Homogenous and Particular Matrix Components

All linear systems can be decomposed into a homogenous and particular solution.

The homogenous solution uses the coefficients of the matrix for \( Ax = 0 \).

The particular solution uses some addition of a constant for \( Ax = b \).

Discussion

You can state the variable components in a matrix in terms some other variable in the matrix, where \( x \) and \( y \) are stated in terms of \( z \), for example.

We can transform the matrix below to achieve this.

\[ \begin{bmatrix} 1 & 2 & -3 & 3 \\ 2 & -2 & -3 & 0 \end{bmatrix} \]

Firstly, we want to transform one of the such rows such that we end up with at least one variable zeroed out and a coefficient equal to the solution. We can do this by subtracting twice the first row of the matrix from the second row.

\[ \begin{bmatrix} 1 & 2 & -3 & 3 \end{bmatrix} \cdot \begin{bmatrix} 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 & -6 & 6 \end{bmatrix} \]

\[ \begin{bmatrix} 2 & -2 & -3 & 0 \end{bmatrix} - \begin{bmatrix} 2 & 4 & -6 & 6 \end{bmatrix} = \begin{bmatrix} 0 & -6 & 3 & -6 \end{bmatrix} \]

And then dividing the new row so that the \( y \) column and the solution equal \( 1 \).

\[ \begin{bmatrix} 0 & -6 & 3 & -6 \end{bmatrix} \div -6 = \begin{bmatrix} 0 & 1 & -\frac{1}{2} & 1 \end{bmatrix} \]

This new final row is equivalent to

\[ 0x + 1y - \frac{1}{2}z = 1 \]

Which can be rearranged to

\[ y = \frac{1}{2}z + 1 \]

With \( y \) stated in terms of \( z \).

This term in \( y \) can be substituted back into the original equation, reducing it to an expression with just 2 unknowns.

\[ x + 2(\frac{1}{2}z + 1) -3z = 3 \]

\[ x = 2(\frac{1}{2}z + 1) + 3z + 3 \]

\[ x = -z - 2 + 3z + 3 \]

\[ x = 2z + 1 \]

By rearranging we have also stated \( x \) in terms of \( z \). Now the whole system of equations can be expressed in terms of this unknown.

\[ x = 2z + 1 \]

\[ y = \frac{1}{2}z + 1 \]

\[ z = z \]

This can be expressed as the product of two matrices, where one matrix represents the coefficients applied to unknown and the other the added term at the end.

\[ z \cdot \begin{bmatrix} 2 \\ \frac{1}{2} \\ 1 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \]

The first column \( \begin{bmatrix} 2 \\ \frac{1}{2} \\ 1 \end{bmatrix} \) is a solution to \( Ax = 0 \) case of the equation.

The first column and the second column are a solution to the \( Ax = b \) case.

These are known as the homogenous and particular solutions to a system of equations. Every system of equations can be split into a homogenous and particular solution.

The unknown \( z \) can also take any real number i.e. \( z \in \mathbb{R} \) as a solution to this system of equations.

For example, if we pick \( z \) to be \( 1 \) and then plug the values for the homogenous solution back into the equation represented by the first row of the original matrix we get

\[ 1 \cdot (2 \cdot 1) + 2 \cdot (\frac{1}{2} \cdot 1) - 3 \cdot 1 \]

\[ 2 + 1 - 3 = 0 \]

For whatever value of \( z \) that we pick, if we just use this homogenous solution as the values for the unknowns the solution will always be \( 0 \).

But even in the particular case you can pick any value?

There is also nothing special about isolating \( z \) here. You could isolate \( x \) and express the other unknowns in terms of it and then use any \( x \) for the homogenous and particular solutions.