1 Introduction to Matrices and Linear Systems

1.1 Linear Geometry: Vectors in Space and Angle & Length

Vectors in Space

Definition: A vector is a magnitude with a direction.

$$\vec{v}=head-tail= \begin{bmatrix} b_1-a_1 \\ b_2-a_2 \end{bmatrix}$$ Vectors or Scalars?

• 60 mph (scalar)
• 60 mph due East (vector)
• 100 lbs (vector)
• 5 kg (scalar)

NOTE: Vectors can exist independently of a coordinate system

Vector Operations

• Add head-to-tail
• Scalar multiplication stretches (scales) a vector’s length

Connection to Free Variables

Example: $$S=\{\begin{bmatrix} 1\\2 \end{bmatrix}t|t \in\mathbb{R}\}$$

   t is free so

$$\begin{bmatrix} 1\\2 \end{bmatrix}$$ can be scaled by any length. We say S generates or spans a line in $\mathbb{R}^2.$

Example: $$S=\{\begin{bmatrix} 1\\2\\1 \end{bmatrix} + \begin{bmatrix} -1\\1\\1 \end{bmatrix}t |t \in\mathbb{R}\}$$

   S spans a line in $\mathbb{R}^3$ as we only have one free variable.

     

Precalculus/Calculus Way: $2x+y+z=4 \rightarrow{x=2-\frac{1}{2}y-\frac{1}{2}z}$ $\Rightarrow (0,0,4), (0,4,0), (2,0,0)$

     

Linear Algebra Way: $$S=\{\begin{bmatrix} 1\\0\\0 \end{bmatrix}s + \begin{bmatrix} 0\\1\\0 \end{bmatrix}t |s, t \in\mathbb{R}\}$$

   S spans a plane in $\mathbb{R}^3$ where all combinations of

$$\begin{bmatrix} 1\\0\\0 \end{bmatrix}$$ and $$\begin{bmatrix} 0\\1\\0 \end{bmatrix}$$ spans the xy-plane.

Vector Length

Definition: The length (magnitude) of a vector $\vec{v}\in\mathbb{R}^n$ is the square root of the sum of the squares of components. Let $\vec{v}=\langle v_1,..., v_n\rangle$, and its magnitude will be $\lVert\vec{v}\rVert=\sqrt{v_1^2+...+v_n^2}$.

Angle between Vectors

Definition: The angle formed by two vectors is defined as the angle formed when they are joined tail-to-tail (subtended), and is denoted as $\theta=\cos^{-1}(\frac{\vec{u}\cdot\vec{v}}{\lVert\vec{u}\rVert\lVert\vec{v}\rVert})$.

Proof:

    Let $\vec{u}=\langle u_1, u_2\rangle$ and $\vec{v}=\langle v_1, v_2\rangle$ be two subtended vectors

    $\Rightarrow \lVert\vec{v}-\vec{u}\rVert^2=\lVert\vec{v}\rVert^2+\lVert\vec{u}\rVert^2-2\lVert\vec{v}\rVert\lVert\vec{u}\rVert\cos(\theta)$ by the law of cosines: $c^2=a^2+b^2-2ab\cos(\theta)$

    $\Rightarrow (v_1-u_1)^2+(v_2-u_2)^2=v_1^2+v_2^2+u_1^2+u_2^2-2\lVert\vec{v}\rVert\lVert\vec{u}\rVert\cos(\theta)$

    $\Rightarrow v_1^2-2u_1v_1+u_1^2+v_2^2-2v_2u_2+u_2^2=v_1^2+v_2^2+u_1^2+u_2^2-2\lVert\vec{v}\rVert\lVert\vec{u}\rVert\cos(\theta)$

    $\Rightarrow -2(u_1v_1+u_2v_2)=-2\lVert\vec{v}\rVert\lVert\vec{u}\rVert\cos(\theta)$

    $\therefore \cos(\theta)=\frac{u_1v_1+u_2v_2}{\lVert\vec{v}\rVert\lVert\vec{u}\rVert}$

Problem Set

1.2 Dot Product

Definition: The dot product, $\vec{u}\cdot\vec{v}$, is $\vec{u}\cdot\vec{v}=u_1v_1+u_2v_2$

Note: if $\vec{u}\cdot\vec{v}=0$, then $\cos(\theta)$ which implies $\theta=\frac{\pi}{2}$ (such vectors are orthogonal).

     

Properties of Dot Product

1. $\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u}\;\;\;\;$ (dot products are commutative)

2. $c(\vec{u}\cdot\vec{v})=(c\vec{u})\cdot\vec{v}=\vec{u}\cdot(c\vec{v})\;\;\;\;$ (dot products are associative)

3. $\vec{u}\cdot(\vec{v}+\vec{w})=\vec{u}\cdot\vec{v}=\vec{u}\cdot\vec{w}\;\;\;\;$ (dot products are distributive)

4. if $\vec{u},\vec{v}\neq\vec{0}$ and $\vec{u}\cdot\vec{v}=0$ then $\vec{u}$ is orthogonal to $\vec{v}$

5. $\vec{u}\cdot\vec{u}=\lVert\vec{u}\rVert^2\;\;\;\;$

Examples: Let $$\vec{u}=\begin{bmatrix} 3\\1 \end{bmatrix}$$ , $$\vec{v}=\begin{bmatrix} -1\\3 \end{bmatrix}$$ , $$\vec{w}=\begin{bmatrix} 2\\5 \end{bmatrix}$$

1. $\vec{u}\cdot\vec{w}=3(2)+1(5)=11=2(3)+5(1)=\vec{w}\cdot\vec{u}$

2. $2\vec{u}\cdot\vec{w}=2(11)=4(3)+10(1)=(2\vec{u})\cdot\vec{v}=2(6)+5(2)=\vec{u}\cdot(2\vec{v})$

3. $\vec{u}\cdot\vec{v}=3(-1)+1(3)=0\;\;\;\;\therefore\theta=\frac{\pi}{2}$

4. $\vec{u}\cdot\vec{u}=3(3)+1(1)=3^2+1^2=\lVert\vec{u}\rVert^2$

     

Parallel Vectors: $\vec{u}\parallel\vec{v}$ if their directions are either same or opposite. In other words, $\vec{u}\parallel\vec{v}\Leftrightarrow c\vec{u}=\vec{v}$ for some $c\in\mathbb{R}$

Example: Let $\vec{u}=\langle2,2,2\rangle$ and $\vec{v}=\langle-1,-1,-1\rangle$, then $vec{u}\parallel\vec{v}$ because $\vec{u}=-2\vec{v}$.

   

Theorem [Cauchy-Schwarz Inequality]: For any $\vec{u},\vec{v}\in\mathbb{R}^n$, $\lvert\vec{u}\cdot\vec{v}\rvert\leq\lVert\vec{u}\rVert\lVert\vec{v}\rVert$.

Proof:

    Start with $\cos(\theta)=\frac{\vec{u}\cdot\vec{v}}{\lVert\vec{v}\rVert\lVert\vec{u}\rVert}\;\;\Rightarrow\;\;\vec{u}\cdot\vec{v}=\lVert\vec{v}\rVert\lVert\vec{u}\rVert\cos({\theta})$

 

    Since $\max\{\cos=\theta\}=1$ and $\min\{\cos=\theta\}=-1$, we have $-\lVert\vec{u}\rVert\lVert\vec{v}\rVert\leq\lVert\vec{u}\rVert\lVert\vec{v}\rVert\cos(\theta)\leq\lVert\vec{u}\rVert\lVert\vec{v}\rVert$.

    $\Rightarrow \lvert\lVert\vec{u}\rVert\lVert\vec{v}\rVert\cos(\theta)\rvert\leq\lVert\vec{u}\rVert\lVert\vec{v}\rVert$

    $\therefore\lvert\vec{u}\cdot\vec{v}\rvert\leq\lVert\vec{u}\rVert\lVert\vec{v}\rVert$

   

Theorem [Triangle Inequality]: For any $\vec{u},\vec{v}\in\mathbb{R}^n$, $\lVert\vec{u}+\vec{v}\rVert\leq\lVert\vec{u}\rVert+\lVert\vec{v}\rVert$.

Proof:

    $(\lVert\vec{u}\rVert+\lVert\vec{v}\rVert)^2$

    $=\lVert\vec{u}\rVert^2+2\lVert\vec{u}\rVert\lVert\vec{v}\rVert+\lVert\vec{v}\rVert^2$

    $=\vec{u}\cdot\vec{u}+2\lVert\vec{u}\rVert\lVert\vec{v}\rVert+\vec{v}\cdot\vec{v}$

 

    $\lVert\vec{u}+\vec{v}\rVert^2=(\vec{u}+\vec{v})\cdot(\vec{u}+\vec{v})$

    $=\vec{u}\cdot\vec{u}+\vec{u}\cdot\vec{v}+\vec{v}\cdot\vec{u}+\vec{v}\cdot\vec{v}$

    $=\vec{u}\cdot\vec{u}+2(\vec{u}\cdot\vec{v})+\vec{v}\cdot\vec{v}$

    $=\vec{u}\cdot\vec{u}+2\lVert\vec{u}\rVert\lVert\vec{v}\rVert\cos(\theta)+\vec{v}\cdot\vec{v}$

    $\leq\vec{u}\cdot\vec{u}+2\lVert\vec{u}\rVert\lVert\vec{v}\rVert+\vec{v}\cdot\vec{v}=(\lVert\vec{u}\rVert+\lVert\vec{v}\rVert)^2$

    Thus, $\lVert\vec{u}+\vec{v}\rVert^2\leq(\lVert\vec{u}\rVert+\lVert\vec{v}\rVert)^2$

    $\lVert\vec{u}+\vec{v}\rVert\leq\lVert\vec{u}\rVert+\lVert\vec{v}\rVert$ by square root propoerty of inequality

Problem Set

1.3 Solving Linear Equations

Gauss’s Method

Definition: A linear combination of $x_1, ... x_n$ is of the form $a_1{x_1} + ... + {a_n{x_n}}$ where $a_i$ are coefficients.

An n-tuple ($S_1, ... , S_n$) is solution of or satisfies a system of equations:

$$\begin{bmatrix} a_{1,1}x_1 + & \dots & + a_{1, n}x_n & | &d_1 \\ \vdots & & \vdots \\ a_{m,1}x_1 + & \dots & + a_{m, n}x_n & | &d_m \end{bmatrix}$$

Example: $S = {\{sinx, cosx\}}$

$4sinx - 3cosx$ is a linear combination of $sinx, cosx$

$3(sinx)^2 - 3cos\sqrt{x}$ is NOT a linear combination of $sinx, cosx$

Gauss’ Method

Elementary row operations are:

1. Swapping Equations

2. Multiplication by nonzero constant $\leftarrow$ (Re-scaling)

3. Replace with linear combination of existing equations $\leftarrow$ (Row Combo)

Theorem: Performing elementary row operations on a system of equations does not change the solution set.

General = Particular + Homogeneous

Theorem: Any linear system has general solution of the form:

$\begin{matrix} \{\vec{p} + c_1\vec{{b_1}}+...+c_1\vec{{b_k}}& c_1,...,c_k \in\mathbb{R}\}\\ \end{matrix}$

where k = Number of free variables, $\vec{\beta_1}, ... , \vec{\beta_k}$ are the vectors associated with free variables, and $\vec{p}$ is a particular solution (vector of constants).

Definition: A linear system is homogeneous if all equations equal to 0.

1. Always guaranteed to have at least one solution. $(x_1 = x_2 = ... = 0)$

2. For homogeneous systems $\vec{p}=\vec{0}$

Theorem: Any linear system has general solution of the form: $\begin{matrix} \{\vec{p} + \vec{h}\} \end{matrix}$

Where $\vec{p}$ is particular solution, $\vec{h}$ satisfies associated homogeneous system. This theorem suggest that $\vec{h} = c_1{\vec{\beta_1}} + .. + c_1\vec{{\beta_k}}i$ in other words, the free variable terms satisfy associated the homogeneous system.

Theorem: Any system of linear equations must have either

1. Exactly 1 solution

2. No solutions

3. Infinitely many solutions

Definitions:

1. A square matrix is a matrix where n = m.

2. A square matrix is non-singular if it is the matrix of coefficients of a homogeneous system with a unique solution.

3. A square matrix is singular if it is not non-singular.

Example:

A = $\begin{bmatrix} 2 & 1\\ -1 & 3\\ \end{bmatrix} \Rightarrow \begin{bmatrix} 2 & 1\\ 0 & 7\\ \end{bmatrix}$

Associated System: $2x+y=0$

$-x+3y=0$

$\therefore y=x=0$

$\therefore$ non singular

Example:

B = $\begin{bmatrix} 2 & 2\\ 1 & 1 \\ \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 2 & 2\\ 0 & 0\\ \end{bmatrix}$ I - II

Problem Set

1.4 Row Echelon Form & Reduced Row Echelon Form

Definition: Occurs when leading variable of an equation is to the right of the row above it AND when all rows of zeros are at the bottom.

Example:

$$x-y=0$$

$$2x+2y+z+2w = 4$$

$$y+w = 0$$ $$2z+w = 5$$

-2I + II, replace II

II swap III

$$x-y =0$$ $$y+w = 0$$ $$z+2w = 4$$ $$2z+w = 5$$

-2III + IV, replace IV

$$x-y =0$$ $$y+w = 0$$ $$z+2w = 4$$ $$-3w = -3$$

w=1

z=2

y=-1

x=-1

Solution: (-1, -1, 2, 1)

1.4.1 Reduced Row Echelon Form

*Gauss-Jordan reduction is an extension of Gauss’ method.

Definition: A matrix is in reduced row echelon form (RREF) if the matrix is in REF and:

1. All nonzero leading entries are 1.

2. Any rows containing all 0’s are below all nonzero rows.

3. Where possible, any non zero entry below a leading variable (column) is 0.

Reminders

Definition: (i) Elementary row opperations are: 1. Swapping, 2. Row Scaling, 3. Row Replacement (linear combination of rows)

Theorem: (ii) Elementary row operations are reversible

Definition: (iii) Two matrices are row equivalent if one can be reduced (changed to) the other by a sequence of elementary row operations.

Pivot Positions: location of a leading entry in RREF matrix.

Example: Compute RREF(A) if A = $\begin{bmatrix} 2 & 6 & 1 & 2 & 5 \\ 0 & 3 & 1 & 4 & 1 \\ 0 & 3 & 1 & 2 & 5 \end{bmatrix}$

-II + III $\Rightarrow$ $\begin{bmatrix} 2 & 6 & 1 & 2 & | & 5 \\ 0 & 3 & 1 & 4 & | & 1 \\ 0 & 0 & 0 & -2 & | & 4 \end{bmatrix}$

(1/2)I + (-1/2)III, (1/3)II $\Rightarrow$ $\begin{bmatrix} 1 & 3 & 1/2 & 1& | & 5/2 \\ 0 & 3 & 1 & 4 & | & 1/3 \\ 0 & 0 & 0 & -2 & | & 4 \end{bmatrix}$

II - (4/3)III, I - III $\Rightarrow$ $\begin{bmatrix} 1 & 3 & 1/2 & 0& | & 9/2 \\ 0 & 1 & 1/3 & 0 & | & 3 \\ 0 & 0 & 0 & 1 & | & -2 \end{bmatrix}$

(-3)II + I $\Rightarrow$ $\begin{bmatrix} 1 & 0 & -1/2 & 0& | & -9/2 \\ 0 & 1 & 1/2 & 0 & | & 3 \\ 0 & 0 & 0 & 1 & | & -2 \end{bmatrix}$ = RREF A

Columns 1,2 & 4 are pivot columns and $a_{11}$, $a_{22}$, $a_{44}$ are the pivots.

$\begin{bmatrix} 2 & 6 & 1 & 2 & | & 5 \\ 0 & 3 & 1 & 4 & | & 1 \\ 0 & 0 & 0 & -2 & | & 4 \end{bmatrix}$ is row equivalent to $\begin{bmatrix} 1 & 0 & -1/2 & 0& | & -9/2 \\ 0 & 1 & 1/2 & 0 & | & 3 \\ 0 & 0 & 0 & 1 & | & -2 \end{bmatrix}$

Corresponding system of equations:

$2x+6y+z+w = 5 \Rightarrow x-(1/2)z = (-9/2)$

$3y+z+4w = 1 \Rightarrow y+(1/3)z = 3$

$3y+z+2w = 5 \Rightarrow w = 2$

Z is a free variable.

S = $\{\begin{bmatrix} (-9/2)\\ 3\\ 0\\ -2 \end{bmatrix}$ + $\begin{bmatrix} (1/2)\\ (1/3)\\ 1\\ 0 \end{bmatrix}\}$ | z $\in\mathbb{R}$}

Important Note: The RREF of a matrix is unique. There’s only 1 RREF of a matrix.

Equivalence Relations

Definition: A relation ~ between a set of elements is an equivalence relation if the relation is reflexive, symmetric, and transitive.

Example 1: Equality:

1. A = A $\Rightarrow$ Reflexive

2. if A = B then B = A $\Rightarrow$ Symmetric

3. if A = B and B = C then A = C $\Rightarrow$ Transitive

Example 2: Is same birthday as, an equivalence relation?

Reflexive, Symmetric, and Transitive work.

Example 3: Is friends with an equivalence relation?

No, any of the conditions could fail.

Example 4: S = $\mathbb{Z}$ relation: R = { x ~ y | x and y same parity}

Reflexive: if x is odd then it is odd (same for even) $\therefore$ x ~ x

Symmetric: if x ~ y then y ~ x

Transitive: if x ~ y and y ~ z then x ~ z (all are odd)

Example 5: Let S be the set of nxm matrices

R = {A ~ B | A can be reduced to B by elementary row opperations}

Linear Combination Lemma

Main Idea: Every matrix is row equivalent to exactly 1 RREF matrix.

Supporting Theorems

Linear Combination Lemma:

a linear combination of linear combination is a linear combination.

 

Proof: Let $S = \{x_1,...,x_n\}$. Consider the following linear combinations of $S$:

$c_{11}x_{1} + ...+c_{1n}x_{n}$

By definition, a linear combination of these combinations is:

$d_{1}(c_{11}x_1+...+c_{1n}x_n) +...+d_m(c_{m1}x_1+...+c_{mn}x_n)$ $= d_1c_{11}x_1+...+d_1c_{1n}x_n +...+d_mc_{m1}x_1+...+d_mc_{mn}x_n$

which is a linear combination of $x_1,...,x_n$

In a REF matrix, no nonzero row is a liner combination of the other rows:

Example: $B = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 3 & 2 \\ 0 & 0 & 5 \end{bmatrix}$ is a REF matrix.

$c_1(II) + c_2(III) \neq (I)$ since $b_21 = 0 = b_31$

$c_1(I) + c_2(II) \neq (III)$ since $c_1(1) = c_1 \ neq 0$, but $b_13 = 0$

$c_1(I) + c_2(III) \neq (II)$ since $c_1(1) = c_1 \neq 0$, but $b_12 = 0$

1.5 Determinants and Inverse Matrices

1.5.1 Determinants

The Determinant as a Function

The determinant of a matrix is a function of the form $f: \mathbb{R}^{m \times n} \rightarrow \mathbb{R}$

Example: $\mathrm{det}(\begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix} = 0$

 

Background of Understanding the function, $\mathrm{det}()$

Permutation: Given $\{1, 2, ..., n\}$ all arrangements without omissions or repetitions.

 

Example: $\{1, 2, 3\} = S$ The permutations of $S$ are

 

Inversion: Any time when a larger integer proceeds a smaller integer in a permutation.

 

Example: The permutation $(6, 1, 3, 4, 5, 2)$ has $5+0+1+1+1=8$ inversions.

 

Odd or Even Permutation: Determine whether the permutation is odd or even based on the number of inversions.

 

Example: perumation $(6, 1, 3, 4, 5, 2)$ is even because there are 8 inversions.

Elementary Prodcuts: Consider permutations of $n$ elements where elements do NOT share some row or columns as products.

 

Let $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a{33} \end{bmatrix}$

 

 

Definition of Determinant Function

Let $A$ be a square matrix. The determinant of $A$ denoted $\mathrm{det}(A)$, is the sum of all signed elementary products.

Example: $\mathrm{det}(\begin{bmatrix} 4 & -1 \\ 3 & 2 \end{bmatrix}) = 8 - (-3) = 11 = a_{11}a_{12} - a_{12}a_{21}$

Use the left expressions to prove the formula on the right

 

Example: $\mathrm{det}(\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}) = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} - a_{13}a_{22}a_{31}$

Trick: Remember $3\times3$ determinant via sum of signed elementary products. (NOTE only valid for $3\times3$ matrices)

Evaluating Determinants by Row Reduction

Theorem 1: If $A$ is a square matrix containing a row of zeros then $\mathrm{det}(A) = 0$

This is so becuase each signed element contains one term from each row, each product is zero.

Theorem 2: If $A$ is $n \times n$ triangular matrix (upper, lower, or diagonal) then $\mathrm{det}(A)= a_{11} \cdot a_{22} \cdot...\cdot a_{nn}$ - which is the product of diagonal elements

This is so since the signed element products are all zero except those along the diagonal.

Theorem 3: Let $A$ be an $n\times n$ matrix

1. Row scale scales $\mathrm{det}(A)$: If $A'$ is a matrix contstructed by multiplying (a connection to row operations) one row of $A$ by $c \in \mathbb{R}$ then $\mathrm{det}(A') = c\mathrm{det}(A)$
2. Row swap swaps sign of $\mathrm{det}(A)$: If $A'$ is a matrix constructed by interchanging 2 rows of $A$ then $\mathrm{det}(A') = -\mathrm{det}(A)$
3. Row replace with linear combination doesn’t change $\mathrm{det}(A)$: If $A'$ is a matrix constructed by replacing a row of $A$ with a linear combination of rows then $\mathrm{det}(A') = \mathrm{det}(A)$

 

Properties of the Determinant Function

Theorem 1: If $A$ is a square matrix then $\mathrm{det}(A) = \mathrm{det}(A^T)$

 

Theorem 2: $\mathrm{det}(kA) = k^n\mathrm{det}(A)$ where $A$ is $n\times n$ and $k \in \mathbb{R}$

 

Theorem 3: $A$ is invertible if and only if $\mathrm{det}{A} \neq 0$

 

NOTE: $\mathrm{det}(A + B) \neq \mathrm{det}(A) + \mathrm{det}(B)$

 

 

1.5.2 Cofactor Expansion

Definition: Cofactor Expansion is a computational method to evaluate determinants.

Let $A$ be a square matrix. The minor of $a_{ij}$, dneoted by $M_{ij}$, is the determinant of the submatrix that remains after deleting the $ith$ and $jth$ row from $A$. The number $(-1)^{i+j}M_{ij}$ is called the cofactor of $a_{ij}$

 

Cofactor Expansion Method

Recall: $\mathrm{det}(A) = \mathrm{det}(\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}) = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} - a_{13}a_{22}a_{31}$

Which we can manipulate to get the following:

$= a_{11}(a_{22}a_{33} - a_{23}a_{32}) + a_{21}(a_{13}a_{32} - a_{12}a_{33}) + a_{31}(a_{12}a_{23} - a_{13}a_{22})$

 

$= a_{11}C_{11} + a_{21}C_{21} + a_{31}C_{31}$ - this is cofactor expansion along the 1st column

NOTE: we can expand along any column OR row

 

Example: $A = \begin{bmatrix} 3 & 1 & 0 \\ -2 & -4 & 3 \\ 5 & 4 & -2 \end{bmatrix}$

$\therefore \mathrm{det}(A) = 3\begin{vmatrix}-4 & 3 \\ 4 & -2 \end{vmatrix} - (-2)\begin{vmatrix} 1 & 0 \\ 4 & -2 \end{vmatrix} + 5\begin{vmatrix} 1 & 0 \\ -4 & 3 \end{vmatrix}$

$= 3(8 - 12) + 2(-2 - 0) + 5(3-0)$

$=3(-4) - 4 + 15$

$=-12 -4 + 15 = -1$

Try expanding along a diffefent row or column

1.5.3 Inverses of Matrices

Definition:

1. An identity matrix, denoted $I_n$, is an $n\times n$ matrix with the value $1$ in every main diagonal entry and zeros everywhere else.

Examples:

$I_2 = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}$

$I_3 = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$

 

Unique Property - identity matrix acts like the number one: $AI = IA = A$

2. The inverse of an $n \times n$ matrix, denoted $A^{-1}$, is a matrix so that $AA^{-1} = I_n$

3. A matrix with an inverse is called invertible

Example:

$A = \begin{bmatrix}1 & 2 \\ 1 & 3 \end{bmatrix}, M = \begin{bmatrix}3 & -2 \\ -1 & 1 \end{bmatrix}$

$AM = \begin{bmatrix}1 & 2 \\ 1 & 3 \end{bmatrix}\begin{bmatrix}3 & -2 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}$

$\therefore M = A^{-1}$

 

 

Theorem: If $B$ and $C$ are inverses of $A$ then $B=C$.

Proof: $BA = I$

$(BA)C = IC$

$(BA)C = C$

$B(AC) = C$

$BI = C$

$B = C$

 

 

Theorem: If $A$ and $B$ are invertible $n\times n$ matrices then

1. $AB$ is invertible

2. $(AB)^{-1} = B^{-1}A^{-1}$

Proof:

$(AB)(B^{-1}A^{-1})$

$=A(BB^{-1})A^{-1}$

$=AIA^{-1}$

$=AA^{-1}$

$=I \therefore (AB)^{-1} = B^{-1}A^{-1}$

 

 

Theorem: If $A$ is invertible then

1. $(A^{-1})^{-1} = A$

2. $(A^n)^{-1} = (A^{-1})^n$ for $n \in \mathbb{N}$

3. $(kA)^{-1} = \frac{1}{k}A^{-1}$ for $k \neq 0$

Proof:

$(kA)(\frac{1}{k} A^{-1}) = \frac{1}{k} (kA) A^{-1}$

$= \frac{1 k}{k}A A^{-1}$

$= 1I$

$= I$

$\therefore (kA)^{-1} = \frac{1}{k}A^{-1}$

 

 

How to Compute the Inverse of A Matrix

Recall: Elementary row operations are reversible.

 

Procedure for Computing the Inverse

1. Adjoin $A$ and $I$

2. Row reduce $A$ to RREF; apply same operations

3. The resultant matrix of the righthand side is $A^{-1}$

 

Example: Compute the inverse of $A = \begin{bmatrix}1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8\end{bmatrix}$

$[A | I]$

$= \begin{bmatrix}1 & 2 & 3 & | & 1 & 0 & 0 \\ 2 & 5 & 3 & | & 0 & 1 & 0\\ 1 & 0 & 8 & | & 0 & 0 & 1\end{bmatrix}$

 

$-2I + III$ and $-I + III \rightarrow$

 

$\begin{bmatrix}1 & 2 & 3 & | & 1 & 0 & 0 \\ 0 & 1 & -3 & | & -2 & 1 & 0\\ 0 & -2 & 5 & | & -1 & 0 & 1\end{bmatrix}$

 

$2II + III \rightarrow$

 

$...A^{-1} = \begin{bmatrix}-40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1\end{bmatrix}$

 

Comment: If row operations result in a row of zeros then $A$ is noninvertible.

1.6 Labs

Geology Lab

1.7 Coding Labs

Linear Geometry and Matrices Lab (1.1-1.2: Linear Geometry and Matrix Definitions and Operations)

Linear Systems Lab (1.3 & 1.4: Solving Linear Systems and REF & RREF)

Determinants, Cofactor Expansion, and Inverse Matrices Lab (1.5: Determinants & Invertibility)

1.8 Projects

Truss Project (1.1-1.3: Vector Spaces Applications)

Encryption Project (1.4-1.5: Matrix Applications)