2.1 Matrix basics

In the equation $2x_1-x_3=5$, the coefficients in the order $(x_1,x_2,x_3)$ are $2,0,-1$. Omitting the zero would lose the distinction between the second and third unknowns. Writing the coefficients of several equations in aligned rows preserves that information: each row belongs to an equation, and each column belongs to a fixed unknown.

What a matrix is

Definition 2.1.1

(Matrix)

A matrix is a rectangular array of numbers arranged in rows and columns.

If a matrix has $m$ rows and $n$ columns, we say it is an $m \times n$ matrix. The entries are usually real numbers in this course, although the formal definitions make sense over other number systems as well.

The point of the rectangular format is not decoration. A row records one line of data, a column records another line of data, and the position of an entry matters. Later, multiplication and row operations will depend on that position.

This is why a matrix should not be treated as a loose list of numbers. The entry in row $1$, column $3$ can play a completely different role from the same number placed in row $3$, column $1$. In a coefficient matrix, for example, columns are tied to the chosen order of the variables. If the variable order is $(x_1, x_2, x_3)$, then the second column records coefficients of $x_2$. If the order is changed to $(x_2, x_1, x_3)$, the displayed matrix must change as well.

Common mistake

(Do not separate a coefficient from its position)

The number $5$ by itself does not say which variable it multiplies. In a coefficient matrix, the column position tells you that information. Moving a column changes the encoded system unless the variable order is changed at the same time.

Interactive · Match up$a_{23}=4$

Read a matrix by its row, column and entry

Select an entry. Match ① its row, ② its column and ③ its value with the same numbered readings.

① Row index i

Row 1Row 2

② Column index j

Column 1Column 2Column 3

③ Entry

$a_{11}$$a_{12}$$a_{13}$$a_{21}$$a_{22}$$a_{23}$

1	2	0
3	−1	4

① Row 2: (3, -1, 4)

② Column 3: (0, 4)^T

③ $a_{23}=4$

① solid row borders · ② dashed column borders · ③ underlined entry

① Row 2

$(a_{21},a_{22},a_{23})=(3, -1, 4)$

② Column 3

$(a_{13},a_{23})^T=(0, 4)^T$

③ Entry and index

$a_{23}=4$
A[1][2] = 4

Size

2 rows by 3 columns, so the size is 2 x 3

Holds: the two indices locate the selected entry

a23 is in row 2, column 3, and equals 4. Array indices start at zero; matrix subscripts here start at one.

NoticeThe first subscript chooses the row; the second chooses the column. Selecting another entry does not change the matrix.

← → Switch within the focused index or entry group.

Read the row-column map

Track one entry through the diagram and identify its row, column, and two indices. Its value and its position are separate pieces of information.

Read the size before you touch the entries

The size of a matrix is written as $m \times n$.

• $m$ is the number of rows.
• $n$ is the number of columns.

If $m = n$, the matrix is square.

Two matrices with different sizes are different kinds of objects. A $2 \times 3$ matrix and a $3 \times 2$ matrix are not even comparable entry by entry, because their row-column positions do not match.

Read entries one by one

If $A$ is a matrix, then $a_{ij}$ means the entry in row $i$ and column $j$. That notation matters because it tells you exactly where a number lives inside the array.

Worked example 2.1.2

(Reading a matrix carefully)

Let

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

This matrix has 2 rows and 3 columns, so its size is $2 \times 3$. Its entry in row 2, column 3 is $4$.

The notation $a_{ij}$ is not optional bookkeeping. It is the language used in definitions such as matrix equality, matrix addition, and matrix multiplication.

Matrix equality is entrywise

Two matrices are equal only when they have the same size and every corresponding entry matches.

Definition 2.1.3

(Matrix equality)

Let $A = [a_{ij}]$ and $B = [b_{ij}]$ be matrices.

Then $A = B$ if and only if:

1. $A$ and $B$ have the same size, and
2. $a_{ij} = b_{ij}$ for every row index $i$ and column index $j$.

This means that proving two matrices are equal is often an entry-by-entry argument.

Worked example 2.1.4

(Using matrix equality to solve for an unknown entry)

Suppose

$$\begin{bmatrix} 1 & 2 \\ 3 & x \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}.$$

The two matrices already agree in three positions. Equality of matrices forces the remaining position to agree as well, so $x = 5$.

A matrix is determined by its ordered columns

The equality definition has a useful consequence: complete column information is enough to reconstruct a matrix. We do not need to compare the whole display at once, but we do need every column in its specified position.

Theorem 2.1.5

(Coordinate reconstruction and equality)

Given an ordered list $c_1,\ldots,c_n$ of column vectors in $\mathbb R^m$, there is exactly one $m\times n$ matrix with these columns. Its entry $a_{ij}$ is the $i$th entry of $c_j$. Consequently, two matrices of this size are equal if and only if their corresponding columns are equal. The analogous statement holds for their ordered rows.

Proof

(Existence and uniqueness come directly from the entries)

For existence, place the $i$th coordinate of the $j$th supplied vector in position $(i,j)$. Every required position receives exactly one real number, and each column then has the prescribed entries in the prescribed order.

For uniqueness, let $B$ be any other matrix with the same ordered columns. For each column index $j$ and each row index $i$, its entry $b_{ij}$ must be the $i$th coordinate of $c_j$, hence must equal $a_{ij}$. The matrices have the same size and all corresponding entries agree, so they are equal by definition. Conversely, equal matrices have equal entries in each column, which makes the corresponding column vectors equal. Grouping the same argument by row instead of by column proves the row version.

Notice that this proof has both an existence part and an at-most-one part. The word “ordered” is also essential. A set of column vectors forgets which column comes first and forgets repetitions. Neither piece of information may be lost when reconstructing a matrix. A row vector and a column vector containing the same numbers likewise have different shapes; the theorem never identifies them merely because the printed numbers agree.

Worked example 2.1.6

(Reconstruct a matrix and test a second description)

Suppose the first, second, and third columns of $A$ are respectively

$$c_1=\begin{bmatrix}2\\-1\end{bmatrix},\qquad c_2=\begin{bmatrix}0\\4\end{bmatrix},\qquad c_3=\begin{bmatrix}-3\\5\end{bmatrix}.$$

There are three columns of length two, so

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

The first row is $[2\ 0\ {-3}]$, and $a_{23}=5$. A second description says that the rows of $A$ are $[2\ 0\ {-3}]$ and $[-1\ 4\ 5]$. Reading each position confirms that both descriptions determine the same matrix. If that second row were $[-1\ 5\ 4]$, the descriptions would conflict in two positions; there would be no matrix satisfying both. Having the same collection of numbers does not resolve the conflict.

If $c_1$ and $c_2$ are exchanged, the new first entry becomes zero instead of two. The result is another matrix, even though the list of available columns has not changed. This is the same positional issue that arises when columns record coefficients of differently ordered variables.

Addition and scalar multiplication come first

Before matrix multiplication appears, there are two basic operations you should already read confidently.

Definition 2.1.7

(Addition and scalar multiplication)

Let $A = [a_{ij}]$ and $B = [b_{ij}]$ be matrices of the same size, and let $c$ be a scalar.

• The sum $A + B$ is the matrix obtained by adding corresponding entries: $(A + B)_{ij} = a_{ij} + b_{ij}$.
• The scalar multiple $cA$ is the matrix obtained by multiplying every entry of $A$ by $c$: $(cA)_{ij} = c a_{ij}$.

The phrase "of the same size" is essential. Matrix addition is not defined for matrices of different sizes.

Worked example 2.1.8

(Compute a sum and a scalar multiple)

Let

$$A = \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}, \qquad B = \begin{bmatrix} 4 & 1 \\ -5 & 2 \end{bmatrix}.$$

Then

$$A + B = \begin{bmatrix} 5 & -1 \\ -5 & 5 \end{bmatrix}, \qquad 2A = \begin{bmatrix} 2 & -4 \\ 0 & 6 \end{bmatrix}.$$

Every entry is handled separately, but the size stays $2 \times 2$.

The zero matrix is the matrix whose entries are all $0$. For each size it plays the role of the additive identity:

$$A + O = A.$$

Which operations can be cancelled?

The entrywise definitions let us justify familiar algebraic moves before matrix multiplication is introduced. An equality between matrix expressions can be solved by scalar arithmetic at each position, provided all the expressions have the required size. Cancellation is a conclusion from those definitions, not a permission to treat every matrix operation like multiplication of numbers.

Theorem 2.1.9

(Cancellation for entrywise operations)

Let $A,B,C$ be real matrices of the same size. If $A+C=B+C$, then $A=B$. For a fixed real scalar $\lambda\ne0$, if $\lambda A=\lambda B$, then $A=B$. The nonzero hypothesis in the second statement is necessary.

Proof

(Reduce cancellation to one arbitrary position)

From $A+C=B+C$, the equality definition gives $a_{ij}+c_{ij}=b_{ij}+c_{ij}$ at every position. Cancelling the real number $c_{ij}$ gives $a_{ij}=b_{ij}$. Since the indices were arbitrary, matrix equality follows. Likewise, $\lambda A=\lambda B$ gives $\lambda a_{ij}=\lambda b_{ij}$; division by the nonzero real number $\lambda$ gives equality at every position. If $\lambda=0$, both scalar multiples are the zero matrix for any $A,B$, so their equality no longer forces the original matrices to agree.

The proofs also explain how to solve a basic matrix equation. In $2X+B=C$, a solution must have entries $x_{ij}=(c_{ij}-b_{ij})/2$. Conversely, assigning those entries produces a matrix satisfying the equation, since substitution recovers every entry of $C$. Thus the formula both constructs a solution and shows there cannot be another one. The matching-size condition is checked before this calculation; subtraction cannot repair incompatible dimensions.

For a concrete check, take

$$B=\begin{bmatrix}1&-2\\3&0\end{bmatrix},\qquad C=\begin{bmatrix}5&4\\-1&6\end{bmatrix}.$$

Then $X=\begin{bmatrix}2&3\\-2&3\end{bmatrix}$, and doubling its entries and adding the corresponding entries of $B$ recovers $C$. If the coefficient of $X$ were zero instead, the equation would reduce to $B=C$: every same-size matrix would solve it when that equality holds, and none would solve it when it fails. This is the matrix version of interpreting a zero-coefficient scalar equation, rather than attempting an illegal division.

A matrix records a linear system compactly

One reason matrices matter so early is that they package a linear system in a form that is easier to transform systematically.

Consider the system

$$\begin{aligned} x_1 + 2x_2 - x_3 &= 4, \\ 3x_1 - x_2 + 5x_3 &= 7. \end{aligned}$$

Its coefficient matrix is

$$A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & -1 & 5 \end{bmatrix},$$

its unknown vector is

$$x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix},$$

and its constant vector is

$$b = \begin{bmatrix} 4 \\ 7 \end{bmatrix}.$$

Use the abbreviation

$$Ax = b.$$

Here $Ax=b$ stands for the two displayed equations: its $i$th equation is $a_{i1}x_1+a_{i2}x_2+a_{i3}x_3=b_i$. The multiplication rule that gives this abbreviation its formal meaning is developed in 3.2 Matrix multiplication, identity matrices, and linear systems.

A missing variable still has a column

Worked example 2.1.10

(From wording to coefficient matrix)

Question: using the variable order $(x_1,x_2,x_3)$, write the coefficient matrix and constant vector for

$$\begin{aligned} 2x_1 - x_3 &= 5,\\ -x_1 + 4x_2 + 3x_3 &= -2. \end{aligned}$$

First, rewrite the first equation with the missing $x_2$ coefficient shown:

$$2x_1 + 0x_2 - x_3 = 5.$$

Now each equation becomes one row, and the columns follow the fixed variable order $(x_1,x_2,x_3)$. Therefore

$$A = \begin{bmatrix} 2 & 0 & -1 \\ -1 & 4 & 3 \end{bmatrix}, \qquad b = \begin{bmatrix} 5 \\ -2 \end{bmatrix}.$$

The most common mistake is to omit the $0$ in the first row. But the missing $x_2$ coefficient is part of the data, and column 2 must still exist.

A short preview of multiplication

Matrix multiplication in 3.2 pairs a row of the left matrix with a column of the right matrix to compute one output entry. The figure previews how those positions match; the next note uses the same fixed positions to record augmented systems and row operations.

Common mistakes

Common mistake

(Do not swap rows and columns)

The first index is the row number, not the column number. $a_{23}$ means row 2, column 3.

Common mistake

(Different sizes cannot be added)

Matrix addition is defined entrywise. If the positions do not line up, there is no operation to perform.

Quick check

Quick check

(Can a $2 \times 3$ matrix be added to a $3 \times 2$ matrix?)

Answer from the definition of matrix addition, not from visual guesswork.

Solution · Answer

No. Matrix addition is only defined for matrices of the same size.

Quick check

(If $A$ is a $4 \times 2$ matrix, what does the symbol $a_{31}$ mean?)

Name both the row and the column.

Solution · Answer

It is the entry in row 3 and column 1 of $A$.

Exercise

Quick check

(Write the coefficient matrix and constant vector for the system $x_1 - x_2 = 3$, $2x_1 + x_2 = 0$.)

Keep the order of the variables fixed.

Solution · Guided solution

Using the variable order $(x_1, x_2)$, the coefficient matrix is

$$\begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix},$$

and the constant vector is

$$\begin{bmatrix} 3 \\ 0 \end{bmatrix}.$$

Exercise: reconcile partial information before calculating

A $2\times3$ matrix has first row $[1\ 2\ 3]$ and second column $\begin{bmatrix}2\\5\end{bmatrix}$. Describe every such matrix. Then decide what changes if the second column is instead $\begin{bmatrix}4\\5\end{bmatrix}$. Explain which answer involves free entries and which involves inconsistent data.

Solution · Solution: shared positions must agree

The two descriptions overlap at position $(1,2)$. In the first case, both give the value two, so they are compatible. The second column fixes the middle entry in the second row, leaving only its first and last entries unspecified. Thus all the matrices, and no others, are

$$\begin{bmatrix}1&2&3\\s&5&t\end{bmatrix},\qquad s,t\in\mathbb R.$$

Every choice of the two real numbers has the required row and column; conversely, any matrix with those data must have exactly this form. In the second case, the same position would have to contain both two and four. There is no matrix satisfying the data. No choice of the unspecified entries can remove that conflict. This separates incomplete information, which may permit many matrices, from incompatible information, which permits none.

This exercise also explains the scope of the reconstruction proposition. A complete ordered list of columns fixes every entry and gives exactly one matrix. A partial row-and-column description need not do so. Before counting unknown entries, first check the positions constrained in two different ways; otherwise a count of apparently free entries can conceal an impossibility.

Related notes

If you want to see how a system becomes a matrix, review 1.1 Equations and solution sets. Matrix multiplication is developed in 3.2 Matrix multiplication, identity matrices, and linear systems.