \vec{u} \) and \( \vec{v} \) satisfy the matrix equation.

Solution verification gives us confidence that our solutions are correct and helps in understanding the nature of the matrix equation being solved.

Vector Spaces

Understanding vector spaces helps in comprehending solutions to matrix equations. A vector space is a collection of vectors that can be added together and multiplied by scalars while still remaining within the set. It includes all possible linear combinations of given vectors and has important applications in solving linear systems.

In the context of this problem, the zero vector \( \vec{b} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \) represents a particular solution in the vector space defined by the matrix equation. The vectors \( \vec{u} \) and \( \vec{v} \) belong to this vector space as they both satisfy the equation \( A\vec{x} = \vec{b} \).

Properties of vector spaces, such as closure under addition and scalar multiplication, provide powerful tools for solving and verifying matrix equations, offering insights into the nature of potential solutions.

• Closure under addition: Adding two solutions yields another solution.
• Closure under scalar multiplication: Multiplying a solution by a scalar still results in a valid solution.

Understanding vector spaces enriches our approach to handling problems in linear algebra effectively.