7.8: The General Solution of a Linear System

Outcomes

1. Use linear transformations to determine the particular solution and general solution to a system of equations.
2. Find the kernel of a linear transformation.

Recall the definition of a linear transformation discussed above. \(T\) is a linear transformation if whenever \(\vec{x}, \vec{y}\) are vectors and \(k,p\) are scalars, \[T\left( k\vec{x}+p\vec{y}\right) =k T \left( \vec{x} \right) +p T\left(\vec{y} \right)\nonumber \] Thus linear transformations distribute across addition and pass scalars to the outside.

It turns out that we can use linear transformations to solve linear systems of equations. Indeed given a system of linear equations of the form \(A\vec{x}=\vec{b}\), one may rephrase this as \(T(\vec{x})=\vec{b}\) where \(T\) is the linear transformation \(T_A\) induced by the coefficient matrix \(A\). With this in mind consider the following definition.

Definition \(\PageIndex{1}\): Particular Solution of a System of Equations

Suppose a linear system of equations can be written in the form \[T\left(\vec{x}\right)=\vec{b}\nonumber \] If \(T\left(\vec{x}_{p}\right)=\vec{b},\) then \(\vec{x}_{p}\) is called a particular solution of the linear system.

Recall that a system is called homogeneous if every equation in the system is equal to \(0\). Suppose we represent a homogeneous system of equations by \(T\left(\vec{x}\right)=0\). It turns out that the \(\vec{x}\) for which \(T \left(\vec{x}\right) = 0\) are part of a special set called the null space of \(T\). We may also refer to the null space as the kernel of \(T\), and we write \(ker\left(T\right)\).

Consider the following definition.

Definition \(\PageIndex{2}\): Null Space or Kernel of a Linear Transformation

Let \(T\) be a linear transformation. Define \[\ker \left( T\right) = \left\{ \vec{x}:T \left(\vec{x} \right)= \vec{0} \right\}\nonumber \] The kernel, \(\ker \left( T\right)\) consists of the set of all vectors \(\vec{x}\) for which \(T (\vec{x}) = \vec{0}\). This is also called the null space of \(T\).

We may also refer to the kernel of \(T\) as the solution space of the equation \(T \left(\vec{x}\right) = \vec{0}\).

Consider the following example.

Example \(\PageIndex{1}\): The Kernel of the Derivative

Let \(\frac{d}{dx}\) denote the linear transformation defined on \(f,\) the functions which are defined on \(\mathbb{R}\) and have a continuous derivative. Find \(\ker \left( \frac{d}{dx}\right) .\)

Solution

The example asks for functions \(f\) which the property that \(\frac{df}{dx} =0.\) As you may know from calculus, these functions are the constant functions. Thus \(\ker \left( \frac{d}{dx}\right)\) is the set of constant functions.

Definition \(\PageIndex{2}\) states that \(\ker \left( T\right)\) is the set of solutions to the equation, \[T\left( \vec{x} \right) = \vec{0}\nonumber\] Since we can write \(T\left( \vec{x} \right)\) as \(A\vec{x}\), you have been solving such equations for quite some time.

We have spent a lot of time finding solutions to systems of equations in general, as well as homogeneous systems. Suppose we look at a system given by \(A\vec{x}=\vec{b}\), and consider the related homogeneous system. By this, we mean that we replace \(\vec{b}\) by \(\vec{0}\) and look at \(A\vec{x}=\vec{0}\). It turns out that there is a very important relationship between the solutions of the original system and the solutions of the associated homogeneous system. In the following theorem, we use linear transformations to denote a system of equations. Remember that \(T\left(\vec{x}\right) = A\vec{x}\).

Theorem \(\PageIndex{1}\): Particular Solution and General Solution

Suppose \(\vec{x}_{p}\) is a solution to the linear system given by , \[T\left( \vec{x} \right) = \vec{b}\nonumber \] Then if \(\vec{y}\) is any other solution to \(T\left(\vec{x}\right)=\vec{b}\), there exists \(\vec{x}_0 \in \ker \left( T\right)\) such that \[\vec{y} = \vec{x}_{p}+ \vec{x}_0\nonumber \] Hence, every solution to the linear system can be written as a sum of a particular solution, \(\vec{x}_p\), and a solution \(\vec{x}_0\) to the associated homogeneous system given by \(T\left(\vec{x}\right)=\vec{0}\).

Proof

Consider \(\vec{y} - \vec{x}_{p}= \vec{y} + \left( -1\right) \vec{x}_{p}.\) Then \(T\left( \vec{y} - \vec{x}_{p}\right) =T\left(\vec{y}\right) -T\left( \vec{x}_{p} \right)\). Since \(\vec{y}\) and \(\vec{x}_{p}\) are both solutions to the system, it follows that \(T\left(\vec{y}\right)= \vec{b}\) and \(T\left(\vec{x}_p\right) = \vec{b}\).

Hence, \(T\left(\vec{y}\right)-T\left( \vec{x}_{p} \right) =\vec{b} - \vec{b} = \vec{0}\). Let \(\vec{x}_0 = \vec{y} - \vec{x}_{p}\). Then, \(T\left(\vec{x}_0\right)= \vec{0}\) so \(\vec{x}_0\) is a solution to the associated homogeneous system and so is in \(\ker \left(T\right)\).

Sometimes people remember the above theorem in the following form. The solutions to the system \(T\left(\vec{x}\right)=\vec{b}\) are given by \(\vec{x}_{p}+\ker \left( T\right)\) where \(\vec{x}_{p}\) is a particular solution to \(T\left(\vec{x}\right)=\vec{b}\).

For now, we have been speaking about the kernel or null space of a linear transformation \(T\). However, we know that every linear transformation \(T\) is determined by some matrix \(A\). Therefore, we can also speak about the null space of a matrix. Consider the following example.

Example \(\PageIndex{2}\): The Null Space of a Matrix

Let \[A=\left[ \begin{array}{rrrr} 1 & 2 & 3 & 0 \\ 2 & 1 & 1 & 2 \\ 4 & 5 & 7 & 2 \end{array} \right]\nonumber \] Find \(\mathrm{null} \left( A\right)\). Equivalently, find the solutions to the system of equations \(A\vec{x}=\vec{0}\).

Solution

We are asked to find \(\left\{ \vec{x} : A\vec{x} = \vec{0}\right\} .\) In other words we want to solve the system, \(A\vec{x}=\vec{0}\). Let \(\vec{x} = \left[ \begin{array}{r} x \\ y \\ z \\ w \end{array} \right].\) Then this amounts to solving \[\left[ \begin{array}{rrrr} 1 & 2 & 3 & 0 \\ 2 & 1 & 1 & 2 \\ 4 & 5 & 7 & 2 \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \\ w \end{array} \right] =\left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right]\nonumber \]

This is the linear system \[\begin{array}{c} x+2y+3z=0 \\ 2x+y+z+2w=0 \\ 4x+5y+7z+2w=0 \end{array}\nonumber \] To solve, set up the augmented matrix and row reduce to find the reduced row-echelon form.

\[\left[ \begin{array}{rrrr|r} 1 & 2 & 3 & 0 & 0 \\ 2 & 1 & 1 & 2 & 0 \\ 4 & 5 & 7 & 2 & 0 \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rrrr|r} 1 & 0 & - \frac{1}{3} & \frac{4}{3} & 0 \\ 0 & 1 & \frac{5}{3} & - \frac{2}{3} & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]\nonumber \]

This yields \(x= \frac{1}{3}z- \frac{4}{3}w\) and \(y= \frac{2}{3}w- \frac{5}{3}z.\) Since \(\mathrm{null} \left( A\right)\) consists of the solutions to this system, it consists vectors of the form, \[\left[ \begin{array}{c} \frac{1}{3}z- \frac{4}{3}w \\ \frac{2}{3}w- \frac{5}{3}z \\ z \\ w \end{array} \right] =z \left[ \begin{array}{r} \frac{1}{3} \\ - \frac{5}{3} \\ 1 \\ 0 \end{array} \right] +w \left[ \begin{array}{r} - \frac{4}{3} \\ \frac{2}{3} \\ 0 \\ 1 \end{array} \right]\nonumber \]

Consider the following example.

Example \(\PageIndex{3}\): A General Solution

The general solution of a linear system of equations is the set of all possible solutions. Find the general solution to the linear system, \[\left[ \begin{array}{rrrr} 1 & 2 & 3 & 0 \\ 2 & 1 & 1 & 2 \\ 4 & 5 & 7 & 2 \end{array} \right] \left[ \begin{array}{r} x \\ y \\ z \\ w \end{array} \right] =\left[ \begin{array}{r} 9 \\ 7 \\ 25 \end{array} \right]\nonumber \]

given that \(\left[ \begin{array}{r} x \\ y \\ z \\ w \end{array} \right]=\left[ \begin{array}{r} 1 \\ 1 \\ 2 \\ 1 \end{array} \right]\) is one solution.

Solution

Note the matrix of this system is the same as the matrix in Example \(\PageIndex{2}\). Therefore, from Theorem \(\PageIndex{1}\), you will obtain all solutions to the above linear system by adding a particular solution \(\vec{x}_p\) to the solutions of the associated homogeneous system, \(\vec{x}\). One particular solution is given above by \[\vec{x}_p = \left[ \begin{array}{r} x \\ y \\ z \\ w \end{array} \right]=\left[ \begin{array}{r} 1 \\ 1 \\ 2 \\ 1 \end{array} \right]\nonumber \]

Using this particular solution along with the solutions found in Example \(\PageIndex{2}\), we obtain the following solutions, \[z\left[ \begin{array}{r} \frac{1}{3} \\ - \frac{5}{3} \\ 1 \\ 0 \end{array} \right] +w\left[ \begin{array}{r} - \frac{4}{3} \\ \frac{2}{3} \\ 0 \\ 1 \end{array} \right] +\left[ \begin{array}{r} 1 \\ 1 \\ 2 \\ 1 \end{array} \right]\nonumber \]

Hence, any solution to the above linear system is of this form.