Math Infinitum

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Intuition

Each entry of a product pairs one row of the first matrix with one column of the second: multiply the entries in pairs and add. This rule and the column-by-column definition are the same product, seen at two scales.

An order lists quantities and a price list lists prices, and the bill multiplies them item by item and adds. A table of bills, for several orders against several price lists, is a matrix product.

The row–column rule

Column $j$ of $AB$ is $A\mathbf{b}_j$ , and component $i$ of $A\mathbf{b}_j$ is $a_{i1}b_{1j}+\cdots+a_{in}b_{nj}$ . So the entry in row $i$ and column $j$ of $AB$ pairs row $i$ of $A$ with column $j$ of $B$ . The pairing needs a row of $A$ exactly as long as a column of $B$ , which is the shape condition met before.

(AB)_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+\cdots+a_{in}b_{nj}

What products obey, and what they do not

$(AB)C=A(BC)$ , so a product of three matrices needs no brackets. The theorem below proves it.
$A(B+C)=AB+AC$ , and .

Multiplication is associative

(AB)C=A(BC)

Two matrices of the same shape are equal when they do the same thing to every vector, because column $j$ of a matrix is what it does to $\mathbf{e}_j$ . Both sides apply $C$ , then $B$ , then , and the rule says so in two steps for each side.

Proof steps

\big((AB)C\big)\mathbf{x}=(AB)(C\mathbf{x})

The product of $AB$ and $C$ applies $C$ first: that is the defining rule of a product.

(AB)(C\mathbf{x})=A\big(B(C\mathbf{x})\big)

The same rule once more, for $AB$ acting on the vector $C\mathbf{x}$ .

\big(A(BC)\big)\mathbf{x}=A\big(B(C\mathbf{x})\big)

The other side unfolds the same way: $A(BC)$ applies $BC$ first, and $BC$ applies $C$ and then $B$ .

\mathbf{x}=\mathbf{e}_j\ \Longrightarrow\ \text{equal columns}

Both sides send every vector to the same place. Taking the standard vectors in turn, their columns agree one by one.

(AB)C=A(BC)

Matrices with the same columns are the same matrix.

Applications

Practice

Row Times Column

To find the entry in row $i$ and column $j$ of $AB$ , run along row $i$ of $A$ and down column $j$ of together, multiply the pairs you meet, and add.

\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}5&6\\7&8\end{pmatrix}=\begin{pmatrix}1\cdot5+2\cdot7&1\cdot6+2\cdot8\\3\cdot5+4\cdot7&3\cdot6+4\cdot8\end{pmatrix}

=\begin{pmatrix}19&22\\43&50\end{pmatrix}

Try it

$A$ is $3\times2$ and $B$ is $2\times4$ , so $AB$ is $3\times4$ . Select the entry of $AB$ that is computed from row 3 of $A$ and column 2 of $B$ .

$AB=\begin{pmatrix}c_{11} & c_{12} & c_{13} & c_{14} \\ c_{21} & c_{22} & c_{23} & c_{24} \\ c_{31} & c_{32} & c_{33} & c_{34}\end{pmatrix}$

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Let $A=\begin{pmatrix}2&-1\\0&3\end{pmatrix}$ and $B=\begin{pmatrix}4&1\\5&-2\end{pmatrix}$ . What is the entry in row 1, column 2 of $AB$ ?

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What is $\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}3&0\\1&2\end{pmatrix}$ ?

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Order Matters

Composing maps in the other order usually gives a different map, so $AB$ and $BA$ usually differ, even for square matrices of the same order.

\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&0\end{pmatrix}

\begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\0&1\end{pmatrix}

Try it

For any two $2\times2$ matrices $A$ and $B$ , $AB=BA$ .

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Two Habits from Numbers That Fail

For numbers, $ab=0$ forces $a=0$ or $b=0$ , and $ab=ac$ with forces . Neither survives for matrices.

\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}1&-1\\-1&1\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}

Try it

$A$ and $B$ are $2\times2$ matrices with $AB=O$ . What follows?

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Powers

For a square matrix, $A^2=AA$ , $A^3=AAA$ , and so on. Because the product is associative, it does not matter how the factors are grouped.

\begin{pmatrix}1&1\\0&1\end{pmatrix}^{2}=\begin{pmatrix}1&2\\0&1\end{pmatrix}

Try it

Let $A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$ . What is the entry in row 1, column 2 of $A^3$ ?

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For square matrices $A$ and $B$ of the same order, what is $(A+B)^2$ ?

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$A$ is a $2\times3$ matrix. Which products are defined and equal to $A$ ?

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What is the single entry of $\begin{pmatrix}1&-2&3\end{pmatrix}\begin{pmatrix}4\\0\\-1\end{pmatrix}$ ?

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If $AB=AC$ for $2\times2$ matrices and $A\neq O$ , then $B=C$ .

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What You Learned

Entry $(i,j)$ of $AB$ pairs row $i$ of $A$ with column $j$ of : multiply in pairs and add.

Final checkpoint

Try it

Let $A=\begin{pmatrix}1&0&2\\-1&3&1\end{pmatrix}$ and $B=\begin{pmatrix}2&1\\0&-1\\4&5\end{pmatrix}$ . What is the entry in row 2, column 2 of $AB$ ?

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Try it

Which statement is true for all square matrices $A$ , $B$ , $C$ of the same order?

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For every square matrix $A$ , each entry of $A^2$ is the square of the corresponding entry of $A$ .

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Try it

The matrix $\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}$ records the roads between three towns: entry $(i,j)$ is $1$ when towns $i$ and $j$ are joined by a road. How many routes of exactly two roads lead from town 1 back to town 1?

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Completion

Lesson complete

Great work! You now know how to:

compute any entry of a product by pairing a row with a column;
multiply matrices of compatible shapes;
use associativity and distributivity, and recognise when order matters;
compute powers of a square matrix.