# Day 15:

Examples of solving $$Ax=b$$

Theorem. Let $$A$$ be an $$m\times n$$ matrix, and let $$b\in\R^{m}$$.

If $$\mathbf{x}_{0}\in\R^{n}$$ is a solution to $$Ax=b$$, then

$\{x\in\R^{n} : Ax=b\} = \{\mathbf{x}_{0} + z : z\in N(A)\}$

Moreover, if $$\{v_{1},v_{2},\ldots,v_{k}\}$$ is a basis for $$N(A)$$, then

$\{x\in\R^{n} : Ax=b\} = \{\mathbf{x}_{0} +a_{1}v_{1}+a_{2}v_{2}+\cdots+a_{k}v_{k} : a_{1},a_{2},\ldots,a_{k}\in\R\}$

Proof. Let $$x$$ be any vector such that $$Ax=b$$. Note that $A(x-\mathbf{x}_{0}) = Ax-A\mathbf{x}_{0} = b-b = \mathbf{0}.$

This shows that $$x-\mathbf{x}_{0}\in N(A)$$. Set $$z = x-\mathbf{x}_{0}$$. Since $x=\mathbf{x}_{0} + (x-\mathbf{x}_{0}),$ we see that $$x\in \{\mathbf{x}_{0} + z : z\in N(A)\}$$.

Next, assume $$y\in \{\mathbf{x}_{0} : z\in N(A)\}$$, that is, $$y=\mathbf{x}_{0}+z$$ for some $$z\in N(A)$$. Then we see that

$Ay = A(\mathbf{x}_{0}+z) = A\mathbf{x}_{0} + Az = b+\mathbf{0} = b.$

Therefore, $$y$$ is a solution to $$Ax=b$$. $$\Box$$

This is a very nice description of the set of solutions to $$Ax=b$$

We call this the vector parametric form of the set of soltuions.

### Solving $$Ax=b$$

Finding a solution to $$Ax=b$$.

Assume $$A$$ is an $$m\times n$$ matrix and $$b\in\R^{m}$$.

1. Find the matrix $$B$$ so that $$BA = \text{rref}(A)$$.
2. Find a solution $$\mathbf{x}_{0}$$ to $$\text{rref}(A)x = Bb$$.
3. The vector $$\mathbf{x}_{0}$$ is also a solution to $$Ax=b$$.

Why?

Because $$B$$ is invertible! $\text{rref}(A)\mathbf{x}_{0} = Bb\quad \Rightarrow\quad BA\mathbf{x}_{0} = Bb$

$\quad \Rightarrow\quad B^{-1}BA\mathbf{x}_{0} = B^{-1}Bb\quad\Rightarrow\quad A\mathbf{x}_{0} = b$

### Finding a "particular" solution to $$Ax=b$$

Example. Consider the matrix equation $$Ax=b$$:

$\begin{bmatrix} 1 & 3 & 0 & -2 & -1\\ 2 & 6 & 2 & 0 & -2\\ 0 & 0 & 1 & 1 & 0\end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end{bmatrix}=\begin{bmatrix} 1\\ 2\\ 1\end{bmatrix}$

First, we find the matrix $$B$$ so that $$BA=\text{rref}(A)$$, and we find $$B^{-1}$$.

$B = \frac{1}{2}\begin{bmatrix} -2 & 2 & -4\\ 2 & -1 & 4\\ -2 & 1 & -2\end{bmatrix} \quad\text{and}\quad B^{-1}=\begin{bmatrix} 1 & 0 & -2\\ 2 & 2 & 0\\ 0 & 1 & 1\end{bmatrix}.$

Multiplying by $$B$$ on both sides of $$Ax=b$$ we obtain

$\begin{bmatrix} 1 & 3 & 0 & 0 & -1\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end{bmatrix} = \begin{bmatrix} -1\\ 2\\ -1\end{bmatrix}$

### Finding a "particular" solution to $$Ax=b$$

Example. Consider the matrix equation $$Ax=b$$:

$\begin{bmatrix} 1 & 3 & 0 & -2 & -1\\ 2 & 6 & 2 & 0 & -2\\ 0 & 0 & 1 & 1 & 0\end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end{bmatrix}=\begin{bmatrix} 1\\ 2\\ 1\end{bmatrix}$

Note that we can compute $$Bb$$ without finding $$B$$. Indeed, we can perform the same row operations on $$b$$ that we preformed on $$A$$ to obtain $$\operatorname{rref}(A)$$.

Multiplying by $$B$$ on both sides of $$Ax=b$$ we obtain

$\begin{bmatrix} 1 & 3 & 0 & 0 & -1\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end{bmatrix} = \begin{bmatrix} -1\\ 2\\ -1\end{bmatrix}$

### Finding a "particular" solution to $$Ax=b$$

Example continued.

$\begin{bmatrix} 1 & 3 & 0 & 0 & -1\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end{bmatrix} = \begin{bmatrix} -1\\ 2\\ -1\end{bmatrix}$

$\mathbf{x}_{0} = \begin{bmatrix}\ &\ &\ &\ &\ &\ &\ \end{bmatrix}^{\top}$

$0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0$

Constructing a solution $$\mathbf{x}_{0}$$ to $$\text{rref}(A)x=Bb$$.

1. Put a zero in each row of $$\mathbf{x}_{0}$$ corresponding to a non-pivot column of $$\text{rref}(A)$$.
2. For each row of $$\mathbf{x}_{0}$$ corresponding to a pivot column of $$\text{rref}(A)$$, if the pivot is in row $$i$$, then the entry in $$\mathbf{x}_{0}$$ is the entry in row $$i$$ of $$Bb.$$

$$-1$$

$$2$$

$$-1$$

Warning: This algorithm might not produce a solution!

### An equation without a solution

$\begin{bmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0& -2\\ 1 & 1 & 1 & -2\end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{bmatrix} = \begin{bmatrix} 0\\ 3\\ 0\end{bmatrix}$

Multiply on the left by the matrix $$B$$ so that $$BA=\text{rref}(A)$$.

$\begin{bmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & -2\\ 0 & 0 & 0 & 0\end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}\quad$

$\Downarrow$

$\begin{bmatrix} x_{1}+x_{3}\\ x_{2}-x_{4}\\ 0\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}$

We cannot find numbers $$x_{1},x_{2},x_{3},x_{4}$$ that satisfy this equation. Hence, the original equation has no solution.

What would be the "particular" solution we would write by the algorithm?

• $$Ax=b$$ has a solution if and only if $$b\in C(A)$$.
• $$Ax=b$$ has exactly one solution if and only if $$b\in C(A)$$ and $$\operatorname{nullity}(A)=0$$
• $$Ax=b$$ has infinitely many solutions if and only if $$b\in C(A)$$ and $$\operatorname{nullity}(A)>0$$.

### Some takeaways

Quiz:

1. For $$A\in\mathbb{R}^{m\times n}$$ and $$b\in\mathbb{R}^{m}$$ such that $$n\geq m$$ and $$\operatorname{nullity}(A)=0$$ it ___________ holds that $$Ax=b$$ has a solution.
2. For $$A\in\mathbb{R}^{n\times n}$$ and $$b\in\mathbb{R}^{n}$$ such that $$Ax=b$$ has infinitely many solutions, it __________ holds that there is some $$c\in\mathbb{R}^{n}$$ such that $$Ax=c$$ has no solutions.
3. For $$A\in\mathbb{R}^{m\times n}$$ and $$b\in\mathbb{R}^{m}$$ such that $$n\geq m$$ and $$\operatorname{nullity}(A)=1$$ it ___________ holds that $$Ax=b$$ has infinitely many solutions.

always

always

sometimes

### Solving $$Ax=b$$

Example. Consider the matrix equation $$Ax=b$$:

$\begin{bmatrix} 2 & 4 & 0 & -2\\ 2 & 4 & 1 & 3\\ 0 & 0 & 1 & 1\end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{bmatrix}=\begin{bmatrix} 0\\ 0\\ -4\end{bmatrix}$

$\text{rref}(A) = \begin{bmatrix} 1 & 2 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}$

We already found (last time) that $$\mathbf{x}_{0} = \begin{bmatrix} 1\\ 0\\ -5\\ 1\end{bmatrix}$$ is a solution.

Hence, the set of solutions to $$Ax=b$$ is $$\left\{\begin{bmatrix} 1\\ 0\\ -5\\ 1\end{bmatrix} + a\begin{bmatrix} -2\\ 1\\ 0\\ 0\end{bmatrix}: a\in\R\right\}$$

$$\Rightarrow\ \left\{\begin{bmatrix} -2\\ 1\\ 0\\ 0\end{bmatrix}\right\}$$ is a basis for $$N(A)$$

### Finding all solutions to $$Ax=b$$

A basis for $$N(A)$$ is $$\left\{\begin{bmatrix} -3\\ 1\\ 0\\ 0\\ 0\end{bmatrix}, \begin{bmatrix} 1\\ 0\\ 0\\ 0\\ 1\end{bmatrix}\right\}$$

A solution is $$\mathbf{x}_{0} = \begin{bmatrix} -1\\ 0\\ 2\\ -1\\ 0 \end{bmatrix}$$

Hence, the set of solutions to $$Ax=b$$ is

$\left\{\begin{bmatrix} -1\\ 0\\ 2\\ -1\\ 0\end{bmatrix} + a_{1}\begin{bmatrix} -3\\ 1\\ 0\\ 0\\ 0\end{bmatrix} + a_{2}\begin{bmatrix} 1\\ 0\\ 0\\ 0\\ 1\end{bmatrix} : a_{1},a_{2}\in\R\right\}$

Example. Consider the matrix equation $$Ax=b$$:

$\begin{bmatrix} 1 & 3 & 0 & -2 & -1\\ 2 & 6 & 2 & 0 & -2\\ 0 & 0 & 1 & 1 & 0\end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end{bmatrix}=\begin{bmatrix} 1\\ 2\\ 1\end{bmatrix}$

Exercise. Which of the following vectors is in $$C(B)$$ where

$B = \left[\begin{array}{rrr} 1 & 0 & 2\\ -2 & 1 & -2\\ 3 & -1 & 4 \end{array}\right]?$

$\text{(a) }\left[\begin{array}{rrr} 0\\ 1\\ -1\end{array}\right]\ \text{(b) }\left[\begin{array}{rrr} 1\\ -1\\ 2\end{array}\right]\ \text{(c) } \left[\begin{array}{rrr} 0\\ -1\\ -1\end{array}\right]\ \text{(d) }\left[\begin{array}{rrr} 2\\ -1\\ 3 \end{array}\right]\ \text{(e) }\left[\begin{array}{rrr} 0\\ 0\\ 0\end{array}\right]\$

$\left[\begin{array}{rrr|c} 1 & 0 & 2 & b_{1}\\ -2 & 1 & -2 & b_{2}\\ 3 & -1 & 4 & b_{3} \end{array}\right] \sim \left[\begin{array}{rrr|c} 1 & 0 & 2 & b_{2}+b_{2}\\ 0 & 1 & 2 & 3b_{2}+2b_{3}\\ 0 & 0 & 0 & b_{1} - b_{2} - b_{3} \end{array}\right]$

Hence, $$Bx=\begin{bmatrix} b_{1}\\ b_{2}\\ b_{3}\end{bmatrix}$$ has a solution if and only if $$b_{1}-b_{2}-b_{3} = 0$$.

By John Jasper

• 414