2.3: Matrix Equations


Objectives
1. Understand the equivalence between a system of linear equations, an augmented matrix, a vector equation, and a matrix equation.
2. Characterize the vectors $$b$$ such that $$Ax=b$$ is consistent, in terms of the span of the columns of $$A$$.
3. Characterize matrices $$A$$ such that $$Ax=b$$ is consistent for all vectors $$b$$.
4. Recipe: multiply a vector by a matrix (two ways).
5. Picture: the set of all vectors $$b$$ such that $$Ax=b$$ is consistent.
6. Vocabulary word: matrix equation.

The Matrix Equation $$Ax=b$$

In this section we introduce a very concise way of writing a system of linear equations: $$Ax=b$$. Here $$A$$ is a matrix and $$x,b$$ are vectors (generally of different sizes), so first we must explain how to multiply a matrix by a vector.

Note $$\PageIndex{1}$$

When we say “$$A$$ is an $$m\times n$$ matrix,” we mean that $$A$$ has $$m$$ rows and $$n$$ columns.

Remark

In this book, we do not reserve the letters $$m$$ and $$n$$ for the numbers of rows and columns of a matrix. If we write “$$A$$ is an $$n\times m$$ matrix”, then $$n$$ is the number of rows of $$A$$ and $$m$$ is the number of columns.

Definition $$\PageIndex{1}$$: Product

Let $$A$$ be an $$m\times n$$ matrix with columns $$v_1,v_2,\ldots,v_n\text{:}$$

$A=\left(\begin{array}{cccc} |&|&\quad &| \\ v_1 &v_2 &\cdots &v_n \\ |&|&\quad &|\end{array}\right)\nonumber$

The product of $$A$$ with a vector $$x$$ in $$\mathbb{R}^n$$ is the linear combination

$Ax=\left(\begin{array}{cccc} |&|&\quad &| \\ v_1 &v_2 &\cdots &v_n \\ |&|&\quad &| \end{array}\right)\:\left(\begin{array}{c}x_1 \\ x_2 \\ \vdots \\ x_n\end{array}\right) =x_1 v_1 +x_2 v_2 +\cdots +x_n v_n .\nonumber$

This is a vector in $$\mathbb{R}^m$$.

Example $$\PageIndex{1}$$

$\left(\begin{array}{ccc}4&5&6 \\ 7&8&9\end{array}\right)\:\left(\begin{array}{c}1\\2\\3\end{array}\right) =1\left(\begin{array}{c}4\\7\end{array}\right) +2\left(\begin{array}{c}5\\8\end{array}\right)+3\left(\begin{array}{c}6\\9\end{array}\right)=\left(\begin{array}{c}32\\50\end{array}\right).\nonumber$

In order for $$Ax$$ to make sense, the number of entries of $$x$$ has to be the same as the number of columns of $$A\text{:}$$ we are using the entries of $$x$$ as the coefficients of the columns of $$A$$ in a linear combination. The resulting vector has the same number of entries as the number of rows of $$A\text{,}$$ since each column of $$A$$ has that number of entries.

Note $$\PageIndex{2}$$

If $$A$$ is an $$m\times n$$ matrix ($$m$$ rows, $$n$$ columns), then $$Ax$$ makes sense when $$x$$ has $$n$$ entries. The product $$Ax$$ has $$m$$ entries.

Proposition $$\PageIndex{1}$$: Properties of the Matrix-Vector Product

Let $$A$$ be an $$m\times n$$ matrix, let $$u,v$$ be vectors in $$\mathbb{R}^n\text{,}$$ and let $$c$$ be a scalar. Then:

• $$A(u+v) = Au + Av$$
• $$A(cu) = cAu$$
Definition $$\PageIndex{2}$$: Matrix Equation

A matrix equation is an equation of the form $$Ax=b\text{,}$$ where $$A$$ is an $$m\times n$$ matrix, $$b$$ is a vector in $$\mathbb{R}^m\text{,}$$ and $$x$$ is a vector whose coefficients $$x_1,x_2,\ldots,x_n$$ are unknown.

In this book we will study two complementary questions about a matrix equation $$Ax=b\text{:}$$

1. Given a specific choice of $$b\text{,}$$ what are all of the solutions to $$Ax=b\text{?}$$
2. What are all of the choices of $$b$$ so that $$Ax=b$$ is consistent?

The first question is more like the questions you might be used to from your earlier courses in algebra; you have a lot of practice solving equations like $$x^2-1=0$$ for $$x$$. The second question is perhaps a new concept for you. Theorem 2.9.1 in Section 2.9, which is the culmination of this chapter, tells us that the two questions are intimately related.

Note $$\PageIndex{3}$$: Matrix Equations and Vector Equations

Let $$v_1,v_2,\ldots,v_n$$ and $$b$$ be vectors in $$\mathbb{R}^m$$. Consider the vector equation

$x_1v_1 + x_2v_2 + \cdots + x_nv_n = b. \nonumber$

This is equivalent to the matrix equation $$Ax=b\text{,}$$ where

$A=\left(\begin{array}{cccc}|&|&\quad &| \\ v_1 &v_2 &\cdots &v_n \\ |&|&\quad &|\end{array}\right)\quad\text{and}\quad x=\left(\begin{array}{c}x_1 \\ x_2 \\ \vdots \\ x_n \end{array}\right).\nonumber$

Conversely, if $$A$$ is any $$m\times n$$ matrix, then $$Ax=b$$ is equivalent to the vector equation

$x_1v_1 + x_2v_2 + \cdots + x_nv_n = b, \nonumber$

where $$v_1,v_2,\ldots,v_n$$ are the columns of $$A\text{,}$$ and $$x_1,x_2,\ldots,x_n$$ are the entries of $$x$$.

Example $$\PageIndex{2}$$

Write the vector equation

$2v_1 +3v_2 -4v_3 =\left(\begin{array}{c}7\\2\\1\end{array}\right)\nonumber$

as a matrix equation, where $$v_1,v_2,v_3$$ are vectors in $$\mathbb{R}^3$$.

Solution

Let $$A$$ be the matrix with columns $$v_1,v_2,v_3\text{,}$$ and let $$x$$ be the vector with entries $$2,3,-4$$. Then

$Ax=\left(\begin{array}{ccc}|&|&| \\ v_1 & v_2 & v_3 \\ |&|&|\end{array}\right)\:\left(\begin{array}{c}2\\3\\-4\end{array}\right) = 2v_1 +3v_2 -4v_3, \nonumber$

so the vector equation is equivalent to the matrix equation $$Ax=\left(\begin{array}{c}7\\2\\1\end{array}\right)$$.

Note $$\PageIndex{4}$$: Four Ways of Writing a Linear System

We now have four equivalent ways of writing (and thinking about) a system of linear equations:

1. As a system of equations:
$\left\{\begin{array}{rrrrrrr} 2x_1 &+& 3x_2 &-& 2x_3 &=& 7\\ x_1 &-& x_2 &-& 3x_3 &=& 5\end{array}\right.\nonumber$
2. As an augmented matrix:
$\left(\begin{array}{ccc|c} 2&3&-2&7 \\ 1&-1&-3&5\end{array}\right)\nonumber$
3. As a vector equation ($$x_1v_1 + x_2v_2 + \cdots + x_nv_n = b$$):
$x_{1}\left(\begin{array}{c}2\\1\end{array}\right)+x_2\left(\begin{array}{c}3\\-1\end{array}\right)+x_3\left(\begin{array}{c}-2\\-3\end{array}\right)=\left(\begin{array}{c}7\\5\end{array}\right)\nonumber$
4. As a matrix equation ($$Ax=b$$):
$\left(\begin{array}{ccc}2&3&-2 \\ 1&-1&-3\end{array}\right)\:\left(\begin{array}{c}x_1 \\ x_2 \\ x_3\end{array}\right)=\left(\begin{array}{c}7\\5\end{array}\right).\nonumber$

In particular, all four have the same solution set.

Note $$\PageIndex{5}$$

We will move back and forth freely between the four ways of writing a linear system, over and over again, for the rest of the book.

Another Way to Compute $$Ax$$

The above definition for Product, Definition $$\PageIndex{1}$$, is a useful way of defining the product of a matrix with a vector when it comes to understanding the relationship between matrix equations and vector equations. Here we give a definition that is better-adapted to computations by hand.

Definition $$\PageIndex{3}$$: Row Vector

A row vector is a matrix with one row. The product of a row vector of length $$n$$ and a (column) vector of length $$n$$ is

$\left(\begin{array}{cccc}a_1 &a_2 &\cdots a_n \end{array}\right)\:\left(\begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array}\right) =a_1 x_1 + a_2 x_2 +\cdots + a_n x_n .\nonumber$

This is a scalar.

Recipe: The Row-Column Rule for Matrix-Vector Multiplication

If $$A$$ is an $$m\times n$$ matrix with rows $$r_1,r_2,\ldots,r_m\text{,}$$ and $$x$$ is a vector in $$\mathbb{R}^n\text{,}$$ then

$Ax=\left(\begin{array}{c} — r_1 — \\ —r_2 — \\ \vdots \\ — r_m —\end{array}\right) x=\left(\begin{array}{c} r_1 x \\ r_2 x \\ \vdots \\ r_m x\end{array}\right).\nonumber$

Example $$\PageIndex{3}$$

$\left(\begin{array}{ccc}4&5&6 \\ 7&8&9\end{array}\right)\:\left(\begin{array}{c}1\\2\\3\end{array}\right)=\left(\begin{array}{cc}{\left(\begin{array}{c}4&5&6\end{array}\right)}&{\left(\begin{array}{c}1\\2\\3\end{array}\right)}\\{\left(\begin{array}{ccc}7&8&9\end{array}\right)}&{\left(\begin{array}{c}1\\2\\3\end{array}\right)}\end{array}\right) =\left(\begin{array}{ccccccccccc} 4 & \cdot &1&+&5 &\cdot & 2&+& 6& \cdot & 3 \\ 7 &\cdot & 1&+&8 & \cdot & 2&+&9 & \cdot &3\end{array}\right)=\left(\begin{array}{c} 32\\50\end{array}\right).\nonumber$

This is the same answer as before:

$\left(\begin{array}{ccc}4&5&6 \\ 7&8&9\end{array}\right)\:\left(\begin{array}{c}1\\2\\3\end{array}\right)=1\left(\begin{array}{c}4\\7\end{array}\right)+2\left(\begin{array}{c}5\\8\end{array}\right)+3\left(\begin{array}{c}6\\9\end{array}\right)=\left(\begin{array}{ccccccccccc} 1 & \cdot & 4&+& 2& \cdot &5&+&3& \cdot &6 \\ 1 & \cdot &7&+& 2& \cdot &8&+&3& \cdot &9\end{array}\right)=\left(\begin{array}{c}32\\50\end{array}\right).\nonumber$

Spans and Consistency

Let $$A$$ be a matrix with columns $$v_1,v_2,\ldots,v_n\text{:}$$

$A=\left(\begin{array}{cccc}|&|&\quad &| \\ v_1 &v_2 &\cdots & v_n \\ |&|&\quad &|\end{array}\right).\nonumber$

Then

$\begin{split} Ax=b&\text{ has a solution} \\ &\iff \text{there exist x_1,x_2,\ldots,x_n such that } A\left(\begin{array}{c}x_1 \\ x_2 \\ \vdots \\ x_n\end{array}\right) = b \\ &\iff \text{there exist x_1,x_2,\ldots,x_n such that } x_1v_1 + x_2v_2 + \cdots + x_nv_n = b \\ &\iff \text{b is a linear combination of } v_1,v_2,\ldots,v_n \\ &\iff \text{b is in the span of the columns of A}. \end{split} \nonumber$

Note $$\PageIndex{6}$$: Spans and Consistency

The matrix equation $$Ax=b$$ has a solution if and only if $$b$$ is in the span of the columns of $$A$$.

This gives an equivalence between an algebraic statement ($$Ax=b$$ is consistent), and a geometric statement ($$b$$ is in the span of the columns of $$A$$).

Example $$\PageIndex{4}$$: An Inconsistent System

Let $$A=\left(\begin{array}{cc}2&1\\ -1&0 \\ 1&-1\end{array}\right)$$. Does the equation $$Ax=\left(\begin{array}{c}0\\2\\2\end{array}\right)$$ have a solution?

Solution

First we answer the question geometrically. The columns of $$A$$ are

$\color{Red}{v_1 =\left(\begin{array}{c}2\\-1\\1\end{array}\right)}\quad\color{black}{\text{and}}\quad\color{blue}{v_2 =\left(\begin{array}{c}1\\0\\-1\end{array}\right)}\color{black}{,}\nonumber$

and the target vector (on the right-hand side of the equation) is $$\color{Green}{w=\left(\begin{array}{c}0\\2\\2\end{array}\right)}$$. The equation $$Ax=w$$ is consistent if and only if $$w$$ is contained in the span of the columns of $$A$$. So we draw a picture:

Figure $$\PageIndex{1}$$

It does not appear that $$w$$ lies in $$\text{Span}\{v_1,v_2\},$$ so the equation is inconsistent.

Let us check our geometric answer by solving the matrix equation using row reduction. We put the system into an augmented matrix and row reduce:

$\left(\begin{array}{cc|c} 2&1&0 \\ -1&0&2 \\ 1&-1&2\end{array}\right) \quad\xrightarrow{\text{RREF}}\quad \left(\begin{array}{cc|c} 1&0&0 \\ 0&1&0 \\ 0&0&1\end{array}\right).\nonumber$

The last equation is $$0=1\text{,}$$ so the system is indeed inconsistent, and the matrix equation

$\left(\begin{array}{cc}2&1\\-1&0\\1&-1\end{array}\right)x=\left(\begin{array}{c}0\\2\\2\end{array}\right)\nonumber$

has no solution.

Example $$\PageIndex{5}$$: A Consistent System

Let $$A=\left(\begin{array}{cc}2&1\\-1&0\\1&-1\end{array}\right)$$. Does the equation $$Ax=\left(\begin{array}{c}1\\-1\\2\end{array}\right)$$ have a solution?

Solution

First we answer the question geometrically. The columns of $$A$$ are

$\color{Red}{v_1=\left(\begin{array}{c}2\\-1\\1\end{array}\right)}\quad\color{black}{\text{and}}\quad\color{blue}{v_2 =\left(\begin{array}{c}1\\0\\-1\end{array}\right)},\nonumber$

and the target vector (on the right-hand side of the equation) is $$\color{Green}{w=\left(\begin{array}{c}1\\-1\\2\end{array}\right)}$$. The equation $$Ax=w$$ is consistent if and only if $$w$$ is contained in the span of the columns of $$A$$. So we draw a picture:

Figure $$\PageIndex{3}$$

It appears that $$w$$ is indeed contained in the span of the columns of $$A\text{;}$$ in fact, we can see

$w = v_1 - v_2 \implies x = \left(\begin{array}{c}1\\-1\end{array}\right). \nonumber$

Let us check our geometric answer by solving the matrix equation using row reduction. We put the system into an augmented matrix and row reduce:

$\left(\begin{array}{cc|c}2&1&1\\-1&0&-1\\1&-1&2\end{array}\right) \quad\xrightarrow{\text{RREF}}\quad \left(\begin{array}{cc|c} 1&0&1 \\ 0&1&-1 \\ 0&0&0\end{array}\right).\nonumber$

This gives us $$x=1$$ and $$y=-1\text{,}$$ which is consistent with the picture:

$1\left(\begin{array}{c}2\\-1\\1\end{array}\right)-1\left(\begin{array}{c}1\\0\\-1\end{array}\right)=\left(\begin{array}{c}1\\-1\\2\end{array}\right)\quad\text{or}\quad A\left(\begin{array}{c}1\\-1\end{array}\right)=\left(\begin{array}{c}1\\-1\\2\end{array}\right).\nonumber$

When Solutions Always Exist.

Building on the Note $$\PageIndex{6}$$: Spans and Consistency, we have the following criterion for when $$Ax=b$$ is consistent for every choice of $$b$$.

Theorem $$\PageIndex{1}$$

Let $$A$$ be an $$m\times n$$ (non-augmented) matrix. The following are equivalent:

1. $$Ax=b$$ has a solution for all $$b$$ in $$\mathbb{R}^m$$.
2. The span of the columns of $$A$$ is all of $$\mathbb{R}^m$$.
3. $$A$$ has a pivot position, Definition 1.2.5 in Section 1.2, in every row.
Proof

The equivalence of 1 and 2 is established by the Note $$\PageIndex{6}$$: Spans and Consistency as applied to every $$b$$ in $$\mathbb{R}^m$$.

Now we show that 1 and 3 are equivalent. (Since we know 1 and 2 are equivalent, this implies 2 and 3 are equivalent as well.) If $$A$$ has a pivot in every row, then its reduced row echelon form looks like this:

$\left(\begin{array}{ccccc}1&0&\star &0&\star \\ 0&1&\star &0&\star \\ 0&0&0&1&\star \end{array}\right),\nonumber$

and therefore $$\left(\begin{array}{c|c}A&b\end{array}\right)$$ reduces to this:

$\left(\begin{array}{ccccc|c} 1&0&\star &0&\star &\star \\ 0&1&\star &0&\star &\star \\ 0&0&0&1&\star &\star\end{array}\right).\nonumber$

There is no $$b$$ that makes it inconsistent, so there is always a solution. Conversely, if $$A$$ does not have a pivot in each row, then its reduced row echelon form looks like this:

$\left(\begin{array}{ccccc}1&0&\star &0&\star \\ 0&1&\star &0&\star \\ 0&0&0&0&0\end{array}\right),\nonumber$

which can give rise to an inconsistent system after augmenting with $$b\text{:}$$

$\left(\begin{array}{ccccc|c} 1&0&\star &0&\star &0 \\ 0&1&\star &0&\star &0 \\ 0&0&0&0&0&16\end{array}\right).\nonumber$

Recall that equivalent means that, for any given matrix $$A\text{,}$$ either all of the conditions of the above Theorem, $$\PageIndex{1}$$, are true, or they are all false.

Note $$\PageIndex{7}$$

Be careful when reading the statement of the above Theorem, $$\PageIndex{1}$$. The first two conditions look very much like this Note $$\PageIndex{6}$$: Spans and Consistency, but they are logically quite different because of the quantifier “for all $$b$$”.

Remark

We will see in this Corollary 2.7.1 in Section 2.7 that the dimension of the span of the columns is equal to the number of pivots of $$A$$. That is, the columns of $$A$$ span a line if $$A$$ has one pivot, they span a plane if $$A$$ has two pivots, etc. The whole space $$\mathbb{R}^m$$ has dimension $$m\text{,}$$ so this generalizes the fact that the columns of $$A$$ span $$\mathbb{R}^m$$ when $$A$$ has $$m$$ pivots.

This page titled 2.3: Matrix Equations is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Dan Margalit & Joseph Rabinoff via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.