Table of Contents
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:
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 $\mat{1&2&3\\4&5&6}^T=\mat{1&4\\2&5\\3&6}$.
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
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 $M_{ij}$.
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
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.
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
If $\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}A=\mat{a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}}$ 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 the Laplace expansion of $\det A$ along the first row, since $a_{11}$, $a_{12}$ and $a_{13}$ make up the first row of $A$.