~~REVEAL~~

===== 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 $A^{-1}=\frac1{\det(A)}\mat{d&-b\\-c&a}$

==== 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 $X=A^{-1}\mat{4&1&0\\0&2&1}=-\frac1{17}\mat{-2&-5\\-3&1}\mat{4&1&0\\0&2&1}$
  * $X=\frac1{17}\mat{8&12&5\\12&1&-1}$

===== The transpose of a matrix =====

{{page>transpose}}

==== 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]] $c$.
  
==== 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)

==== 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*}

==== 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 $(AB)^T=B^TA^T$. ■ 

====== $n\times n$ 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

==== Step 1: minors ====

{{page>minor}}

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

==== Examples of minors (1) ====

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


Example: for $A=\mat{3&5\\-4&7}$:
  * $M_{11}=\det[7]=7$
  * $M_{12}=\det[-4]=-4$
  * $M_{21}=5$
  * $M_{22}=3$.

==== 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

==== Step 2: cofactors ====

{{page>cofactor}}

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).
====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 $C_{22}=3$.

====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

==== 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)

==== 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!)

==== 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*}

==== 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)

==== 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*}