Contents

11. General solutions

Let’s focus on the linear system \(\bfA\bfx=\bfb\) for an \(n\times n\) coefficient matrix \(\bfA\). If \(\bfA\) is invertible, then \(\bfx=\bfA^{-1}\bfb\) is the unique solution. What about when \(\bfA\) is singular? All we know at the moment is that there are no solutions, or infinitely many of them.

11.1. Homogeneous systems

We say that the linear system is homogeneous if \(\bfb=\bfzero\), that is, the system is \(\bfA\bfx = \bfzero\). Here is one of those old habits you must learn to break.

Warning

A homogeneous system \(\bfA\bfx = \bfzero\) can have nonzero solutions for \(\bfx\), even if \(\bfA\) is not the zero matrix.

To be clear, nonzero homogeneous solutions can occur only when \(\bfA\) is singular, because otherwise the solution \(\bfx=\bfA^{-1}\bfzero=\bfzero\) is unique.

A nice feature of zero is that adding multiples of zero to it is still zero. So, if \(\mathbf{R}\) is the RREF of \(\bfA\), then the RREF of the augmented matrix \([\bfA;\,\bfzero]\) is just \([\mathbf{R};\,\bfzero]\). Each free column of \(\mathbf{R}\) therefore introduces a free variable in the homogeneous solution.

Example

The RREF of

\[\begin{split}\bfA = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\end{split}\]

is

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

Find all solutions of the homogeneous system \(\bfA\bfx=\bfzero\).

Solution

The RREF of the augmented matrix is

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

Column 3 is free, so the solution starts with \(x_3=s\). Rows 1 and 2 then yield

\[x_1 = s, \quad x_2 = -2s,\]

so \(\bfx =\threevec{1}{-2}{1}\) for any value of \(s\).

Example

Given the equivalence

\[\begin{split}\bfA = \begin{bmatrix} 1 & 2 & 0 & -4 \\ -2 & -4 & 1 & 9 \\ -3 & -6 & 1 & 13 \\ -2 & -4 & 0 & 8 \end{bmatrix} \quad \overset{\text{RREF}}{\Longrightarrow} \quad \begin{bmatrix} 1 & 2 & 0 & -4 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix},\end{split}\]

find all solutions of the homogeneous system \(\bfA\bfx=\bfzero\).

Solution

The RREF of the augmented matrix just tacks on a column of zeros. Hence columns 2 and 4 are free, so we set \(x_2=s\), \(x_4=t\). The rest of the solution is

\[x_1 = -2s+4t, \quad x_3 = -t.\]

Note that we can also write

\[\begin{split}\bfx = s \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t \begin{bmatrix} 4 \\ 0 \\ -1 \\ 1 \end{bmatrix}.\end{split}\]

The preceding examples are typical in that the homogeneous solution always takes the form

(11.1)\[\bfx = c_1 \bfu_1 + c_2 \bfu_2 + \cdots + c_k \bfu_k,\]

where the \(c_j\) are arbitrary, one for each free column of the RREF, and the vectors are constant. It’s no accident that this has the form of a linear combination!

Definition 11.1

The formula (11.1) is called the general solution of the homogeneous problem \(\bfA\bfx=\bfzero\). We say that \(\bfu_1,\ldots,\bfu_k\) form a basis for the general solution.

Note

The general homogeneous solution is actually a set of vectors. In linear algebra, more common names for this set are nullspace and kernel. We use “general homogeneous solution” because that terminology appears in differential equations for an analogous situation.

In the new terminology, we can unify the case of an invertible \(\bfA\) by saying that it has an empty basis, and a general solution of the single vector \(\bfzero\).

11.2. Particular solution

Now, let’s consider the nonhomogeneous system \(\bfA\bfx=\bfb\). We introduce here one of the simplest yet most confusing terms in the course. A particular solution of the linear system \(\bfA\bfx=\bfb\) is just any one solution of the problem. The only reason the term exists is to distinguish it from the general solution, which (as above) is an expression for every possible solution of the system.

Theorem 11.2 (General solution of a linear system)

All solutions of \(\bfA\bfx=\bfb\) may be written as

\[\bfx = \bfx_h + \bfx_p,\]

where \(\bfx_h\) is the general solution of \(\bfA\bfx=\bfzero\) and \(\bfx_p\) is any particular solution of \(\bfA\bfx=\bfb\). We call this the general solution of the linear system \(\bfA\bfx=\bfb\).

Proof

Let \(\bfx_p\) be a particular solution, and suppose \(\bfv\) is another solution. Then

\[\bfA(\bfv-\bfx_p)=\bfA\bfv - \bfA\bfx_p = \bfb - \bfb = \bfzero.\]

Hence \(\bfv-\bfx_p\) is a homogeneous solution, which means that we can write it as \(\bfx_h\).

Hence every linear system is closely tied to its homogeneous counterpart. The general solution of \(\bfA\bfx=\bfb\) requires just one solution, plus the general homogeneous solution.

Example

Find the general solution of

\[\begin{split}\begin{bmatrix} 0 & -2 & 4 \\ 0 & 1 & -2 \\ 0 & 3 & -6 \end{bmatrix} \bfx = \threevec{-8}{4}{12}.\end{split}\]
Solution

The RREF of the augmented matrix is found to be

\[\begin{split}\begin{bmatrix} 0 & 1 & -2 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}. \end{split}\]

If we replace the last column with zeros, we get the RREF of the associated homogeneous problem. Columns 1 and 3 are free, so we would get the homogeneous solution

\[\bfx_h = \threevec{s}{2t}{t} = s \threevec{1}{0}{0} + t \threevec{0}{2}{1}.\]

Returning to the original nonhomogeneous problem, we only need to find any one particular \(\bfx_p\). It’s easiest for us if we set all the free variables to be zero, in which case we get \(x_1=x_3=0\) and \(x_2 = 4\). Thus the general solution is

\[\bfx = \bfx_p + \bfx_h = \threevec{0}{4}{0} + c_1 \threevec{1}{0}{0} + c_2 \threevec{0}{2}{1}.\]

Note

The basis vectors in \(\bfx_h\) and the particular solution \(\bfx_p\) aren’t unique, so it’s possible to write down two equivalent forms of the solution that look rather different.

The homogeneous \(+\) particular approach will come up again, and often, for differential equations.

10. Detecting singularity 12. Complex numbers