Chapter 3 · Matrices

Exercise 3.2 · Q3: Compute six products of matrices

Exercise 3.2Multiplication of matrices · Section 3.4.5

The question

Compute the indicated products.

\text{(i)}\ \begin{bmatrix}a&b\\-b&a\end{bmatrix}\begin{bmatrix}a&-b\\b&a\end{bmatrix}

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

\text{(iii)}\ \begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}

\text{(iv)}\ \begin{bmatrix}2&3&4\\3&4&5\\4&5&6\end{bmatrix}\begin{bmatrix}1&-3&5\\0&2&4\\3&0&5\end{bmatrix}

\text{(v)}\ \begin{bmatrix}2&1\\3&2\\-1&1\end{bmatrix}\begin{bmatrix}1&0&1\\-1&2&1\end{bmatrix}

\text{(vi)}\ \begin{bmatrix}3&-1&3\\-1&0&2\end{bmatrix}\begin{bmatrix}2&-3\\1&0\\3&1\end{bmatrix}

Exam answer

In each part let $P$ be the product. Multiply row by column.

(i)

\begin{aligned} P &= \begin{bmatrix} a & b \\ -b & a \end{bmatrix}\begin{bmatrix} a & -b \\ b & a \end{bmatrix} \\ &= \begin{bmatrix} a^2 + b^2 & -ab + ab \\ -ab + ab & b^2 + a^2 \end{bmatrix} \\ &= \begin{bmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{bmatrix} \end{aligned}

(ii)

\begin{aligned} P &= \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \end{bmatrix} \\ &= \begin{bmatrix} 1\cdot2 & 1\cdot3 & 1\cdot4 \\ 2\cdot2 & 2\cdot3 & 2\cdot4 \\ 3\cdot2 & 3\cdot3 & 3\cdot4 \end{bmatrix} \\ &= \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix} \end{aligned}

(iii)

\begin{aligned} P &= \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3 \end{bmatrix} \\ &= \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix} \end{aligned}

(iv)

\begin{aligned} &P \\ &= \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}\begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix} \end{aligned}

With $P = [c_{ij}]$ ⁠,

\begin{aligned} c_{11} &= 2\cdot1 + 3\cdot0 + 4\cdot3 = 14 \\ c_{12} &= 2(-3) + 3\cdot2 + 4\cdot0 = 0 \\ c_{13} &= 2\cdot5 + 3\cdot4 + 4\cdot5 = 42 \\ c_{21} &= 3\cdot1 + 4\cdot0 + 5\cdot3 = 18 \\ c_{22} &= 3(-3) + 4\cdot2 + 5\cdot0 = -1 \\ c_{23} &= 3\cdot5 + 4\cdot4 + 5\cdot5 = 56 \\ c_{31} &= 4\cdot1 + 5\cdot0 + 6\cdot3 = 22 \\ c_{32} &= 4(-3) + 5\cdot2 + 6\cdot0 = -2 \\ c_{33} &= 4\cdot5 + 5\cdot4 + 6\cdot5 = 70 \end{aligned}

\therefore\ P = \begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix}

(v)

\begin{aligned} P &= \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 2-1 & 0+2 & 2+1 \\ 3-2 & 0+4 & 3+2 \\ -1-1 & 0+2 & -1+1 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix} \end{aligned}

(vi)

\begin{aligned} P &= \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 6-1+9 & -9+0+3 \\ -2+0+6 & 3+0+2 \end{bmatrix} \end{aligned}

$\therefore\ P = \begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}$

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

The one idea this problem is built on

One rule does all six parts. A product $PQ$ exists when the number of columns of $P$ equals the number of rows of $Q$ ⁠; an $m\times n$ matrix times an $n\times p$ matrix is $m\times p$ ⁠. Its entry in row $i$ ⁠, column $k$ is row $i$ of $P$ times column $k$ of $Q$ : multiply the entries in pairs (first with first, second with second, and so on) and add. The six parts only differ in shape, which changes how many entries there are and how long each sum is.

Given, to find, and the plan

Given: Six pairs of matrices, written side by side, meaning "multiply the left one by the right one" in that order.
To find: Each product.
Plan: For each part: (1) write both orders, check the inner numbers agree, and read off the order of the product; (2) list the rows of the left matrix and the columns of the right matrix; (3) compute every entry as a row-times-column sum and assemble.

Basics you need

1Rows, columns, order

A matrix with $m$ rows and $n$ columns has order $m\times n$ ⁠: rows first, columns second. A single column is $m\times1$ ⁠; a single row is $1\times n$ ⁠. Basics: rows, columns and $a_{ij}$

2When a product exists, and its order

$(m\times n)(n\times p)$ gives $m\times p$ ⁠. The two inside numbers must match; the two outside numbers give the order of the answer.

3Row times column

A row $(r_1,\dots,r_n)$ times a column $(s_1,\dots,s_n)$ is the single number $r_1s_1+r_2s_2+\dots+r_ns_n$ ⁠. Each entry of a product is one such number. Basics: pair up, multiply, add

Part (i): a $2\times2$ product with letters

Orders

Left: $2\times2$ ⁠. Right: $2\times2$ ⁠. Inner numbers 2 and 2 agree, so the product is defined and is $2\times2$ ⁠.

Square matrices of the same order can always be multiplied, in either order.

Lesson: when a product is defined
Every entry, row times column

Rows of the left matrix: $(a,\ b)$ and $(-b,\ a)$ ⁠. Columns of the right matrix: $(a,\ b)$ and $(-b,\ a)$ ⁠.
$\begin{aligned} c_{11}&=a\cdot a+b\cdot b=a^2+b^2\\ c_{12}&=a\cdot(-b)+b\cdot a=-ab+ab=0\\ c_{21}&=(-b)\cdot a+a\cdot b=-ab+ab=0\\ c_{22}&=(-b)(-b)+a\cdot a=b^2+a^2 \end{aligned}$ $\begin{bmatrix}a&b\\-b&a\end{bmatrix}\begin{bmatrix}a&-b\\b&a\end{bmatrix}=\begin{bmatrix}a^2+b^2&0\\0&a^2+b^2\end{bmatrix}$
Read the columns of the right matrix downward: its first column is $a$ over $b$ ⁠, not $a,-b$ (that is its first row). The off-diagonal entries cancel because $ab=ba$ for numbers. The result is $(a^2+b^2)$ times the identity matrix, a scalar matrix.

Basics: pair up, multiply, add

Part (ii): a column times a row

Orders

Left: a column with 3 entries, order $3\times1$ ⁠. Right: a row with 3 entries, order $1\times3$ ⁠. Inner numbers 1 and 1 agree, so the product is defined and its order is the outer numbers, $3\times3$ ⁠.

This surprises many students: two small matrices with 3 entries each give a product with 9 entries. The outer numbers decide the size, and here they are 3 and 3.
Each "row times column" is a single product

Each row of the left matrix has just one entry (1, then 2, then 3), and each column of the right matrix has just one entry (2, then 3, then 4). So each entry is one multiplication, with nothing to add:
$\begin{aligned} \text{row }1&:\ 1\cdot2=2,\ 1\cdot3=3,\ 1\cdot4=4\\ \text{row }2&:\ 2\cdot2=4,\ 2\cdot3=6,\ 2\cdot4=8\\ \text{row }3&:\ 3\cdot2=6,\ 3\cdot3=9,\ 3\cdot4=12 \end{aligned}$ $\begin{bmatrix}1\\2\\3\end{bmatrix}\begin{bmatrix}2&3&4\end{bmatrix}=\begin{bmatrix}2&3&4\\4&6&8\\6&9&12\end{bmatrix}$

A column times a row is a multiplication table: the entry in row $i$ ⁠, column $k$ is (entry $i$ of the column) times (entry $k$ of the row).

The rule is unchanged; the sums are just one term long. Writing the answer row by row (row 1 uses the 1, row 2 uses the 2, row 3 uses the 3) keeps the nine entries in their right places.

Order of the product = outer numbers of $(3\times1)(1\times3)$ ⁠. Basics: pair up, multiply, add

Part (iii): $(2\times2)(2\times3)$

Orders

Left: $2\times2$ ⁠. Right: $2\times3$ ⁠. Inner numbers 2 and 2 agree, so the product is defined and is $2\times3$ ⁠: six entries.

The reverse product would be $(2\times3)(2\times2)$ ⁠: inner numbers 3 and 2 differ, so it does not exist. Order matters even for whether a product exists.
Every entry, row times column

Rows of the left matrix: $(1,\ -2)$ and $(2,\ 3)$ ⁠. Columns of the right matrix: $(1,\ 2)$ ⁠, $(2,\ 3)$ ⁠, $(3,\ 1)$ ⁠.
$\begin{aligned} c_{11}&=1\cdot1+(-2)\cdot2=1-4=-3\\ c_{12}&=1\cdot2+(-2)\cdot3=2-6=-4\\ c_{13}&=1\cdot3+(-2)\cdot1=3-2=1\\ c_{21}&=2\cdot1+3\cdot2=2+6=8\\ c_{22}&=2\cdot2+3\cdot3=4+9=13\\ c_{23}&=2\cdot3+3\cdot1=6+3=9 \end{aligned}$ $\begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}=\begin{bmatrix}-3&-4&1\\8&13&9\end{bmatrix}$
Each sum has two terms because the inner number is 2: a row of the left matrix has two entries, and so does a column of the right matrix. The $-2$ in the first row makes every entry of the first row of the answer a subtraction.

Part (iv): two $3\times3$ matrices

Orders

Both are $3\times3$ ⁠, so the product is defined and is $3\times3$ ⁠: nine entries, each a sum of three terms.
Every entry, row times column

Rows of the left matrix: $(2,3,4)$ ⁠, $(3,4,5)$ ⁠, $(4,5,6)$ ⁠. Columns of the right matrix: $(1,0,3)$ ⁠, $(-3,2,0)$ ⁠, $(5,4,5)$ ⁠.
$\begin{aligned} c_{11}&=2\cdot1+3\cdot0+4\cdot3\\ &=2+0+12=14\\ c_{12}&=2(-3)+3\cdot2+4\cdot0\\ &=-6+6+0=0\\ c_{13}&=2\cdot5+3\cdot4+4\cdot5\\ &=10+12+20=42\\ c_{21}&=3\cdot1+4\cdot0+5\cdot3\\ &=3+0+15=18\\ c_{22}&=3(-3)+4\cdot2+5\cdot0\\ &=-9+8+0=-1\\ c_{23}&=3\cdot5+4\cdot4+5\cdot5\\ &=15+16+25=56\\ c_{31}&=4\cdot1+5\cdot0+6\cdot3\\ &=4+0+18=22\\ c_{32}&=4(-3)+5\cdot2+6\cdot0\\ &=-12+10+0=-2\\ c_{33}&=4\cdot5+5\cdot4+6\cdot5\\ &=20+20+30=70 \end{aligned}$ $\begin{aligned}&\begin{bmatrix}2&3&4\\3&4&5\\4&5&6\end{bmatrix}\begin{bmatrix}1&-3&5\\0&2&4\\3&0&5\end{bmatrix}\\&=\begin{bmatrix}14&0&42\\18&-1&56\\22&-2&70\end{bmatrix}\end{aligned}$
Nine entries is where errors creep in, so work systematically: fix a row of the left matrix and run along the three columns of the right matrix, then move to the next row. The column $(-3,2,0)$ contains a negative entry and a zero; the zero kills the third term in the whole middle column of the answer.

Basics: pair up, multiply, add

Part (v): $(3\times2)(2\times3)$

Orders

Left: 3 rows, 2 columns, so $3\times2$ ⁠. Right: $2\times3$ ⁠. Inner numbers 2 and 2 agree, so the product is defined and is $3\times3$ ⁠.

The answer is bigger than either factor because the outer numbers are 3 and 3. (Multiplying in the other order would give a $2\times2$ matrix, a completely different object.)
Every entry, row times column

Rows of the left matrix: $(2,1)$ ⁠, $(3,2)$ ⁠, $(-1,1)$ ⁠. Columns of the right matrix: $(1,-1)$ ⁠, $(0,2)$ ⁠, $(1,1)$ ⁠.
$\begin{aligned} c_{11}&=2\cdot1+1\cdot(-1)=2-1=1\\ c_{12}&=2\cdot0+1\cdot2=0+2=2\\ c_{13}&=2\cdot1+1\cdot1=2+1=3\\ c_{21}&=3\cdot1+2\cdot(-1)=3-2=1\\ c_{22}&=3\cdot0+2\cdot2=0+4=4\\ c_{23}&=3\cdot1+2\cdot1=3+2=5\\ c_{31}&=(-1)\cdot1+1\cdot(-1)=-2\\ c_{32}&=(-1)\cdot0+1\cdot2=0+2=2\\ c_{33}&=(-1)\cdot1+1\cdot1=-1+1=0 \end{aligned}$ $\begin{bmatrix}2&1\\3&2\\-1&1\end{bmatrix}\begin{bmatrix}1&0&1\\-1&2&1\end{bmatrix}=\begin{bmatrix}1&2&3\\1&4&5\\-2&2&0\end{bmatrix}$
Each sum has only two terms (inner number 2), but there are nine of them (outer numbers 3 and 3). The third row $(-1,1)$ has a negative first entry, which matters in every entry of that row: in $c_{31}$ both terms are negative.

Part (vi): $(2\times3)(3\times2)$

Orders

Left: $2\times3$ ⁠. Right: $3\times2$ ⁠. Inner numbers 3 and 3 agree, so the product is defined and is $2\times2$ ⁠: four entries, each a sum of three terms.
Every entry, row times column

Rows of the left matrix: $(3,-1,3)$ and $(-1,0,2)$ ⁠. Columns of the right matrix: $(2,1,3)$ and $(-3,0,1)$ ⁠.
$\begin{aligned} c_{11}&=3\cdot2+(-1)\cdot1+3\cdot3\\ &=6-1+9=14\\ c_{12}&=3(-3)+(-1)\cdot0+3\cdot1\\ &=-9+0+3=-6\\ c_{21}&=(-1)\cdot2+0\cdot1+2\cdot3\\ &=-2+0+6=4\\ c_{22}&=(-1)(-3)+0\cdot0+2\cdot1\\ &=3+0+2=5 \end{aligned}$ $\begin{bmatrix}3&-1&3\\-1&0&2\end{bmatrix}\begin{bmatrix}2&-3\\1&0\\3&1\end{bmatrix}=\begin{bmatrix}14&-6\\4&5\end{bmatrix}$
Compare with part (v): there the outer numbers made the answer large; here they make it small ( $2\times2$ ⁠), while each entry is a longer, three-term sum. In $c_{22}$ ⁠, $(-1)(-3)=+3$ ⁠: a negative times a negative is positive.

Lesson: matrix multiplication

Result

\text{(i)}\ \begin{bmatrix}a^2+b^2&0\\0&a^2+b^2\end{bmatrix}

\text{(ii)}\ \begin{bmatrix}2&3&4\\4&6&8\\6&9&12\end{bmatrix}

\text{(iii)}\ \begin{bmatrix}-3&-4&1\\8&13&9\end{bmatrix}

\text{(iv)}\ \begin{bmatrix}14&0&42\\18&-1&56\\22&-2&70\end{bmatrix}

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

\text{(vi)}\ \begin{bmatrix}14&-6\\4&5\end{bmatrix}

(Textbook answer: the same six matrices.)

Check it with numbers

(i) With $a=3$ ⁠, $b=4$ ⁠: rows $(3,4),(-4,3)$ times columns $(3,4),(-4,3)$ give $9+16=25$ ⁠, $-12+12=0$ ⁠, $-12+12=0$ ⁠, $16+9=25$ ⁠, so the product is 25 times the identity, and $a^2+b^2=25$ ⁠. ✓

(ii) The nine entries total $54$ ⁠, and $(1+2+3)(2+3+4)=6\cdot9=54$ ⁠, as every product of a column entry with a row entry appears exactly once. ✓

(iii) Column totals of a product equal (column totals of the left matrix) times the right matrix. The left matrix has column totals $(3,\ 1)$ ⁠; then $3\cdot1+1\cdot2=5$ ⁠, $3\cdot2+1\cdot3=9$ ⁠, $3\cdot3+1\cdot1=10$ ⁠. The answer has column totals $-3+8=5$ ⁠, $-4+13=9$ ⁠, $1+9=10$ ⁠. ✓

(iv) Each row of the left matrix is the previous row plus $(1,1,1)$ ⁠, so each row of the answer should be the previous row plus the column totals of the right matrix, $(4,-1,14)$ ⁠. Indeed $18-14=4$ ⁠, $-1-0=-1$ ⁠, $56-42=14$ ⁠, and $22-18=4$ ⁠, $-2-(-1)=-1$ ⁠, $70-56=14$ ⁠. ✓

(v) In the right matrix, column 1 + column 2 = column 3, since $(1,-1)+(0,2)=(1,1)$ ⁠. So the same must hold in the answer: $1+2=3$ ⁠, $1+4=5$ ⁠, $-2+2=0$ ⁠. ✓

(vi) Column totals of the left matrix are $(2,-1,5)$ ⁠. Times the right matrix: $4-1+15=18$ and $-6+0+5=-1$ ⁠. Column totals of the answer: $14+4=18$ ⁠, $-6+5=-1$ ⁠. ✓

Common mistakes

Reading a row of the right matrix instead of a column. In part (i) the first column of the right matrix is $(a,\ b)$ ⁠, not $(a,\ -b)$ ⁠.
Getting the order of the product wrong. In part (ii) the answer is $3\times3$ ⁠, not a single number; the single number $20$ would be the product in the other order, $(1\times3)(3\times1)$ ⁠.
Sign slips with negative entries, such as $(-1)(-3)$ in part (vi), which is $+3$ ⁠.
Swapping the factors. In parts (iii), (v) and (vi) the reverse product either does not exist or has a different order; never reorder a product.

What to take away

Before multiplying, write $(m\times n)(n\times p)$ ⁠: the inner numbers must match and the outer numbers give the size of the answer.
The inner number is how many terms are in each sum; the outer numbers are how many entries you must compute.
Work row by row, running along the columns, and check with column totals.

Every fact used, and where it comes from

Fact	Used in	Source
$PQ$ is defined when columns of $P$ = rows of $Q$ ⁠; $(m\times n)(n\times p)$ is $m\times p$	In every Part, Step 1	Section 3.4.5 (Lesson: multiplication)
Entry $(i,k)$ of $PQ$ is $\sum_j p_{ij}q_{jk}$ ⁠, row $i$ times column $k$	In every Part, Step 2	Section 3.4.5
If $PQ$ is defined, $QP$ need not be	Part (iii), Step 1	Section 3.4.5, Remark after Example 12
Column matrix $m\times1$ ⁠, row matrix $1\times n$ ⁠; scalar and identity matrices	Part (i), Step 2; Part (ii), Step 1	Section 3.3 (Lesson: types of matrices)

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Exercise 3.2 · Q3: Compute six products of matrices

Basics you need

Part (i): a 2×22\times22×2 product with letters

Orders

Every entry, row times column

Part (ii): a column times a row

Orders

Each "row times column" is a single product

Part (iii): (2×2)(2×3)(2\times2)(2\times3)(2×2)(2×3)

Orders

Every entry, row times column

Part (iv): two 3×33\times33×3 matrices

Orders

Every entry, row times column

Part (v): (3×2)(2×3)(3\times2)(2\times3)(3×2)(2×3)

Orders

Every entry, row times column

Part (vi): (2×3)(3×2)(2\times3)(3\times2)(2×3)(3×2)

Orders

Every entry, row times column

Every fact used, and where it comes from

Part (i): a $2\times2$ product with letters

Part (iii): $(2\times2)(2\times3)$

Part (iv): two $3\times3$ matrices

Part (v): $(3\times2)(2\times3)$

Part (vi): $(2\times3)(3\times2)$