Example: $n=3$

Let $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$.

• Matrix of signs: $\mat{+&-&+\\-&+&-\\+&-&+}$
• Matrix of cofactors: $C=\def\vm#1{\left|\begin{smallmatrix}#1\end{smallmatrix}\right|}\mat{\vm{-4&3\\4&-2}&-\vm{-2&3\\5&-2}&\vm{-2&-4\\5&4}\\-\vm{1&0\\4&-2}&\vm{3&0\\5&-2}&-\vm{3&1\\5&4}\\\vm{1&0\\-4&3}&-\vm{3&0\\-2&3}&\vm{3&1\\-2&-4}}= \mat{-4&11&12\\2&-6&-7\\3&-9&-10}$
• So the adjoint of $A$ is $J=C^T=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$.

• Adjoint of $A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$ is $J=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$
• $AJ=\mat{3&1&0\\-2&-4&3\\5&4&-2}\mat{-4&2&3\\11&-6&-9\\12&-7&-10}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$
• $JA=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}\mat{3&1&0\\-2&-4&3\\5&4&-2}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$
• So $A^{-1}=-J$.
• And $\det(A)=-1$.
• So $A^{-1}=\frac1{\det(A)}J$ again.

Theorem: key property of the adjoint of a square matrix

If $A$ is any $n\times n$ matrix and $J$ is its adjoint, then $AJ=(\det A)I_n=JA$.

Proof

• Omitted

Corollary: a formula for the inverse of a square matrix

If $A$ is any $n\times n$ matrix with $\det(A)\ne 0$, then $A$ is invertible, and $A^{-1}=\frac1{\det A}J$ where $J$ is the adjoint of $A$.

Proof

• Divide the equation $AJ=(\det A)I_n=JA$ by $\det A$.
• $A(\frac1{\det A}J)=I_n=(\frac1{\det A})JA$
• So $A^{-1}=\frac1{\det A} J$. ■

Example ($n=4$)

Let $A=\mat{1&0&0&0\\1&2&0&0\\1&2&3&0\\1&2&3&4}$.

• Reminder: repeated row or zero row gives determinant zero
• $C=\mat{+\vm{2&0&0\\2&3&0\\2&3&4}&-\vm{1&0&0\\1&3&0\\1&3&4}&+0&-0\\-0&+\vm{1&0&0\\1&3&0\\1&3&4}&-\vm{1&0&0\\1&2&0\\1&2&4}&+0\\+0&-0&+\vm{1&0&0\\1&2&0\\1&2&4}&-\vm{1&0&0\\1&2&0\\1&2&3}\\-0&+0&-0&+\vm{1&0&0\\1&2&0\\1&2&3}}=\mat{24&-12&0&0\\0&12&-8&0\\0&0&8&-6\\0&0&0&6}$

Example ($n=4$)

Let $A=\mat{1&0&0&0\\1&2&0&0\\1&2&3&0\\1&2&3&4}$.

• $C=\mat{24&-12&0&0\\0&12&-8&0\\0&0&8&-6\\0&0&0&6}$ so $J=C^T=\mat{24&0&0&0\\-12&12&0&0\\0&-8&8&0\\0&0&-6&6}$.
• So $A^{-1}=\frac1{\det A}J = \frac1{24}\mat{24&0&0&0\\-12&12&0&0\\0&-8&8&0\\0&0&-6&6}=\mat{1&0&0&0\\-1/2&1/2&0&0\\0&-1/3&1/3&0\\0&0&-1/4&1/4}$.
• (Easy to check that $AA^{-1}=I_4=A^{-1}A$.)

A more efficient way to find $A^{-1}$

• Given an $n\times n$ matrix $A$, form the $n\times 2n$ matrix $\def\m#1{\small\left[\begin{array}{@{} c|c {}@}#1\end{array}\right]}\m{A&I_n}$.
• Use EROs to put this into RREF.
• One of two things can happen:
  • you get a row of the form $[0~0~\dots~0~|~*~*~\dots~*]$ which starts with $n$ zeros. You can then conclude that $A$ is not invertible.
  • OR you get a matrix of the form $\m{I_n&B}$ for some $n\times n$ matrix $B$. You can then conclude that $A$ is invertible, and $A^{-1}=B$.

Examples

• Consider $A=\def\mat#1{\begin{matrix}#1\end{matrix}}\left[\mat{1&3\\2&6}\right]$. \begin{align*}\m{A&I_2}&=\m{\mat{1&3\\2&6}&\mat{1&0\\0&1}} \def\go#1#2{\m{\mat{#1}&\mat{#2}}} \def\ar#1{\\[6pt]\xrightarrow{#1}&} \ar{R2\to R2-2R1}\go{1&3\\0&0}{1&0\\-2&1} \end{align*} Conclusion: $A$ is not invertible.

• Consider $A=\left[\mat{1&3\\2&7}\right]$.\begin{align*}\m{A&I_2}&=\m{\mat{1&3\\2&7}&\mat{1&0\\0&1}} \ar{R2\to R2-2R1}\go{1&3\\0&1}{1&0\\-2&1} \ar{R1\to R1-3R1}\go{1&0\\0&1}{7&-3\\-2&1} \end{align*} Conclusion: $A$ is invertible and $A^{-1}=\left[\mat{7&-3\\-2&1}\right]$.

• Consider $A=\left[\mat{3&1&0\\-2&-4&3\\5&4&-2}\right]$.\begin{align*}\m{A&I_3}&=\go{3&1&0\\-2&-4&3\\5&4&-2}{1&0&0\\0&1&0\\0&0&1} \ar{R1\to R1+R2} \go{1&-3&3\\-2&-4&3\\5&4&-2}{1&1&0\\0&1&0\\0&0&1}\end{align*}

\begin{align*} \ar{R2\to R2+2R1,\ R3\to R3-5R1} \go{1&-3&3\\0&-10&9\\0&19&-17}{1&1&0\\2&3&0\\-5&-5&1} \ar{R3\leftrightarrow R2} \go{1&-3&3\\0&19&-17\\0&-10&9}{1&1&0\\-5&-5&1\\2&3&0}\end{align*}

\begin{align*} \ar{R2\to R2+2R3} \go{1&-3&3\\0&-1&1\\0&-10&9}{1&1&0\\-1&1&1\\2&3&0} \ar{R1\to R1+3R2,\ R3\to R3-10R2} \go{1&0&0\\0&-1&1\\0&0&-1}{4&-2&3\\-1&1&1\\12&-7&-10}\end{align*}

\begin{align*} \ar{R2\to R2+R3} \go{1&0&0\\0&-1&0\\0&0&-1}{4&-2&3\\11&-6&-9\\12&-7&-10} \ar{R2\to -R2,\ R3\to -R3} \go{1&0&0\\0&1&0\\0&0&1}{4&-2&3\\-11&6&9\\-12&7&10}\end{align*}

• So $\m{A&I_3}\xrightarrow{\text{EROs}}\go{1&0&0\\0&1&0\\0&0&1}{4&-2&3\\-11&6&9\\-12&7&10}$
• Conclusion: $A$ is invertible, and $A^{-1}=\left[\mat{4&-2&3\\-11&6&9\\-12&7&10}\right]$.

Chapter 3: Vectors and geometry

Vectors

• $\def\m#1{\begin{bmatrix}#1\end{bmatrix}}\m{4\\3}$ is a $2\times 1$ column vector
• i.e., a pair of numbers written in a column
• We also use pairs of numbers to write points in the plane $\mathbb R^2$
• e.g., $(4,3)$ is a point
  • you get there by starting from the origin, moving $4$ units to the right and $3$ units up.
• We think $\vec v=\m{4\\3}$ as an instruction to move $4$ units to the right and $3$ units up.
• This movement is called “translation by $\vec v$”.

Translation by $\vec v$

The vector $\vec v=\m{4\\3}$ moves:

• $(0,0)$ to $(4,3)$
• $(-2,6)$ to $(2,9)$
• $(x,y)$ to $(x+4,y+3)$.

• We're not too fussy about the difference between points like $(4,3)$ and vectors like $\m{4\\3}$.
• If we write points as column vectors, we can perform algebra (addition, subtraction, scalar multiplication) using points and column vectors.

For example, we could rewrite the examples above by saying that $\vec v=\m{4\\3}$ moves:

• $\m{0\\0}$ to $\m{0\\0}+\m{4\\3}=\m{4\\3}$
• $\m{-2\\6}$ to $\m{-2\\6}+\m{4\\3}=\m{2\\9}$
• $\m{x\\y}$ to $\m{x\\y}+\m{4\\3}=\m{x+4\\y+3}$.

• More generally: a column vector $\vec v$ moves a point $\vec x$ to $\vec x+\vec v$.

Example

Which vector $\vec v$ moves the point $A=(-1,3)$ to $B=(5,-4)$?

• We need a vector $\vec v$ with $A+\vec v=B$
• So $\vec v=B-A = \m{5\\-4}-\m{-1\\3}=\m{6\\-7}$.

• We write $\vec{AB}=\m{6\\-7}$, since this is the vector which moves $A$ to $B$.

Definition of $\vec{AB}$

If $A$ and $B$ are any points in $\mathbb{R}^n$, then the vector $\vec{AB}$ is defined by \[ \vec{AB}=B-A\] (on the right, we interpret points as column vectors so we can subtract them to get a column vector).

• $\vec{AB}$ is the vector which moves $A$ to $B$.

Example

In $\mathbb{R}^3$,

• if $A=(3,-4,5)$
• and $B=(11,6,-2)$
• then $\vec{AB}=\m{11\\6\\-2}-\m{3\\-4\\5}=\m{8\\10\\-7}$.

The uses of vectors

Vectors are used in geometry and science to represent quantities with both a magnitude (size/length) and a direction. For example:

• displacements (in geometry)
• velocities
• forces

Recall that a column vector moves points. Its magnitude, or length, is how far it moves points.

Definition: the length of a vector

If $\vec v=\m{v_1\\v_2\\\vdots\\v_n}$ is a column vector in $\mathbb{R}^n$, then its magnitude, or length, or norm, is the number \[ \|\vec v\|=\sqrt{v_1^2+v_2^2+\dots+v_n^2}.\]

Examples

• $\left\|\m{4\\3}\right\|=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$

• \begin{align*}\left\|\m{1\\0\\-2\\3}\right\|&=\sqrt{1^2+0^2+(-2)^2+3^2}\\&=\sqrt{1+0+4+9}=\sqrt{14}\end{align*}

Exercise

Prove that if $c\in \mathbb{R}$ is a scalar and $\vec v$ is a vector in $\mathbb{R}^n$, then \[ \|c\vec v\|=|c|\,\|\vec v\|.\] That is, multiplying a vector by a scalar $c$ scales its length by $|c|$, the absolute value of $c$.

Distance between two points

$\|\vec{AB}\|$ is the distance from point $A$ to point $B$

• since this is the length of vector which takes point $A$ to point $B$.

• e.g. how far from from $A=(1,2)$ to $B=(-3,4)$?
• $\small\|\def\m#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\vec{AB}\|=\left\|\m{-3\\4}-\m{1\\2}\right\|=\left\|\m{-4\\2}\right\|=\sqrt{(-4)^2+2^2}=\sqrt{20}$.

• e.g. what's the length of the main diagonal of the unit cube in $\mathbb{R}^3$?
• = distance from $0=(0,0,0)$ to $A=(1,1,1)$
• $\|\vec{0A}\|=\left\|\m{1\\1\\1}\right\|=\sqrt{1^2+1^2+1^2}=\sqrt3$.