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

Short answer

(a) \(\vec{x}=\begin{bmatrix}0 \\ 0\end{bmatrix}\). (b) \(\vec{x}=\begin{bmatrix}-2 \\ -1\end{bmatrix}\).

Step-by-step solution

Find the solution to the homogeneous equation

The homogeneous equation is given by \( A \vec{x} = \vec{O} \). To solve this, we set up the system of equations derived from the matrix \( A \):\[0x + 2y = 0\]\[-x + 3y = 0\]From the first equation, we immediately find that \( y = 0 \). Substituting \( y = 0 \) into the second equation gives \( -x + 3(0) = 0 \), so \( x = 0 \). Hence, the only solution in vector format is:\[\vec{x} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]

Check for particular solutions to the homogeneous equation

Since a homogeneous linear equation of non-zero vectors usually has a unique solution (zero vector), it is clear there are no non-trivial solutions. Thus, the solution found in step 1 is the only particular solution.

Set up the system for the non-homogeneous equation

The non-homogeneous equation needed is \( A \vec{x} = \vec{b} \), leading to:\[0x + 2y = -2\]\[-x + 3y = -1\]Solve these using substitution or elimination.

Solve the non-homogeneous system

From the first equation, we find \( y = -1 \). Substituting \( y = -1 \) into the second equation:\[-x + 3(-1) = -1 \Rightarrow -x - 3 = -1 \Rightarrow -x = 2 \Rightarrow x = -2\]Thus, one particular solution in vector format is:\[\vec{x} = \begin{bmatrix} -2 \ -1 \end{bmatrix}\]

Check for additional particular solutions

As the system is non-homogeneous and the coefficient matrix is non-singular, the solution found in step 4 is unique to this particular \(\vec{b}\). If another solution exists, it follows over specific dependent rows, which is not the case here. Thus, this is the sole vector solution.

Key concepts

Homogeneous Equations

A homogeneous matrix equation is a special type of matrix equation where the output vector, or the vector on the right side of the equation, is the zero vector \(\vec{O}\). This type of problem asks if there are any non-zero solutions that will satisfy the equation \(A \vec{x} = \vec{O}\). Often, such equations have one solution: the trivial solution where all variables equal zero.
For example, if you have the equation where \(A \vec{x} = \vec{O}\) with the matrix and vector \(A = \begin{bmatrix} 0 & 2 \ -1 & 3 \end{bmatrix}\) and \(\vec{b} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\), you reach the system:\

• \(0x + 2y = 0\)
\
• \(-x + 3y = 0\).

\
In our solution, inputs lead to the conclusion where \( y = 0 \) and therefore \( x = 0 \) are the only solutions, resulting in the solution vector expressed as \(\vec{x} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\). This verifies the nature of homogeneous equations, tending toward trivial solutions.

Non-Homogeneous Equations

A non-homogeneous equation involves a constant vector on the right side of the equation that is not the zero vector, resulting in an equation like \(A \vec{x} = \vec{b}\). These are more complex than homogeneous equations because they frequently have a unique solution or may not have any at all.
Consider the case of solving \(A \vec{x} = \vec{b}\) where given \(A = \begin{bmatrix} 0 & 2 \ -1 & 3 \end{bmatrix}\) and \(\vec{b} = \begin{bmatrix} -2 \ -1 \end{bmatrix}\). This results in the system of equations:

• \(0x + 2y = -2\)
• \(-x + 3y = -1\).

Upon solving, we found that \(y = -1\) and \(x = -2\), leading to the solution vector represented as \(\vec{x} = \begin{bmatrix} -2 \ -1 \end{bmatrix}\). Non-homogeneous equations often yield a specific solution like this, distinguished by not producing purely zero vectors when solutions exist.

Vector Format Solutions

In linear algebra, presenting solutions in vector format is crucial for a comprehensive understanding of matrix equations. Solutions to equations like \(A \vec{x} = \vec{b}\) are often expressed as vectors, efficiently communicating the solutions for each variable involved.
Our solutions are displayed as vectors to offer a clear and concise overview. For example, the solution to the homogeneous equation was \(\vec{x} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\), while the non-homogeneous solution was \(\vec{x} = \begin{bmatrix} -2 \ -1 \end{bmatrix}\). This format provides the solution values in a neat package, emphasizing each variable's resolution and ensuring clarity when shared with others. When you're dealing with multiple equations or larger systems, using vectors helps simplify complex solutions.

System of Equations

Understanding systems of equations is foundational for solving matrix equations. A system of equations refers to multiple equations applied simultaneously to find unknown values for variables. The matrix equation \(A \vec{x} = \vec{b}\) translates to such a system.
For the matrix \(A = \begin{bmatrix} 0 & 2 \ -1 & 3 \end{bmatrix}\), this was broken down into:

• \(0x + 2y = 0\) for the homogeneous system
• \(-x + 3y = 0\), similarly for the homogeneous case.
• \(0x+2y = -2\) for the non-homogeneous equation
• \(-x+3y = -1\), representing the non-homogeneous system.

These equations help define how variations in variables like \(x\) and \(y\) must demonstrate, depending on the context of the equations. Systems of equations solve by methods such as substitution, elimination, or matrix row operations, illustrating a broad range of numerical solutions. Grasping this concept is key for developing an analytical approach to difficult algebraic problems.