Chapter 3 · Matrices

Example 3: Construct a $3\times2$ matrix from $a_{ij}=\frac12|i-3j|$

ExamplesOrder of a matrix · Section 3.2.1

The question

Construct a $3\times2$ matrix whose elements are given by $a_{ij} = \frac12\,|i - 3j|$ ⁠.

Exam answer

$a_{ij} = \frac12|i - 3j|$ ⁠, where $i = 1, 2, 3$ and $j = 1, 2$ ⁠.

\begin{aligned} a_{11} &= \tfrac12|1 - 3| = 1 \\ a_{12} &= \tfrac12|1 - 6| = \tfrac52 \\ a_{21} &= \tfrac12|2 - 3| = \tfrac12 \\ a_{22} &= \tfrac12|2 - 6| = 2 \\ a_{31} &= \tfrac12|3 - 3| = 0 \\ a_{32} &= \tfrac12|3 - 6| = \tfrac32 \end{aligned}

$\therefore\ A = \begin{bmatrix} 1 & \dfrac52 \\[2pt] \dfrac12 & 2 \\[2pt] 0 & \dfrac32 \end{bmatrix}$

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A formula for $a_{ij}$ is a recipe that takes an address (row $i$ ⁠, column $j$ ⁠) and returns the number stored there. "Construct the matrix" means: list every address the order allows, feed each one into the recipe, and put the answer in that place.

Given, to find, and the plan

Given: The order $3\times2$ and the rule $a_{ij} = \frac12|i-3j|$ ⁠.
To find: The matrix $A = [a_{ij}]_{3\times2}$ ⁠.
Plan: The order tells us $i = 1, 2, 3$ and $j = 1, 2$ ⁠: six addresses. Compute $i - 3j$ ⁠, take its absolute value, halve it, for each address. Then assemble.

Basics you need

1The address $a_{ij}$

$a_{ij}$ is the entry in row $i$ ⁠, column $j$ ⁠. A $3\times2$ matrix has rows $i = 1, 2, 3$ and columns $j = 1, 2$ ⁠, so its entries are $a_{11}, a_{12}, a_{21}, a_{22}, a_{31}, a_{32}$ ⁠. Basics: rows, columns and the address a_ij

2Absolute value

$|t|$ is the distance of $t$ from 0, so it is never negative: $|t| = t$ if $t\ge 0$ and $|t| = -t$ if $t < 0$ ⁠. For example $|-5| = 5$ ⁠, $|0| = 0$ ⁠, $|3| = 3$ ⁠.

3Order of operations

In $\frac12|i-3j|$ ⁠: first multiply $3j$ ⁠, then subtract from $i$ ⁠, then take the absolute value, then halve.

Solution

Write down the shape and the six addresses

A $3\times2$ matrix has 3 rows and 2 columns:
$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix}.$
So we need $a_{ij}$ for $i = 1, 2, 3$ and $j = 1, 2$ ⁠.

Writing the empty shape first shows exactly which entries we owe. The order decides the range of each index: $i$ runs over the 3 rows and $j$ over the 2 columns, so there are $3\times2 = 6$ entries to compute.

First index = row, second = column. Basics: the address a_ij
Row 1 ( $i = 1$ ⁠)
$\begin{aligned} a_{11} &= \tfrac12|1 - 3\cdot1| = \tfrac12|-2| = \tfrac12\cdot 2 = 1,\\ a_{12} &= \tfrac12|1 - 3\cdot2| = \tfrac12|-5| = \tfrac52. \end{aligned}$
For each entry we substitute its own row number for $i$ and its own column number for $j$ ⁠. The quantity inside the bars is negative here, and the absolute value turns it positive before halving.

$|-2| = 2$ and $|-5| = 5$ ⁠: the absolute value of a negative number is its negative, $-(-2) = 2$ ⁠.
Row 2 ( $i = 2$ ⁠)
$\begin{aligned} a_{21} &= \tfrac12|2 - 3\cdot1| = \tfrac12|-1| = \tfrac12,\\ a_{22} &= \tfrac12|2 - 3\cdot2| = \tfrac12|-4| = \tfrac12\cdot 4 = 2. \end{aligned}$
Same recipe, now with $i = 2$ ⁠. Only $i$ changes as we move down the column; only $j$ changes as we move along the row.
Row 3 ( $i = 3$ ⁠)
$\begin{aligned} a_{31} &= \tfrac12|3 - 3\cdot1| = \tfrac12|0| = 0,\\ a_{32} &= \tfrac12|3 - 3\cdot2| = \tfrac12|-3| = \tfrac32. \end{aligned}$
$a_{31}$ is the only address where $i = 3j$ ⁠, so it is the only zero entry. An entry of 0 is still an entry and must be written in its place.

$|0| = 0$ ⁠: zero is its own absolute value.
Assemble the matrix

Placing each value at its address:
$A = \begin{bmatrix} 1 & \tfrac52 \\[2pt] \tfrac12 & 2 \\[2pt] 0 & \tfrac32 \end{bmatrix}.$
The matrix is the collection of all six values, each in the row and column given by its indices. It has 3 rows and 2 columns, as the question demands.

Result

A = \begin{bmatrix} 1 & \tfrac52 \\[2pt] \tfrac12 & 2 \\[2pt] 0 & \tfrac32 \end{bmatrix}_{3\times2}

This agrees with the textbook's solution.

Check it with numbers

The values inside the bars, $i - 3j$ ⁠, are $-2, -5$ (row 1), $-1, -4$ (row 2), $0, -3$ (row 3). Along each row the value drops by 3 (because $j$ goes up by 1 and is multiplied by 3), and down each column it rises by 1 (because $i$ goes up by 1). Halving the absolute values gives $1, \tfrac52, \tfrac12, 2, 0, \tfrac32$ ⁠, matching the matrix.

Common mistakes

Dropping the absolute value. Without it, $a_{11}$ would be $-1$ ⁠. The bars make every entry non-negative.
Swapping $i$ and $j$ ⁠. $a_{12}$ uses $i = 1$ ⁠, $j = 2$ ⁠: $\tfrac12|1-6| = \tfrac52$ ⁠, not $\tfrac12|2-3| = \tfrac12$ ⁠.
Building a $2\times3$ matrix. The order $3\times2$ means 3 rows and 2 columns.

What to take away

To build a matrix from a formula, list every address $(i, j)$ the order allows and evaluate the formula at each.
Work row by row so no entry is skipped, and keep $i$ as the row number and $j$ as the column number throughout.

Every fact used, and where it comes from

Fact	Used in	Source
$A = [a_{ij}]_{m\times n}$ ⁠, $1\le i\le m$ ⁠, $1\le j\le n$	Step 1	Section 3.2.1 (Lesson: what a matrix is)
$\|t\| = t$ for $t\ge0$ ⁠, $\|t\| = -t$ for $t<0$	Steps 2–4	Definition of absolute value

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Example 3: Construct a 3×23\times23×2 matrix from aij=12∣i−3j∣a_{ij}=\frac12|i-3j|aij​=21​∣i−3j∣

Basics you need

Solution

Write down the shape and the six addresses

Row 1 (i=1i = 1i=1⁠)

Row 2 (i=2i = 2i=2⁠)

Row 3 (i=3i = 3i=3⁠)

Assemble the matrix

Every fact used, and where it comes from

Example 3: Construct a $3\times2$ matrix from $a_{ij}=\frac12|i-3j|$

Row 1 ( $i = 1$ ⁠)

Row 2 ( $i = 2$ ⁠)

Row 3 ( $i = 3$ ⁠)