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} \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*} Conclusion: $A$ is invertible, and $A^{-1}=\left[\mat{4&-2&3\\-11&6&9\\-12&7&10}\right]$.

Chapter 3: Vectors and geometry

Recall that a $2\times 1$ column vector such as $\def\m#1{\begin{bmatrix}#1\end{bmatrix}}\m{4\\3}$ is a pair of numbers written in a column. We are also used to writing points in the plane $\mathbb R^2$ as a pair of numbersl; for example $(4,3)$ is the point obtained by starting from the origin, and moving $4$ units to the right and $3$ units up.

We think of a (column) vector like $\vec v=\m{4\\3}$ as an instruction to move $4$ units to the right and $3$ units up. This movement is called “translation by $\vec v$”.

Examples

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

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

It is convenient to not be too fussy about the difference between a point like $(4,3)$ and the vector $\m{4\\3}$. If we agree to write points as column vectors, then we can perform algebra (addition, subtraction, scalar multiplication) as discussed in Chapter 2, 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 moves the point $A=(-1,3)$ to $B=(5,-4)$?

Answer: 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 \[ \vec{AB}=B-A\] (where on the right hand side, we interpret the points as column vectors so we can subtract them to get a column vector).

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

Example

In $\mathbb{R}^3$, the points $A=(3,-4,5)$ and $B=(11,6,-2)$ have $\vec{AB}=\m{11\\6\\-2}-\m{3\\-4\\5}=\m{8\\10\\-7}$.

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

• $\left\|\m{4\\3}\right\|=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$
• $\left\|\m{1\\0\\-2\\3}\right\|=\sqrt{1^2+0^2+(-2)^2+3^2}=\sqrt{1+0+4+9}=\sqrt{14}$

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 $c$.