→ Slide 1

Recap: $2\times 2$ inverses

Let $A=\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\mat{a&b\\c&d}$
$\det(A)=ad-bc$ (a number)
$A$ is invertible if and only if $\det(A)\ne 0$
and we then have $A^{-1}=\frac1{\det(A)}\mat{d&-b\\-c&a}$

↓ Slide 2

Using the inverse to solve a matrix equation

Solve $\mat{1&5\\3&-2}X=\mat{4&1&0\\0&2&1}$ for $X$
Reminder: if $A$ invertible, solution to $AX=B$ is $X=A^{-1}B$
Write $A=\mat{1&5\\3&-2}$
$\det(A)=1(-2)-5(3)=-2-15=-17$
So $A$ is invertible, and $A^{-1}=-\frac1{17}\mat{-2&-5\\-3&1}$
Solution is $X=A^{-1}\mat{4&1&0\\0&2&1}=-\frac1{17}\mat{-2&-5\\-3&1}\mat{4&1&0\\0&2&1}$
$X=\frac1{17}\mat{8&12&5\\12&1&-1}$

→ Slide 3

The transpose of a matrix

The transpose of an $n\times m$ matrix $A$ is the $m\times n$ matrix $A^T$ whose $(i,j)$ entry is the $(j,i)$ entry of $A$. In other words, to get $A^T$ from $A$, you write the rows of $A$ as columns, and vice versa; equivalently, you reflect $A$ in its main diagonal.

For example, $\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\mat{a&b\\c&d}^T=\mat{a&c\\b&d}$ and $\mat{1&2&3\\4&5&6}^T=\mat{1&4\\2&5\\3&6}$.

↓ Slide 4

Simple properties of the transpose

$(A^T)^T=A$; and
$(A+B)^T=A^T+B^T$ if $A$ and $B$ are matrices of the same size; and
$(cA)^T=c(A^T)$ for any scalar $c$.

↓ Slide 5

Lemma: transposes and row-column multiplication

If $a$ is a $1\times m$ row vector and $b$ is an $m\times 1$ column vector, then \[ ab=b^Ta^T.\]

Proof is a calculation (see Tutorial worksheet 4)

↓ Slide 6

Observation: the transpose swaps rows with columns

Formally, for any matrix $A$ and any $i,j$, we have \begin{align*}\def\col#1{\text{col}_{#1}}\def\row#1{\text{row}_{#1}} \row i(A^T)&=\col i(A)^T\\\col j(A^T)&=\row j(A)^T .\end{align*}

↓ Slide 7

Theorem: the transpose reverses the order of matrix multiplication

If $A$, $B$ are matrices and the matrix product $AB$ is defined, then $B^TA^T$ is also defined, and $(AB)^T=B^TA^T$.

Proof

$A$ is $n\times m$ and $B$ is $m\times k$ for some $n,m,k$
- $AB$ is an $n\times k$ matrix, so $(AB)^T$ is $k\times n$.
- $B^T$ is $k\times m$ and $A^T$ is $m\times n$
- So $B^TA^T$ is defined, and is $k\times n$.
So $(AB)^T$ and $B^TA^T$ have the same sizes.
To show that $(AB)^T=B^TA^T$, we have to check their entries all agree.

$(i,j)$ entry of $(AB)^T$
- =$(j,i)$ entry of $AB$
- = $\row j(A)\cdot\col i(B)$
- = $\col i(B)^T\cdot \row j(A)^T$ since $ab=b^Ta^T$ for row-col multiplication
- = $\row i(B^T)\cdot \col j(A^T)$ by the Observation about transposes
- = the $(i,j)$ entry of $B^TA^T$.
Hence $(AB)^T$ and $B^TA^T$ have the same sizes and the same entries
So $(AB)^T=B^TA^T$. ■

→ Slide 8

$n\times n$ determinants

For $A$: $n\times n$ we'll define a number $\det(A)$ so that \[ A\text{ is invertible} \iff \det(A)\ne0.\]

If $A=[a]$ is a $1\times 1$ matrix, then $\det[a]=a$.
If $A=\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\mat{a&b\\c&d}$ is a $2\times2$ matrix, then $\det(A)=ad-bc$.
If $A=\mat{a&b&c\\d&e&f\\g&h&i}$ is a $3\times 3$ matrix, then it turns out that $\det(A)=aei-afh+bfg-bdi+cdh-ceg$.
If $A$ is $4\times 4$, the formula for $\det(A)$ has $24$ terms.
If $A$ is $5\times 5$, the formula for $\det(A)$ has $120$ terms.
etc

To understand $n\times n$ determinants, we'll use several steps:

Define the minors of a matrix
Use minors to define the cofactors of a matrix
Use cofactors to find the determinant of a $3\times 3$ matrix
Use cofactors to find the determinant of an $n\times n$ matrix

After that: some other ways to calculate the determinant

↓ Slide 9

Step 1: minors

If $A$ is an $n\times n$ matrix, then the $(i,j)$ minor of $A$ is defined to be the determinant of the $(n-1)\times (n-1)$ matrix formed by removing row $i$ and column $j$ from $A$. We will write this number as $M_{ij}$.

Short version: to find $M_{ij}$: delete the row and col containing $(i,j)$ entry, then take determinant.

↓ Slide 10

Examples of minors (1)

Short version: to find $M_{ij}$: delete the row and col containing $(i,j)$ entry, then take determinant.

Example: for $A=\mat{3&5\\-4&7}$:

$M_{11}=\det[7]=7$
$M_{12}=\det[-4]=-4$
$M_{21}=5$
$M_{22}=3$.

↓ Slide 11

Examples of minors (2)

Short version: to find $M_{ij}$: delete the row and col containing $(i,j)$ entry, then take determinant.

Example: for $A=\mat{1&2&3\\7&8&9\\11&12&13}$:

$M_{23}=\det\mat{1&2\\11&12}=1\cdot 12-2\cdot 11=-10$
$M_{32}=\det\mat{1&3\\7&9}=-12$
etc

↓ Slide 12

Step 2: cofactors

The $(i,j)$ cofactor of an $n\times n$ matrix $A$ is $(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the (i,j) minor of $A$.

Note that $(-1)^{i+j}$ is $+1$ or $-1$, and can looked up in the matrix of signs: $\mat{+&-&+&-&\dots\\-&+&-&+&\dots\\+&-&+&-&\dots\\\vdots&\vdots&\vdots&\vdots&\ddots}$. This matrix starts with a $+$ in the $(1,1)$ entry (corresponding to $(-1)^{1+1}=(-1)^2=+1$) and the signs then alternate.

Short version: minors with some sign changes (according to matrix of signs).

↓ Slide 13

Examples of cofactors (1)

Short version: minors with sign changes $\mat{+&-&+&-&\dots\\-&+&-&+&\dots\\+&-&+&-&\dots\\\vdots&\vdots&\vdots&\vdots&\ddots}$

If $A=\mat{3&5\\-4&7}$, then

$C_{11}=+M_{11}=\det[7]=7$
$C_{12}=-M_{12}=-\det[-4]=4$
$C_{21}=-5$, and $C_{22}=3$.

↓ Slide 14

Examples of cofactors (2)

Short version: minors with sign changes $\mat{+&-&+&-&\dots\\-&+&-&+&\dots\\+&-&+&-&\dots\\\vdots&\vdots&\vdots&\vdots&\ddots}$

If $A=\mat{1&2&3\\7&8&9\\11&12&13}$, then

$C_{23}=-M_{23}=-(-10)=10$
$C_{33}=+M_{33}=\det\mat{1&2\\7&8}=-6$
etc

↓ Slide 15

Step 3: the determinant of a $3\times 3$ matrix

If $A=\mat{a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}}$ is a $3\times 3$ matrix, then \[\det A=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}.\] Here $C_{ij}$ are the cofactors of $A$.

This formula is called Laplace expansion of $\det A$ along the first row
Short version: sum along first row of (entries $\times$ cofactors)

↓ Slide 16

Example

Short version: sum along first row of (entries $\times$ cofactors)

$\det\mat{1&2&3\\7&8&9\\11&12&13}$
= $1\cdot C_{11} + 2 C_{12} + 3 C_{13}$
= $M_{11} - 2 M_{12} + 3 M_{13}$
= $\det\mat{8&9\\12&13} -2\det\mat{7&9\\11&13} + 3\det\mat{7&8\\11&12}$
= $-4 -2(-8)+3(-4)$
= $0$.
(So this matrix isn't invertible!)

↓ Slide 17

Notation

Writing $\det$ all the time is annoying
Sometimes we write the entries of a matrix inside vertical bars $|\ |$ instead
e.g.\begin{align*}\def\vm#1{\left|\begin{smallmatrix}#1\end{smallmatrix}\right|}\vm{1&2&3\\7&8&9\\11&12&13} &= 1\vm{8&9\\12&13} -2\vm{7&9\\11&13} + 3\vm{7&8\\11&12}\\ &=-4 -2(-8)+3(-4)\\ &=0.\end{align*}

↓ Slide 18

Step 4: the determinant of an $n\times n$ matrix

If $A$ is an $n\times n$ matrix, then \[\det A=a_{11}C_{11}+a_{12}C_{12}+\dots+a_{1n}C_{1n}.\] Here $C_{ij}$ are the cofactors of $A$.

Called Laplace expansion of $\det A$ along the first row since $a_{11}, a_{12},\dots,a_{1n}$ make up the first row of $A$
Short version: sum along first row of (entries $\times$ cofactors)

↓ Slide 19

Example

\begin{align*} \def\vm#1{\left|\begin{smallmatrix}#1\end{smallmatrix}\right|} \vm{\color{red}1&\color{red}0&\color{red}2&\color{red}3\\0&2&1&-1\\2&0&0&1\\3&0&4&2} &= \color{red}1\vm{\color{blue}2&\color{blue}1&\color{blue}-1\\0&0&1\\0&4&2}-\color{red}0\vm{0&1&-1\\2&0&1\\3&4&2}+\color{red}2\vm{\color{orange}0&\color{orange}2&\color{orange}{-1}\\2&0&1\\3&0&2}-\color{red}3\vm{\color{purple}0&\color{purple}2&\color{purple}1\\2&0&0\\3&0&4} \\&= 1\left(\color{blue}2\vm{0&1\\4&2}-\color{blue}1\vm{0&1\\0&2}\color{blue}{-1}\vm{0&0\\0&4}\right) \\&\quad -0+2\left(\color{orange}0-\color{orange}{2}\vm{2&1\\3&2}\color{orange}{-1}\vm{2&0\\3&0}\right) \\&\quad -3\left(\color{purple}0-\color{purple}2\vm{2&0\\3&4}+\color{purple}1\vm{2&0\\3&0}\right) \\&=1(2(-4)-0-0)+2(-2(1)-0)-3(-2(8)+0) \\&=-8-4+48\\ &=36. \end{align*}