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. $$ A=\left[\begin{array}{ccc} -1 & -2 & -2 \\ 3 & 4 & -2 \end{array}\right], \vec{b}=\left[\begin{array}{l} -4 \\ -4 \end{array}\right] $$

Short answer

Homogeneous solutions: \( x_3 \begin{bmatrix} 6 \\ -4 \\ 1 \end{bmatrix} \), Non-homogeneous: \( \begin{bmatrix} 12 \\ 8 - 4t \\ t \end{bmatrix} \). Particulars: Homogeneous \( \begin{bmatrix} 6 \\ -4 \\ 1 \end{bmatrix}, \begin{bmatrix} -6 \\ 4 \\ -1 \end{bmatrix} \); Non-homogeneous \( \begin{bmatrix} 12 \\ 8 \\ 0 \end{bmatrix}, \begin{bmatrix} 12 \\ 4 \\ 1 \end{bmatrix} \).

Step-by-step solution

Define the Equations

We have two equations to solve:1. \( A \vec{x} = \vec{O} \) where \( \vec{O} \) is the zero vector.2. \( A \vec{x} = \vec{b} \) where \( A \) is the given 2x3 matrix and \( \vec{b} \) is a given vector. Our task is to solve both equations for \( \vec{x} \).

Solve Homogeneous Equation

To solve \( A \vec{x} = \vec{O} \), write it explicitly:\[\begin{bmatrix}-1 & -2 & -2 \3 & 4 & -2\end{bmatrix}\begin{bmatrix}x_1 \x_2 \x_3\end{bmatrix}=\begin{bmatrix}0 \0\end{bmatrix}\]This leads to a system of linear equations:- \( -x_1 - 2x_2 - 2x_3 = 0 \)- \( 3x_1 + 4x_2 - 2x_3 = 0 \).Simplify using Gaussian elimination to find the solution set in terms of a free variable.

Apply Gaussian Elimination

Rewrite the system of equations in augmented matrix form:\[\begin{bmatrix}-1 & -2 & -2 & | & 0 \3 & 4 & -2 & | & 0\end{bmatrix}\]To eliminate \( x_1 \) from the second equation, multiply the first row by 3 and add it to the second row:\[\begin{bmatrix}-1 & -2 & -2 & | & 0 \0 & -2 & -8 & | & 0\end{bmatrix}\]Solve for leading variables and express in terms of free variable.

Derive the Solution for Homogeneous Equation

From the reduced matrix, express the system:- \( -x_1 - 2x_2 - 2x_3 = 0 \)- \( -2x_2 - 8x_3 = 0 \).Solving these:\( x_2 = -4x_3 \)Substitute \( x_2 \) in the first equation:\( x_1 = 6x_3 \)Thus, the solution is \( \vec{x} = x_3 \begin{bmatrix} 6 \ -4 \ 1 \end{bmatrix} \). This shows infinite solutions, parameterized by \( x_3 \). Choose two particular solutions by setting \( x_3 = 1 \) and \( x_3 = -1 \).

Solve the Non-Homogeneous Equation

Now solve \( A \vec{x} = \vec{b} \):\[\begin{bmatrix}-1 & -2 & -2 \3 & 4 & -2\end{bmatrix}\begin{bmatrix}x_1 \x_2 \x_3\end{bmatrix}=\begin{bmatrix}-4 \-4\end{bmatrix}\]Use Gaussian elimination to bring the matrix to row-echelon form and find particular solutions.

Apply Elimination for Non-Homogeneous Equation

Construct the augmented matrix:\[\begin{bmatrix}-1 & -2 & -2 & | & -4 \3 & 4 & -2 & | & -4\end{bmatrix}\]Use elimination:Multiply the first row by 3 and add to the second row:\[\begin{bmatrix}-1 & -2 & -2 & | & -4 \0 & -2 & -8 & | & -16\end{bmatrix}\]Now solve for \( x_1, x_2, x_3 \).

Derive Solution for Non-Homogeneous Equation

From the reduced form:- Solve \( -2x_2 - 8x_3 = -16 \): Simplifying, \( x_2 + 4x_3 = 8 \). Let \( x_3 = t \), then \( x_2 = 8 - 4t \).- Substitute back \( x_2 \) in equation for \( x_1 \): \( x_1 = -4 + 2(8 - 4t) + 2t \), \( x_1 = 12 \).Thus, the particular solution is \( \vec{x} = \begin{bmatrix} 12 \ 8 - 4t \ t \end{bmatrix} \). Choose two solutions by setting distinct values for \( t \).

Conclusion

For the homogeneous equation, one particular solution can be \( \begin{bmatrix} 6 \ -4 \ 1 \end{bmatrix} \) and another \( \begin{bmatrix} -6 \ 4 \ -1 \end{bmatrix} \). For the non-homogeneous, solutions like \( \begin{bmatrix} 12 \ 8 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 12 \ 4 \ 1 \end{bmatrix} \) depending on the choice for \( t \).

Key concepts

Gaussian Elimination

Gaussian elimination is a fundamental technique used to solve systems of linear equations. It involves performing a series of row operations to transform a given matrix into its row-echelon form, making it easier to solve for variables.
The method utilizes three types of row operations:

• Swapping two rows.
• Multiplying a row by a non-zero scalar.
• Adding a multiple of one row to another row.

These operations help in eliminating variables step-by-step from the matrix, eventually leading to a simpler system that can be solved by back-substitution.
The ultimate goal is to achieve an upper triangular matrix, which greatly simplifies the process of finding solutions for the unknowns in the system of equations.

Homogeneous Systems

A homogeneous system of equations is one where all the constant terms are zero, that is, a system in the form of \(A \vec{x} = \vec{O}\). These systems always have at least one solution known as the trivial solution where all variables are zero. Nontrivial solutions, however, exist if the system has more variables than equations, indicating at least one free variable.
Solving homogeneous systems involves finding the general solution expressed in terms of free variables, illustrating the infinite number of solutions due to the parameterization by these free variables.
In our exercise, setting values for the free variable gave us different particular solutions such as \( \begin{bmatrix} 6 \ -4 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} -6 \ 4 \ -1 \end{bmatrix} \), showing the range of solutions available.

Non-Homogeneous Systems

Non-homogeneous systems involve matrix equations in the form of \(A \vec{x} = \vec{b} \), where \( \vec{b} \) is a non-zero vector. Solving these requires finding at least one particular solution that satisfies the equation.
Gaussian elimination is again employed to row-reduce the augmented matrix of a non-homogeneous system. The aim is to arrive at a point where back-substitution can help identify the variables' values. However, unlike homogeneous systems, non-homogeneous ones may have a unique, no, or infinite number of solutions based on the rank of the matrix compared to the augmented matrix.
In the provided example, particular solutions such as \( \begin{bmatrix} 12 \ 8 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 12 \ 4 \ 1 \end{bmatrix} \) demonstrate how adjusting the free variable yields different outcomes depending on the system's constraints.

Particular Solutions

In both homogeneous and non-homogeneous systems, particular solutions represent valid solutions for the equation set identified by assigning specific values to the free variables.
For homogeneous systems, starting with the general solution, specific values for the free variables are chosen to find distinct particular solutions. For example, setting \( x_3 = 1 \) yielded a particular solution \( \begin{bmatrix} 6 \ -4 \ 1 \end{bmatrix} \).
In non-homogeneous systems, a particular solution is found that satisfies the initial equation \(A \vec{x} = \vec{b}\). It serves as a base from which other solutions can be constructed by adding the general solution of the associated homogeneous system.
Thus, particular solutions are crucial for understanding the nature of the solution set for any given system, especially in verifying the applicability of solutions.