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 \(A \vec{u}=A \vec{v}=\vec{b}\). $$ \begin{array}{l} A=\left[\begin{array}{cc} 1 & -2 \\ -3 & 6 \end{array}\right] \\ \vec{b}=\left[\begin{array}{c} 2 \\ -6 \end{array}\right], \vec{u}=\left[\begin{array}{c} 0 \\ -1 \end{array}\right], \vec{v}=\left[\begin{array}{l} 2 \\ 0 \end{array}\right] \end{array} $$

Short answer

Both \(\vec{u}\) and \(\vec{v}\) satisfy \(A\vec{x}=\vec{b}\), so they are solutions.

Step-by-step solution

Verify A\vec{u}=\vec{b}

To verify if \(\vec{u}\) is a solution, calculate \(A\vec{u}\). First, perform the matrix-vector multiplication: \[A\vec{u} = \begin{bmatrix} 1 & -2 \ -3 & 6 \end{bmatrix} \begin{bmatrix} 0 \ -1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 0 + (-2) \cdot (-1) \ -3 \cdot 0 + 6 \cdot (-1) \end{bmatrix} = \begin{bmatrix} 2 \ -6 \end{bmatrix}\] The result \(\begin{bmatrix} 2 \ -6 \end{bmatrix}\) matches \(\vec{b}\), confirming \(\vec{u}\) is a solution.

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

Similarly, verify if \(\vec{v}\) is a solution by calculating \(A\vec{v}\). Perform the matrix-vector multiplication: \[A\vec{v} = \begin{bmatrix} 1 & -2 \ -3 & 6 \end{bmatrix} \begin{bmatrix} 2 \ 0 \end{bmatrix} = \begin{bmatrix} 1 \cdot 2 + (-2) \cdot 0 \ -3 \cdot 2 + 6 \cdot 0 \end{bmatrix} = \begin{bmatrix} 2 \ -6 \end{bmatrix}\] The result \(\begin{bmatrix} 2 \ -6 \end{bmatrix}\) also matches \(\vec{b}\), confirming \(\vec{v}\) is a solution.

Key concepts

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra, performed often when dealing with systems of linear equations. It's essential to understand that matrix multiplication isn't the same as regular number multiplication. Instead, it involves a series of dot products.

When you multiply a matrix by a vector, you're taking each row of the matrix and computing its dot product with the vector. Here's a simple step-by-step process for multiplying a matrix \(A\) by a vector \(\vec{x}\):

• Take the first row of the matrix \(A\) and multiply it by each corresponding element of the vector \(\vec{x}\), then sum the results. This gives you the first element of the resulting vector.
• Repeat the process for each row of the matrix \(A\).

The result is a new vector that, depending on the context, can represent many things, like a transformed version of the input vector \(\vec{x}\).

An example is shown in the exercise where matrix \(A\) is multiplied by vectors \(\vec{u}\) and \(\vec{v}\) to check for solutions to a linear equation. The multiplication yields a vector that is either equal to \(\vec{b}\) (confirming a solution) or not.

Vector Solutions

Vector solutions are the values that a vector can take, which satisfy a given equation or system, such as \(A\vec{x} = \vec{b}\). In our exercise, we verify whether both \(\vec{u}\) and \(\vec{v}\) satisfy this condition.

This involves checking that the result of multiplying the matrix \(A\) with each vector individually produces the vector \(\vec{b}\). If \(A\vec{u} = \vec{b}\) and \(A\vec{v} = \vec{b}\) hold true, then both vectors \(\vec{u}\) and \(\vec{v}\) are solutions to the equation. This tells us that applying the transformation described by the matrix \(A\) to these vectors places them at \(\vec{b}\) in the output vector space.

The concept of vector solutions is crucial in understanding linear mappings and transformations, as it allows us to find all possibile input vectors that produce a desired output vector when transformed via a linear operation like matrix multiplication.

Homogeneous Systems

A homogeneous system of linear equations is a set of linear equations where all the constant terms are zero, typically expressed as \(A\vec{x} = \vec{0}\). Even though our exercise deals with a non-homogeneous system (where the constants form the vector \(\vec{b}\), not zero), understanding homogeneous systems is essential when analyzing vector solutions.

In homogeneous systems, the zero vector \(\vec{0}\) is always a solution, as it satisfies \(A\vec{x} = \vec{0}\) when \(\vec{x} = \vec{0}\). More importantly, they often have infinitely many solutions which form a vector space known as the null space of a matrix \(A\).

This concept can be expanded to non-homogeneous systems through the idea of finding a particular solution (like \(\vec{u}\) or \(\vec{v}\) in our exercise) and then examining the associated homogeneous system \(A\vec{x} = \vec{0}\) to find what are known as the null solutions. The full solution set to the non-homogeneous problem is a translation of the null space by any particular solution, providing insight into the structure and nature of solutions to linear equations.