# 5.9: The General Solution of a Linear System

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## 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.

Below is a video on finding the solution as a sum of a particular solution and the homogeneous solution.

This page titled 5.9: The General Solution of a Linear System is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) .