Chapter 3 · Matrices

Exercise 3.1 · Q5: Construct a $3\times4$ matrix from a rule for $a_{ij}$

Exercise 3.1Order of a matrix · Section 3.2

The question

Construct a $3\times4$ matrix, whose elements are given by:

(i) $a_{ij} = \dfrac{1}{2}\,|-3i + j|$ (ii) $a_{ij} = 2i - j$

Exam answer

(i) $a_{ij} = \frac12|-3i + j|$

\begin{aligned} a_{11} &= \tfrac12|-3+1| = 1 \\ a_{12} &= \tfrac12|-3+2| = \tfrac12 \\ a_{13} &= \tfrac12|-3+3| = 0 \\ a_{14} &= \tfrac12|-3+4| = \tfrac12 \\ a_{21} &= \tfrac12|-6+1| = \tfrac52 \\ a_{22} &= \tfrac12|-6+2| = 2 \\ a_{23} &= \tfrac12|-6+3| = \tfrac32 \\ a_{24} &= \tfrac12|-6+4| = 1 \\ a_{31} &= \tfrac12|-9+1| = 4 \\ a_{32} &= \tfrac12|-9+2| = \tfrac72 \\ a_{33} &= \tfrac12|-9+3| = 3 \\ a_{34} &= \tfrac12|-9+4| = \tfrac52 \end{aligned}

\therefore\ A = \begin{bmatrix} 1 & \frac12 & 0 & \frac12 \\[1mm] \frac52 & 2 & \frac32 & 1 \\[1mm] 4 & \frac72 & 3 & \frac52 \end{bmatrix}

(ii) $a_{ij} = 2i - j$

\begin{aligned} &A \\ &= \begin{bmatrix} 2-1 & 2-2 & 2-3 & 2-4 \\ 4-1 & 4-2 & 4-3 & 4-4 \\ 6-1 & 6-2 & 6-3 & 6-4 \end{bmatrix} \end{aligned}

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

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

The one idea this problem is built on

A $3\times4$ matrix has 3 rows and 4 columns, so it has $3\times4 = 12$ addresses $(i, j)$ with $i\in\{1,2,3\}$ and $j\in\{1,2,3,4\}$ ⁠. The rule turns each address into a number. We go through the rows one at a time, substituting $j = 1, 2, 3, 4$ along each row, and then assemble the matrix.

Given, to find, and the plan

Given: Two rules for $a_{ij}$ ⁠; order $3\times4$ ⁠, so $i = 1, 2, 3$ (row) and $j = 1, 2, 3, 4$ (column).
To find: The $3\times4$ matrix $[a_{ij}]$ for each rule.
Plan: For each rule, compute row 1 ( $i = 1$ ⁠), row 2 ( $i = 2$ ⁠) and row 3 ( $i = 3$ ⁠), four entries each, then assemble. In Part (i), find $-3i + j$ first, then take its absolute value, then halve.

Basics you need

1The general $3\times4$ matrix

$\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}$ Row $i$ holds $a_{i1}, a_{i2}, a_{i3}, a_{i4}$ ⁠. Basics: rows, columns and the address

2Absolute value

$|x|$ is the distance of $x$ from 0, so it is never negative: $|x| = x$ if $x\ge0$ and $|x| = -x$ if $x < 0$ ⁠. For example $|-5| = 5$ ⁠, $|3| = 3$ ⁠, $|0| = 0$ ⁠.

3Substitution with negatives

$-3i$ with $i = 2$ is $-3\cdot2 = -6$ ⁠. Work out the inside of $|\ |$ completely before taking the absolute value. Basics: working with equations

4The worked model

Example 3 builds a $3\times2$ matrix from $a_{ij} = \frac12|i - 3j|$ ⁠, a close cousin of Part (i). Lesson: what a matrix is

Part (i): $a_{ij} = \frac{1}{2}|-3i + j|$

Row 1 ( $i = 1$ ⁠): here $-3i = -3$
$\begin{aligned} a_{11} &= \tfrac12|-3+1| = \tfrac12|-2| = 1 \\ a_{12} &= \tfrac12|-3+2| = \tfrac12|-1| = \tfrac12 \\ a_{13} &= \tfrac12|-3+3| = \tfrac12|0| = 0 \\ a_{14} &= \tfrac12|-3+4| = \tfrac12|1| = \tfrac12 \end{aligned}$
The inside $-3 + j$ is negative for $j = 1, 2$ ⁠, zero for $j = 3$ and positive for $j = 4$ ⁠. The absolute value makes every result non-negative, so $-2$ becomes $2$ and $-1$ becomes $1$ ⁠. Halving comes last.

$|x| = -x$ when $x < 0$ ⁠, e.g. $|-2| = 2$ ⁠.
Row 2 ( $i = 2$ ⁠): here $-3i = -6$
$\begin{aligned} a_{21} &= \tfrac12|-6+1| = \tfrac12|-5| = \tfrac52 \\ a_{22} &= \tfrac12|-6+2| = \tfrac12|-4| = 2 \\ a_{23} &= \tfrac12|-6+3| = \tfrac12|-3| = \tfrac32 \\ a_{24} &= \tfrac12|-6+4| = \tfrac12|-2| = 1 \end{aligned}$
For $i = 2$ and $j\le4$ ⁠, the inside $-6 + j$ is at most $-2$ ⁠, so it is always negative here and $|-6 + j| = 6 - j$ ⁠. That gives the values $5, 4, 3, 2$ ⁠, each then halved.

Work inside the bars first, then the bars, then the $\frac12$ ⁠.
Row 3 ( $i = 3$ ⁠): here $-3i = -9$
$\begin{aligned} a_{31} &= \tfrac12|-9+1| = \tfrac12|-8| = 4 \\ a_{32} &= \tfrac12|-9+2| = \tfrac12|-7| = \tfrac72 \\ a_{33} &= \tfrac12|-9+3| = \tfrac12|-6| = 3 \\ a_{34} &= \tfrac12|-9+4| = \tfrac12|-5| = \tfrac52 \end{aligned}$
Again the inside $-9 + j$ is negative for every $j\le4$ ⁠, so $|-9+j| = 9 - j$ ⁠, giving $8, 7, 6, 5$ before halving.
Assemble the matrix

Rows 1, 2, 3 from Part (i), Steps 1, 2, 3, in that order:
$A = \begin{bmatrix} 1 & \frac12 & 0 & \frac12 \\[1mm] \frac52 & 2 & \frac32 & 1 \\[1mm] 4 & \frac72 & 3 & \frac52 \end{bmatrix}.$
The matrix has 3 rows and 4 columns, as the question requires. Row 1 is the only row with a 0 and with a rise at the end, because it is the only row where $-3i + j$ changes sign (at $j = 3$ ⁠).

Row $i$ of the matrix lists $a_{i1}, \dots, a_{i4}$ ⁠. Basics: the address

Part (ii): $a_{ij} = 2i - j$

Compute the three rows
$\begin{aligned} i = 1:&\ \ 2-1,\ 2-2,\ 2-3,\ 2-4 \\ &= 1,\ 0,\ -1,\ -2 \\ i = 2:&\ \ 4-1,\ 4-2,\ 4-3,\ 4-4 \\ &= 3,\ 2,\ 1,\ 0 \\ i = 3:&\ \ 6-1,\ 6-2,\ 6-3,\ 6-4 \\ &= 5,\ 4,\ 3,\ 2 \end{aligned}$
For row $i$ ⁠, $2i$ is fixed ( $2, 4, 6$ ⁠) and we subtract the column numbers $j = 1, 2, 3, 4$ in turn. There is no absolute value here, so negative entries such as $a_{13} = -1$ and $a_{14} = -2$ stay negative.

$a_{ij} = 2i - j$ ⁠: row number doubled, minus the column number. Basics: substitution
Assemble the matrix
$A = \begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}.$
The pattern is a good self-check: along each row the entries fall by 1 ( $j$ goes up by 1), and down each column they rise by 2 ( $i$ goes up by 1, so $2i$ goes up by 2).

Result

\text{(i)}\ \begin{bmatrix} 1 & \frac12 & 0 & \frac12 \\[1mm] \frac52 & 2 & \frac32 & 1 \\[1mm] 4 & \frac72 & 3 & \frac52 \end{bmatrix}

\text{(ii)}\ \begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}

(Textbook answer: the same two matrices. In some PDF viewers the first row of (ii) in the answer key shows as "1 0 1 2", because its minus signs use a font that is not embedded in the file; the key itself has the minus signs. The entries are $a_{13} = 2 - 3 = -1$ and $a_{14} = 2 - 4 = -2$ ⁠, as in Part (ii), Step 1.)

Check it with numbers

Part (i): $a_{32} = 3.5$ should satisfy $2a_{32} = |-3\cdot3 + 2| = |-7| = 7$ ✓. Every entry is $\ge 0$ ⁠, as an absolute value must give ✓. Part (ii): add all twelve entries row by row, then add them again straight from the rule (each $2i$ occurs 4 times, once per column; each $j$ occurs 3 times, once per row):

\begin{aligned} &(1+0-1-2) + (3+2+1+0) \\ &\quad + (5+4+3+2) = -2 + 6 + 14 = 18, \\ &4\cdot2\cdot(1+2+3) - 3\cdot(1+2+3+4) \\ &\quad = 48 - 30 = 18. \end{aligned}

Both totals are 18 ✓.

Common mistakes

Dropping the absolute value in Part (i), giving negative entries such as $-1$ for $a_{11}$ ⁠.
Taking $|-3i|$ and $|j|$ separately. $|-3 + 1| = 2$ ⁠, but $|-3| + |1| = 4$ ⁠. The bars apply to the whole sum.
Swapping rows and columns, producing a $4\times3$ matrix. Row $i$ runs over 3 values and column $j$ over 4.
Sign slips in Part (ii), e.g. writing $a_{14} = 2$ ⁠. It is $2 - 4 = -2$ ⁠.

What to take away

Work row by row: fix $i$ ⁠, then run $j$ through all column numbers.
With an absolute value, simplify the inside first, then remove the sign.
Look for patterns along rows and down columns; they catch arithmetic slips.

Every fact used, and where it comes from

Fact	Used in	Source
$A = [a_{ij}]_{3\times4}$ has 3 rows, 4 columns, $a_{ij}$ in row $i$ ⁠, column $j$	Both parts	Section 3.2.1 (Lesson: what a matrix is)
Building a matrix from a rule with an absolute value	Part (i)	Example 3
$\|x\| = x$ for $x\ge0$ ⁠, $\|x\| = -x$ for $x<0$	Part (i), Steps 1–3	Definition of absolute value
All 24 entries	Both parts	Recomputed in the site's verification script

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Exercise 3.1 · Q5: Construct a 3×43\times43×4 matrix from a rule for aija_{ij}aij​

Basics you need

Part (i): aij=12∣−3i+j∣a_{ij} = \frac{1}{2}|-3i + j|aij​=21​∣−3i+j∣

Row 1 (i=1i = 1i=1⁠): here −3i=−3-3i = -3−3i=−3

Row 2 (i=2i = 2i=2⁠): here −3i=−6-3i = -6−3i=−6

Row 3 (i=3i = 3i=3⁠): here −3i=−9-3i = -9−3i=−9

Assemble the matrix

Part (ii): aij=2i−ja_{ij} = 2i - jaij​=2i−j

Compute the three rows

Assemble the matrix

Every fact used, and where it comes from

Exercise 3.1 · Q5: Construct a $3\times4$ matrix from a rule for $a_{ij}$

Part (i): $a_{ij} = \frac{1}{2}|-3i + j|$

Row 1 ( $i = 1$ ⁠): here $-3i = -3$

Row 2 ( $i = 2$ ⁠): here $-3i = -6$

Row 3 ( $i = 3$ ⁠): here $-3i = -9$

Part (ii): $a_{ij} = 2i - j$