Question

A matrix \(A\) and vector \(\vec{b}\) are given. (a) Solve the equation \(A \vec{x}=\vec{O}\) (b) Solve the equation \(A \vec{x}=\vec{b}\). In each of the above, be sure to write your answer in vector format. Also, when possible, give 2 particular solutions to each equation. $$ A=\left[\begin{array}{cc} 1 & 0 \\ 5 & -4 \end{array}\right], \vec{b}=\left[\begin{array}{l} -2 \\ -1 \end{array}\right] $$

Short answer

(a) \( \begin{bmatrix} 0 \\ 0 \end{bmatrix} \). (b) \( \begin{bmatrix} -2 \\ -\frac{9}{4} \end{bmatrix} \).

Step-by-step solution

Identify the Matrix and Vectors

Matrix \( A \) is \( \begin{bmatrix} 1 & 0 \ 5 & -4 \end{bmatrix} \) and vector \( \vec{b} \) is \( \begin{bmatrix} -2 \ -1 \end{bmatrix} \). To solve \( A \vec{x} = \vec{O} \), our vector \( \vec{O} \) is the zero vector \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \). In part (b), we solve \( A \vec{x} = \vec{b} \).

Solve for \( A \vec{x} = \vec{O} \)

We aim to find vector \( \vec{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \) such that \( A \vec{x} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). Set the equations from the matrix multiplication: \( 1x_1 + 0x_2 = 0 \) and \( 5x_1 - 4x_2 = 0 \). From the first equation: \( x_1 = 0 \). Substituting into the second equation: \(-4x_2 = 0\) gives \( x_2 = 0 \). Therefore, the solution is \( \vec{x} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \).

Find Two Particular Solutions for \( A \vec{x} = \vec{O} \)

Since we found the general solution \( \vec{x} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \), the particular solutions are \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \) (only one because it's the zero vector, thus no additional options).

Solve for \( A \vec{x} = \vec{b} \)

We need to find \( \vec{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \) such that \( A \vec{x} = \vec{b} \). Write the system: \( 1x_1 + 0x_2 = -2 \) and \( 5x_1 - 4x_2 = -1 \). From the first equation, \( x_1 = -2 \). Substituting into the second equation gives \( 5(-2) - 4x_2 = -1 \) which simplifies to: \(-10 - 4x_2 = -1\), thus \(-4x_2 = 9\) leading to \( x_2 = -\frac{9}{4} \). The solution vector is \( \begin{bmatrix} -2 \ -\frac{9}{4} \end{bmatrix} \).

Find Two Particular Solutions for \( A \vec{x} = \vec{b} \)

For particular solutions to \( A \vec{x} = \vec{b} \), we can consider only the solution found, \( \begin{bmatrix} -2 \ -\frac{9}{4} \end{bmatrix} \), plus variations involving any free variables. In this case, no additional variations, since the system is fully determined from \( A \). Hence, the particular solution is unique.

Key concepts

Homogeneous System

A homogeneous system of linear equations is a system in which all the constant terms are zero. The general form of such a system can be written as \( A \mathbf{x} = \mathbf{0} \), where \( A \) is a matrix and \( \mathbf{0} \) is the zero vector. This setup leads to one particular type of solution: the trivial solution.

• The trivial solution is when all variables are equal to zero, meaning \( \mathbf{x} = \mathbf{0} \).
• Such systems are always consistent because the zero vector \( \mathbf{0} \) is always a solution.

In our case, solving \( A \mathbf{x} = \mathbf{0} \) led directly to the trivial solution \( \mathbf{x} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \).

When working with a homogeneous system, if the system possesses more unknowns than equations, it can have non-trivial solutions, which are an infinite set of solutions. However, our exercise does not exhibit such a scenario, as it resulted in the strictly trivial solution.

Non-Homogeneous System

A non-homogeneous system differs from a homogeneous system by having at least one non-zero constant term. This can be represented as \( A \mathbf{x} = \mathbf{b} \), where \( \mathbf{b} eq \mathbf{0} \). Non-homogeneous systems can either be consistent or inconsistent, depending on whether a solution exists.

For the equation \( A \mathbf{x} = \mathbf{b} \), you need to analyze if \( \mathbf{b} \) can be expressed as a linear combination of the columns of \( A \).

• If it can be expressed, the system is consistent and has a solution.
• If not, the system is inconsistent and no solutions exist.

In our exercise, the system \( A \mathbf{x} = \begin{bmatrix} -2 \ -1 \end{bmatrix} \) was consistent, and we found the solution vector \( \mathbf{x} = \begin{bmatrix} -2 \ -\frac{9}{4} \end{bmatrix} \), illustrating that \( \mathbf{b} \) can indeed be obtained as a combination of the matrix \( A \)'s columns.

Particular Solution

In the context of linear systems, a particular solution refers to a specific solution to either a homogeneous or non-homogeneous system. For homogeneous systems, the trivial solution is such a case. However, when considering a non-homogeneous system, this refers to any specific set of values for the variables that satisfy the equation.

• For homogeneous equations, the particular solution can often be a zero vector.
• For non-homogeneous systems, there might be one or several particular solutions.

In our scenario, the solutions derived for \( A \mathbf{x} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \) and \( A \mathbf{x} = \begin{bmatrix} -2 \ -1 \end{bmatrix} \) were \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} -2 \ -\frac{9}{4} \end{bmatrix} \) respectively. For the non-homogeneous system, this particular solution represents the only solution since the system equations precisely define the values without variables to alter.

Vector Format

Expressing solutions in vector format provides a clear and organized way to represent the results of a linear system. It enables easy manipulation and understanding of the components involved. In this format, the solution is neatly written as a vector or a set of vectors.

• Each vector component corresponds to a variable in the system, helping directly represent the values that satisfy the equations.
• Vector format is essential for advanced matrix operations and higher-dimensional space representations.

Utilizing vectors in our solutions facilitated easier reading and comprehension, as seen with \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} -2 \ -\frac{9}{4} \end{bmatrix} \). Using vector format also aids in systems with more variables, where aligning solutions with variables is crucial.