==== Example ====

Let's solve the matrix equation $\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\mat{1&5\\3&-2}X=\mat{4&1&0\\0&2&1}$ for $X$.

Write $A=\mat{1&5\\3&-2}$. Then $\det(A)=1(-2)-5(3)=-2-15=-17$ which isn't zero, so $A$ is invertible. And $A^{-1}=\frac1{-17}\mat{-2&-5\\-3&1}=\frac1{17}\mat{2&5\\3&-1}$.


Hence the solution is $X=A^{-1}\mat{4&1&0\\0&2&1}=\frac1{17}\mat{2&5\\3&-1}\mat{4&1&0\\0&2&1}=\frac1{17}\mat{8&12&5\\12&1&-1}$.


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

We defined this in tutorial sheet 4:

{{page>transpose}}
==== Exercise: simple properties of the transpose ====
Prove that for any matrix $A$:

  * $(A^T)^T=A$; 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]] $c$.
  
In tutorial sheet 4, we proved:

==== 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
\[ ab=b^Ta^T.\]

==== 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 $A$ and $B$ are matrices and the [[matrix product]] $AB$ is defined, then $B^TA^T$ is also defined. Moreover, in this case we have
\[ (AB)^T=B^TA^T.\]

=== Proof ===

If $AB$ is defined, then $A$ is $n\times m$ and $B$ is $m\times k$ for some $n,m,k$, so $B^T$ is $k\times m$ and $A^T$ is $m\times n$, so $B^TA^T$ is defined. Moreover, in this case $B^TA^T$ is an $k\times n$ matrix, and $AB$ is an $n\times k$ matrix, so $(AB)^T$ is a $k\times n$ matrix. Hence $B^TA^T$ has the same size as $(AB)^T$. To show that they are equal, we calculate, using the fact that the transpose swaps rows with columns:
\begin{align*}
\text{the }(i,j)\text{ entry of }(AB)^T&= \text{the }(j,i)\text{ entry of }AB
\\&= \row j(A)\cdot\col i(B)
\\&=\col i(B)^T\cdot \row j(A)^T \quad\text{by the previous Lemma}
\\&=\row i(B^T)\cdot \col j(A^T) \quad\text{by the Observation}
\\&=\text{the }(i,j)\text{ entry of }B^TA^T
\end{align*}
Hence $(AB)^T=B^TA^T$. ■ 

====== Determinants of $n\times n$ matrices ======

Given any $n\times n$ matrix $A$, it is possible to define a number $\det(A)$  (as a formula using the entries of $A$) so that 
\[ A\text{ is invertible} \iff \det(A)\ne0.\]

  - If $A$ is a $1\times 1$ matrix, say $A=[a]$, then we just define $\det[a]=a$.
  - If $A$ is a $2\times 2$ matrix, say $A=\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\mat{a&b\\c&d}$, then we've seen that $\det(A)=ad-bc$.
  - If $A$ is a $3\times 3$ matrix, say $A=\mat{a&b&c\\d&e&f\\g&h&i}$, then it turns out that $\det(A)=aei-afh+bfg-bdi+cdh-ceg$.
  - If $A$ is a $4\times 4$ matrix, then the formula for $\det(A)$ is more complicated still, with $24$ terms.
  - If $A$ is a $5\times 5$ matrix, then the formula for $\det(A)$ has $120$ terms.

Trying to memorise a formula in every case (or even in the $3\times 3$ case!) isn't convenient unless we understand it somehow. We will approach this is several steps.

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

=== Definition ===
{{page>minor}}

=== Examples ===

  * If $A=\mat{3&5\\-4&7}$, then $M_{11}=\det[7]=7$, $M_{12}=\det[-4]=-4$, $M_{21}=5$, and $M_{22}=3$.
  * If $A=\mat{1&2&3\\7&8&9\\11&12&13}$, then $M_{23}=\det\mat{1&2\\11&12}=1\cdot 12-2\cdot 11=-10$ and $M_{32}=\det\mat{1&3\\7&9}=-12$.

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

===Definition===
{{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.

===Examples===

  * 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 $C_{22}=3$.
  * If $A=\mat{1&2&3\\7&8&9\\11&12&13}$, then $C_{23}=-M_{23}=-(-10)=10$ and $C_{33}=+M_{33}=\det\mat{1&2\\7&8}=-6$.
==== Step 3: the determinant of a $3\times 3$ matrix using Laplace expansion along the first row ====

===Definition===

{{page>determinant of a 3x3 matrix}}