Matrix Algebra

A Matrix is an ordered rectangular array of numbers or variables enclosed within brackets.

Rows are the horizontal lines in a matrix.

A Row matrix or Row vector is a matrix with only one row of elements.
Examples are: $ \begin{bmatrix} 1 & 2 \end{bmatrix} \:\:\:and $ $ \begin{bmatrix} 3 & -3 & 7 \end{bmatrix} $

Columns are the vertical lines in a matrix.

A Column matrix or Column vector is a matrix with only one column of elements.
Examples are: $ \begin{bmatrix} 1 \\[3ex] 3 \end{bmatrix} \:\:\:and\:\: \begin{bmatrix} 10 \\[3ex] -12 \\[3ex] 9 \end{bmatrix} $

The Order or Size or Dimension of a matrix is the number of rows "by" the number of columns in the matrix.
It is read as number of rows by number of columns
For example: if a matrix has $2$ rows and $3$ columns, the order is read as $2$ by $3$ matrix
But, the order is written as $2 * 3$ matrix

A Zero matrix or Null matrix is a matrix whose elements are zeros.
Examples are: $2 * 2$ null matrix and $2 * 3$ zero matrix among others
$ \begin{bmatrix} 0 & 0 \\[3ex] 0 & 0 \end{bmatrix} \:\:\:and\:\: \begin{bmatrix} 0 & 0 & 0 \\[3ex] 0 & 0 & 0 \end{bmatrix} $

A Rectangular matrix is a matrix that does not have the same number of rows as columns.
In other words, the number of rows is not equal to the number of columns.
Compare with a rectangle. A rectangle does not have the same length as the width.
Examples are: $3 * 2$ matrix and $3 * 4$ matrix among others
$ \begin{bmatrix} 3 & -1 \\[3ex] -5 & 0 \\[3ex] 2 & 2 \end{bmatrix} \:\:\:and\:\: \begin{bmatrix} 3 & -1 & 3 & -7 \\[3ex] -5 & 0 & -3 & -1 \\[3ex] 2 & 2 & 4 & -1 \end{bmatrix} $

A Square matrix is a matrix with an equal number of rows and columns.
In other words, the number of rows is equal to the number of columns.
Compare with a square. A square has the same length as the width.
Examples are: $2 * 2$ matrix and $3 * 3$ matrix among others
$ \begin{bmatrix} 3 & -1 \\[3ex] -5 & 0 \end{bmatrix} \:\:\:and\:\: \begin{bmatrix} 3 & -1 & 3 \\[3ex] -5 & 0 & -3 \\[3ex] 2 & 2 & 4 \end{bmatrix} $

A Diagonal matrix is a square matrix whose:
(1.) diagonal elements (elements along the principal diagonal) are non-zeros
(2.) all other elements are zeros.
Diagonal elements or Principal diagonal elements are elements that are in the principal diagonal of a square matrix.
The Principal diagonal of a square matrix is the diagonal that goes from the top left to the bottom right.

Examples are the $2 * 2$ diagonal matrix and the $3 * 3$ diagonal matrix among others
$ \begin{bmatrix} -2 & 0 \\[3ex] 0 & 3 \end{bmatrix} \:\:\:and\:\: \begin{bmatrix} 4 & 0 & 0 \\[3ex] 0 & 3 & 0 \\[3ex] 0 & 0 & 7 \end{bmatrix} \\[5ex] $ In the $2 * 2$ diagonal matrix, the principal diagonal elements are $-2$ and $3$
In the $3 * 3$ diagonal matrix, the principal diagonal elements are $4$, $3$ and $7$

All diagonal matrices are square matrices. However, all square matrices are not diagonal matrices.
Discuss.

A Scalar matrix is a diagonal matrix whose diagonal elements are equal.
Examples are the $2 * 2$ scalar matrix and the $3 * 3$ scalar matrix among others
$ \begin{bmatrix} -3 & 0 \\[3ex] 0 & -3 \end{bmatrix} \:\:\:and\:\: \begin{bmatrix} 4 & 0 & 0 \\[3ex] 0 & 4 & 0 \\[3ex] 0 & 0 & 4 \end{bmatrix} $

All scalar matrices are diagonal matrices. However, all diagonal matrices are not scalar matrices.
Discuss.

An Identity matrix or a Unit matrix is a diagonal matrix whose principal diagonal elements are $1$.
Examples are the $2 * 2$ identity matrix and the $3 * 3$ identity matrix among others
$ \begin{bmatrix} 1 & 0 \\[3ex] 0 & 1 \end{bmatrix} \:\:\:and\:\: \begin{bmatrix} 1 & 0 & 0 \\[3ex] 0 & 1 & 0 \\[3ex] 0 & 0 & 1 \end{bmatrix} \\[5ex] $ A $2 * 2$ identity matrix is also written as $I_{2}$
A $3 * 3$ identity matrix is also written as $I_{3}$

An Triangular matrix is a square matrix in which the elements above the diagonal elements are only zeros or the elements below the diagonal elements are only zeros while all other elements are non-zeros.
If you remove the zeros, it looks like a triangle.
A Lower triangular matrix is a square matrix in which the elements above the diagonal elements are only zeros while all other elements are non-zeros.
If you remove the zeros, it looks like a "lower" right triangle.
An Upper triangular matrix is a square matrix in which the elements below the diagonal elements are only zeros while all other elements are non-zeros.
If you remove the zeros, it looks like an "upper" right triangle.
Let us review these examples
$ \begin{bmatrix} 3 & -3 \\[3ex] -5 & 7 \end{bmatrix} \\[5ex] 2 * 2 \:\:square\:\: matrix \\[5ex] \begin{bmatrix} 3 & 0 \\[3ex] 0 & 7 \end{bmatrix} \\[5ex] 2 * 2 \:\:diagonal\:\: matrix \\[5ex] \begin{bmatrix} 3 & 0 \\[3ex] -5 & 7 \end{bmatrix} \\[5ex] 2 * 2 \:\:lower\:\: triangular\:\: matrix \\[5ex] \begin{bmatrix} 3 & -3 \\[3ex] 0 & 7 \end{bmatrix} \\[5ex] 2 * 2 \:\:upper\:\: triangular\:\: matrix $

$ \begin{bmatrix} 3 & -3 & 7 \\[3ex] -5 & 12 & -1 \\[3ex] 2 & -4 & 5 \end{bmatrix} \\[5ex] 3 * 3 \:\:square\:\: matrix \\[5ex] \begin{bmatrix} 3 & 0 & 0 \\[3ex] 0 & 12 & 0 \\[3ex] 0 & 0 & 5 \end{bmatrix} \\[5ex] 3 * 3 \:\:diagonal\:\: matrix \\[5ex] \begin{bmatrix} 3 & 0 & 0 \\[3ex] -5 & 12 & 0 \\[3ex] 2 & -4 & 5 \end{bmatrix} \\[5ex] 3 * 3 \:\:lower\:\: triangular\:\: matrix \\[5ex] \begin{bmatrix} 3 & -3 & 7 \\[3ex] 0 & 12 & -1 \\[3ex] 0 & 0 & 5 \end{bmatrix} \\[5ex] 3 * 3 \:\:upper\:\: triangular\:\: matrix $

The transpose of a matrix, say $A$ is:
(1.) denoted by $A^T$
(2.) is read as $A-transpose$ or the $transpose\:\: of\:\: A$
(3.) is a matrix formed by interchanging the rows and columns of matrix $A$. This means that:
rows of matrix $A$ are the columns of $A-transpose$
columns of matrix $A$ are the rows of $A-transpose$

If a matrix has a size of $m * n$, the transpose of that matrix has a size of $n * m$

Say we have:
$ A = \begin{bmatrix} a_{11} & a_{12} \\[3ex] a_{21} & a_{22} \end{bmatrix} \\[5ex] 2 * 2 \:\:matrix \\[5ex] A^T = \begin{bmatrix} a_{11} & a_{21} \\[3ex] a_{12} & a_{22} \end{bmatrix} \\[5ex] 2 * 2 \:\:matrix \\[5ex] D = \begin{bmatrix} d_{11} & d_{12} & d_{13} \\[3ex] d_{21} & d_{22} & d_{23} \end{bmatrix} \\[5ex] 2 * 3 \:\:matrix \\[5ex] D^T = \begin{bmatrix} d_{11} & d_{21} \\[3ex] d_{12} & d_{22} \\[3ex] d_{13} & d_{23} \end{bmatrix} \\[5ex] 3 * 2 \:\:matrix $

Some properties of matrix transpose are:
(1.) The transpose of the transpose of a matrix is the matrix.
$(A^T)^T = A$

(2.) The transpose of the sum of two matrices is the sum of the transpose of each matrix.
$(A + D)^T = A^T + D^T$

(3.) The transpose of the product of two matrices is the product of the transpose of the second matrix and the transpose of the first matrix.
$(AD)^T = D^T * A^T$

See Examples

Critical Thinking Questions: Ask students to develop more properties of the transpose of matrices and test each property.

A Symmetric matrix is a matrix that is equal to its transpose.
Say we have a matrix, $A$; matrix $A$ is a symmetric matrix if $A = A^T$
Ask students to write some examples of symmetric matrices before you inform them about the trick.
That is part of critical thinking techniques.
For this condition to be satisfied:
(1.) the matrix must be a square matrix.
(2.) the "other diagonal(s)" elements must be equal. See colored elements in the examples
Examples are the $2 * 2$ symmetric matrix and the $3 * 3$ symmetric matrix among others
$ A = \begin{bmatrix} 3 & \color{red}{10} \\[3ex] \color{red}{10} & -7 \end{bmatrix} \\[7ex] A^T = \begin{bmatrix} 3 & 10 \\[3ex] 10 & -7 \end{bmatrix} \\[5ex] A = A^T \\[5ex] D = \begin{bmatrix} 1 & \color{red}{5} & \color{red}{-3} \\[3ex] \color{red}{5} & 2 & \color{red}{-4} \\[3ex] \color{red}{-3} & \color{red}{-4} & 6 \end{bmatrix} \\[10ex] D^T = \begin{bmatrix} 1 & 5 & -3 \\[3ex] 5 & 2 & -4 \\[3ex] -3 & -4 & 6 \end{bmatrix} \\[7ex] D = D^T \\[3ex] $ Do you understand what I mean?

A Skew symmetric matrix is a matrix that is equal to the negative of its transpose.
Say we have a matrix, $A$; matrix $A$ is a symmetric matrix if $A = -A^T$
Ask students to write some examples of symmetric matrices before you inform them about the trick.
That is part of critical thinking techniques.
For this condition to be satisfied:
(1.) the matrix must be a square matrix.
(2.) the principal diagonal elements must be zero.
(3.) the "other diagonal(s)" elements must be opposites. See colored elements in the examples
Examples are the $2 * 2$ skew symmetric matrix and the $3 * 3$ skew symmetric matrix among others
$ A = \begin{bmatrix} 0 & \color{red}{-1} \\[3ex] \color{red}{1} & 0 \end{bmatrix} \\[7ex] A^T = \begin{bmatrix} 0 & 1 \\[3ex] -1 & 0 \end{bmatrix} \\[7ex] -A^T = -1 * A^T = \begin{bmatrix} -1 * 0 & -1 * 1 \\[3ex] -1 * -1 & -1 * 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\[3ex] 1 & 0 \end{bmatrix} \\[7ex] A = -A^T \\[5ex] D = \begin{bmatrix} 0 & \color{red}{2} & \color{red}{-5} \\[3ex] \color{red}{-2} & 0 & \color{red}{-4} \\[3ex] \color{red}{5} & \color{red}{4} & 0 \end{bmatrix} \\[10ex] D^T = \begin{bmatrix} 0 & 2 & -5 \\[3ex] -2 & 0 & -4 \\[3ex] 5 & 4 & 0 \end{bmatrix} \\[10ex] -D^T = -1 * D^T = \begin{bmatrix} -1 * 0 & -1 * 2 & -1 * -5 \\[3ex] -1 * -2 & -1 * 0 & -1 * -4 \\[3ex] -1 * 5 & -1 * 4 & -1 * 0 \end{bmatrix} = \begin{bmatrix} 0 & -2 & 5 \\[3ex] 2 & 0 & 4 \\[3ex] -5 & -4 & 0 \end{bmatrix} \\[7ex] D = -D^T \\[3ex] $ Do you understand what I mean?
Why must the principal diagonal elements of skew symmetric matrices be zero?

The Minor of a $\boldsymbol{2 * 2}$ matrix element is the "only" element remaining after the elements in the row and column of that matrix element are deleted.

The Minor of a $\boldsymbol{3 * 3}$ matrix element is the determinant of the $2 * 2$ matrix of the four remaining elements remaining after the elements in the row and column of that matrix element are deleted.

Cofactors are signed minors (minors with signs).

An Augmented matrix is the matrix formed by joining the matrix of the coefficients of the variables of a linear system with the matrix of constants.
A vertical line separates the coefficient matrix from the matrix of constants.
An augmented matrix modifies a linear system into a "compact form".
It is the matrix used to represent a system of linear equations.
In writing the augmented matrix of a​ system, the variables of each equation must be on the left side of the equal sign and the constants on the right side.
A variable that does not appear in an equation has a coefficient of 0.

The rank is the number of non-zero rows of a row echelon matrix.
In other words, the rank is the number of pivots of a row echelon matrix.
Based on the row echelon matrices above;

Cryptology is the science or the study of secret messages.
It is the process of:
encoding and decoding OR
encrypting and decrypting OR
enciphering and deciphering OR
writing and cracking OR
making and breaking secret messages/codes/texts.

Cryptology consists of:
Cryptography: This is the process of enciphering/encoding secret messages.
Cryptanalysis: This is the process of deciphering/decoding secret messages.

Student: Is this someone who is less than $18$ years old?
Teacher: Oh no... ☺☺☺
Student: Okay, what is a minor?
Teacher: What is the minor of a matrix?

Minor of a Matrix

Let us begin with a $2 * 2$ matrix

Let:

$ A = \begin{bmatrix} a_{11} & a_{12} \\[3ex] a_{21} & a_{22} \end{bmatrix} $

Recall:
A matrix is denoted by an uppercase letter, say $A$
The element of a matrix is denoted by a lowercase letter with subscript, say: $a_{11}$, etc

Learn:
The minor of a matrix element is denoted by an uppercase letter with subscript, say: $A_{11}$
Did you notice the different representations?

This means that:

$ A_{11} \:\:is\:\: the\:\: minor\:\: of\:\: a_{11} \\[3ex] A_{12} \:\:is\:\: the\:\: minor\:\: of\:\: a_{12} \\[3ex] A_{21} \:\:is\:\: the\:\: minor\:\: of\:\: a_{21} \\[3ex] A_{22} \:\:is\:\: the\:\: minor\:\: of\:\: a_{22} $

If we delete the elements in the row and column of $a_{11}$, we have $a_{22}$
The minor of $a_{11}$ is $a_{22}$
$A_{11} = a_{22}$

If we delete the elements in the row and column of $a_{12}$, we have $a_{21}$
The minor of $a_{12}$ is $a_{21}$
$A_{12} = a_{21}$

If we delete the elements in the row and column of $a_{21}$, we have $a_{12}$
The minor of $a_{21}$ is $a_{12}$
$A_{21} = a_{12}$

If we delete the elements in the row and column of $a_{22}$, we have $a_{11}$
The minor of $a_{22}$ is $a_{11}$
$A_{22} = a_{11}$

Therefore, the matrix of the minor of $A$ is:

$ minor\:\: A = \begin{bmatrix} A_{11} & A_{12} \\[3ex] A_{21} & A_{22} \end{bmatrix} = \begin{bmatrix} a_{22} & a_{21} \\[3ex] a_{12} & a_{11} \end{bmatrix} $

Can we make this easier?
Let:

$ A = \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \\[5ex] minor\:\: A = \begin{bmatrix} d & c \\[3ex] b & a \end{bmatrix} $

The Minor of a $\boldsymbol{2 * 2}$ matrix element is the only element remaining after the elements in the row and column of that matrix element are deleted.

Cofactor of a Matrix

Cofactors are signed minors (minors with signs).

Rules for Assigning Signs
(1.) Add the subscripts of the minor
(2.) If the sum is an even number, assign a positive sign to the cofactor
(3.) If the sum is an odd number, assign a negative sign to the cofactor

Let:

$ C_{11} = cofactor\:\: of\:\: a_{11} \\[3ex] 1 + 1 = 2 ...even \\[3ex] C_{11} = +A_{11} = A_{11} = a_{22} \\[3ex] C_{12} = cofactor\:\: of\:\: a_{12} \\[3ex] 1 + 2 = 3 ...odd \\[3ex] C_{12} = -A_{12} = -a_{21} \\[3ex] C_{21} = cofactor\:\: of\:\: a_{21} \\[3ex] 2 + 1 = 3 ...odd \\[3ex] C_{21} = -A_{21} = -a_{12} \\[3ex] C_{22} = cofactor\:\: of\:\: a_{22} \\[3ex] 2 + 2 = 4 ...even \\[3ex] C_{22} = +A_{22} = A_{22} = a_{11} \\[3ex] $

The matrix of the cofactor of $A$ is:

$ C = \begin{bmatrix} C_{11} & C_{12} \\[3ex] C_{21} & C_{22} \end{bmatrix} = \begin{bmatrix} A_{11} & -A_{12} \\[3ex] -A_{21} & A_{22} \end{bmatrix} = \begin{bmatrix} a_{22} & -a_{21} \\[3ex] -a_{12} & a_{11} \end{bmatrix} $

Can we make this easier?
Let:

$ A = \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \\[5ex] cofactor\:\: A = \begin{bmatrix} d & -c \\[3ex] -b & a \end{bmatrix} $

Adjoint of a Matrix

The Adjoint is the transpose of the matrix of the cofactors.
It is denoted by adj $A$

The adjoint matrix of $A$ is:

$ adj\;A = C^T = \begin{bmatrix} C_{11} & C_{21} \\[3ex] C_{12} & C_{22} \end{bmatrix} = \begin{bmatrix} A_{11} & -A_{21} \\[3ex] -A_{12} & A_{22} \end{bmatrix} = \begin{bmatrix} a_{22} & -a_{12} \\[3ex] -a_{21} & a_{11} \end{bmatrix} $

Can we make this easier?
Let:

$ A = \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \\[5ex] adj\:\: A = \begin{bmatrix} d & -b \\[3ex] -c & a \end{bmatrix} $

Determinant of a Matrix

The determinant of a matrix, say $A$:

(1.) is denoted by det $A$ or $|A|$

(2.) can be positive, negative, or zero

(3.) is used to determine whether a matrix has an inverse or not.
If the determinant is zero, the matrix does not have an inverse.
If the determinant is non-zero, the matrix has an inverse.

(4.) is one of the methods used to classify linear systems based on Cramer's Rule.
See Cramer's Rule

(5.) is one of the methods used in solving linear systems based on Cramer's Rule.
Cramer's Rule

Formula for Finding the Determinant of a Matrix
We can use any of the two rows of the $2 * 2$ matrix to calculate the determinant of the matrix

$ \underline{First\:\: Row} \\[3ex] det A = a_{11}C_{11} + a_{12}C_{12} \\[3ex] det A = a_{11} * a_{22} + a_{12} * -a_{21} \\[3ex] det A = a_{11}a_{22} + -a_{21} * a_{12} \\[3ex] det A = a_{11}a_{22} - a_{21}a_{12} \\[3ex] OR \\[3ex] \underline{Second\:\: Row} \\[3ex] det A = a_{21}C_{21} + a_{22}C_{22} \\[3ex] det A = a_{21} * -a_{12} + a_{22} * a_{11} \\[3ex] det A = -a_{21}a_{12} + a_{22}a_{11} \\[3ex] det A = a_{11}a_{22} - a_{21}a_{12} $

Can we make this easier?
Let:

$ A = \begin{bmatrix} \color{red}{a} & \color{darkblue}{b} \\[3ex] \color{darkblue}{c} & \color{red}{d} \end{bmatrix} \\[5ex] det A = a * d - c * b \\[3ex] det A = ad - cb \\[5ex] $ Notice the two colors: red color and darkblue color.
Notice the two diagonals: $ad$ with the red color and $cb$ with the darkblue color.
We shall use this approach in the Second Method for Calculating the Determinant of a Matrix

Student: What about the minor of a $3 * 3$ matrix?
How do we deal with the remaining "elements"? Several elements remain.
Teacher: Good question. Let us discuss it.

Say we have $3 * 3$ matrix

Let $ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\[3ex] a_{21} & a_{22} & a_{23} \\[3ex] a_{31} & a_{32} & a_{33} \end{bmatrix} $

$ A_{11} \:\:is\:\: the\:\: minor\:\: of\:\: a_{11} \\[3ex] A_{12} \:\:is\:\: the\:\: minor\:\: of\:\: a_{12} \\[3ex] A_{13} \:\:is\:\: the\:\: minor\:\: of\:\: a_{13} \\[3ex] A_{21} \:\:is\:\: the\:\: minor\:\: of\:\: a_{21} \\[3ex] A_{22} \:\:is\:\: the\:\: minor\:\: of\:\: a_{22} \\[3ex] A_{23} \:\:is\:\: the\:\: minor\:\: of\:\: a_{23} \\[3ex] A_{31} \:\:is\:\: the\:\: minor\:\: of\:\: a_{31} \\[3ex] A_{32} \:\:is\:\: the\:\: minor\:\: of\:\: a_{32} \\[3ex] A_{33} \:\:is\:\: the\:\: minor\:\: of\:\: a_{33} $

If we delete the elements in the row and column of $a_{11}$, we have four remaining elements:
$a_{22}, a_{23}, a_{32}, a_{33}$
The four remaining elements form a $2 * 2$ matrix
The minor of $a_{11}$ is the determinant of the $2 * 2$ matrix

The Minor of a $\boldsymbol{3 * 3}$ matrix element is the determinant of the $2 * 2$ matrix of the four remaining elements after the elements in the row and column of that matrix element are deleted.

Minor of a $3 * 3$ Matrix

$ A_{11} = \begin{vmatrix} a_{22} & a_{23} \\[3ex] a_{32} & a_{33} \end{vmatrix} \\[5ex] A_{11} = a_{22}a_{33} - a_{32}a_{23} \\[3ex] $

$ A_{12} = \begin{vmatrix} a_{21} & a_{23} \\[3ex] a_{31} & a_{33} \end{vmatrix} \\[5ex] A_{12} = a_{21}a_{33} - a_{31}a_{23} \\[3ex] $

$ A_{13} = \begin{vmatrix} a_{21} & a_{22} \\[3ex] a_{31} & a_{32} \end{vmatrix} \\[5ex] A_{13} = a_{21}a_{32} - a_{31}a_{22} \\[3ex] $

$ A_{21} = \begin{vmatrix} a_{12} & a_{13} \\[3ex] a_{32} & a_{33} \end{vmatrix} \\[5ex] A_{21} = a_{12}a_{33} - a_{32}a_{13} \\[3ex] $

$ A_{22} = \begin{vmatrix} a_{11} & a_{13} \\[3ex] a_{31} & a_{33} \end{vmatrix} \\[5ex] A_{22} = a_{11}a_{33} - a_{31}a_{13} \\[3ex] $

$ A_{23} = \begin{vmatrix} a_{11} & a_{12} \\[3ex] a_{31} & a_{32} \end{vmatrix} \\[5ex] A_{23} = a_{11}a_{32} - a_{31}a_{12} \\[3ex] $

$ A_{31} = \begin{vmatrix} a_{12} & a_{13} \\[3ex] a_{22} & a_{23} \end{vmatrix} \\[5ex] A_{31} = a_{12}a_{23} - a_{22}a_{13} \\[3ex] $

$ A_{32} = \begin{vmatrix} a_{11} & a_{13} \\[3ex] a_{21} & a_{23} \end{vmatrix} \\[5ex] A_{32} = a_{11}a_{23} - a_{21}a_{13} \\[3ex] $

$ A_{33} = \begin{vmatrix} a_{11} & a_{12} \\[3ex] a_{21} & a_{22} \end{vmatrix} \\[5ex] A_{33} = a_{11}a_{22} - a_{21}a_{12} \\[3ex] $

Cofactor of a $3 * 3$ Matrix

Let:

$ C_{11} = cofactor\:\: of\:\: a_{11} \\[3ex] 1 + 1 = 2 ...even \\[3ex] C_{11} = +A_{11} = A_{11} \\[3ex] C_{11} = a_{22}a_{33} - a_{32}a_{23} \\[5ex] C_{12} = cofactor\:\: of\:\: a_{12} \\[3ex] 1 + 2 = 3 ...odd \\[3ex] C_{12} = -A_{12} \\[3ex] C_{12} = -(a_{22}a_{33} - a_{32}a_{23}) \\[5ex] C_{13} = cofactor\:\: of\:\: a_{13} \\[3ex] 1 + 3 = 4 ...even \\[3ex] C_{13} = +A_{13} = A_{13} \\[3ex] C_{13} = a_{21}a_{32} - a_{31}a_{22} \\[5ex] C_{21} = cofactor\:\: of\:\: a_{21} \\[3ex] 2 + 1 = 3 ...odd \\[3ex] C_{21} = -A_{21} \\[3ex] C_{21} = -(a_{12}a_{33} - a_{32}a_{13}) \\[5ex] C_{22} = cofactor\:\: of\:\: a_{22} \\[3ex] 2 + 2 = 4 ...even \\[3ex] C_{22} = +A_{22} = A_{22} \\[3ex] C_{22} = a_{11}a_{33} - a_{31}a_{13} \\[5ex] C_{23} = cofactor\:\: of\:\: a_{23} \\[3ex] 2 + 3 = 5 ...odd \\[3ex] C_{23} = -A_{23} \\[3ex] C_{23} = -(a_{11}a_{32} - a_{31}a_{12}) \\[5ex] C_{31} = cofactor\:\: of\:\: a_{31} \\[3ex] 3 + 1 = 4 ...even \\[3ex] C_{31} = +A_{31} = A_{31} \\[3ex] C_{31} = a_{12}a_{23} - a_{22}a_{13} \\[5ex] C_{32} = cofactor\:\: of\:\: a_{32} \\[3ex] 3 + 2 = 5 ...odd \\[3ex] C_{32} = -A_{32} \\[3ex] C_{32} = -(a_{11}a_{23} - a_{21}a_{13}) \\[5ex] C_{33} = cofactor\:\: of\:\: a_{33} \\[3ex] 3 + 3 = 6 ...even \\[3ex] C_{33} = +A_{33} = A_{33} \\[3ex] C_{33} = a_{11}a_{22} - a_{21}a_{12} \\[3ex] $

Can we make all these easier?
Let:

$ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\[3ex] a_{21} & a_{22} & a_{23} \\[3ex] a_{31} & a_{32} & a_{33} \end{bmatrix} = \begin{bmatrix} a & b & c \\[3ex] d & e & f \\[3ex] g & h & i \end{bmatrix} \\[5ex] $ This implies that:

$ a_{11} = a \\[3ex] a_{12} = b \\[3ex] a_{13} = c \\[3ex] a_{21} = d \\[3ex] a_{22} = e \\[3ex] a_{23} = f \\[3ex] a_{31} = g \\[3ex] a_{32} = h \\[3ex] a_{33} = i \\[3ex] minor\:\: A = \begin{bmatrix} ei - hf & di - gf & dh - ge \\[3ex] bi - hc & ai - gc & ah - gb \\[3ex] bf - ec & af - dc & ae - db \end{bmatrix} \\[10ex] cofactor\:\: A = \begin{bmatrix} ei - hf & -(di - gf) & dh - ge \\[3ex] -(bi - hc) & ai - gc & -(ah - gb) \\[3ex] bf - ec & -(af - dc) & ae - db \end{bmatrix} \\[3ex] $

Can we make the Cofactor easier?

$ -(di - gf) = -di + gf = gf - di \\[3ex] -(bi - hc) = -bi + hc = hc - bi \\[3ex] -(ah - gb) = -ah + gb = gb - ah \\[3ex] -(af - dc) = -af + dc = dc - af \\[3ex] $ Therefore,

$ cofactor\:\: A = \begin{bmatrix} ei - hf & gf - di & dh - ge \\[3ex] hc - bi & ai - gc & gb - ah \\[3ex] bf - ec & dc - af & ae - db \end{bmatrix} \\[10ex] adj\: A = \begin{bmatrix} ei - hf & hc - bi & bf - ec \\[3ex] gf - di & ai - gc & dc - af \\[3ex] dh - ge & gb - ah & ae - db \end{bmatrix} \\[5ex] $

Determinant of a $3 * 3$ Matrix

We can use any of the three rows of the $3 * 3$ matrix to calculate the determinant of the matrix

$ \underline{First\:\: Row} \\[3ex] det A = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} \\[3ex] det A = a_{11}A_{11} + a_{12} * -A_{12} + a_{13}A_{13} \\[3ex] det A = a_{11}A_{11} - a_{12}A_{12} + a_{13}A_{13} \\[3ex] OR \\[3ex] \underline{Second\:\: Row} \\[3ex] det A = a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23} \\[3ex] det A = a_{21} * -A_{21} + a_{22}A_{22} + a_{23} * - A_{23} \\[3ex] det A = -a_{21}A_{21} + a_{22}A_{22} - a_{23}A_{23} \\[3ex] OR \\[3ex] \underline{Third\:\: Row} \\[3ex] det A = a_{31}C_{31} + a_{32}C_{32} + a_{33}C_{33} \\[3ex] det A = a_{31}A_{31} + a_{32} * -A_{32} + a_{33}A_{33} \\[3ex] det A = a_{31}A_{31} - a_{32}A_{32} + a_{33}A_{33} \\[3ex] $

The signs to remember cofactors when determining the determinant of a matrix is:

$ \begin{vmatrix} + & - & + \\[3ex] - & + & - \\[3ex] + & - & + \end{vmatrix} \\[3ex] $

Can we make this easier?
We shall be using only the First Row this time to find the determinant
Let:

$ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\[3ex] a_{21} & a_{22} & a_{23} \\[3ex] a_{31} & a_{32} & a_{33} \end{bmatrix} = \begin{bmatrix} a & b & c \\[3ex] d & e & f \\[3ex] g & h & i \end{bmatrix} \\[5ex] $ This implies that:

$ a_{11} = a \\[3ex] a_{12} = b \\[3ex] a_{13} = c \\[3ex] a_{21} = d \\[3ex] a_{22} = e \\[3ex] a_{23} = f \\[3ex] a_{31} = g \\[3ex] a_{32} = h \\[3ex] a_{33} = i \\[3ex] det\: A = a(ei - hf) - b(di - gf) + c(dh - ge) \\[3ex] det\: A = aei - ahf - bdi + bgf + cdh - cge \\[3ex] det\: A = aei + bgf + cdh - ahf - bdi - cge $

Row operations are arithmetic operations performed on rows to reduce the matrix to a row echelon form or a reduced row echelon form.
Typically, we use one row to work on another row or other rows.
The arithmetic operations are mainly addition, multiplication, and *division*.

*NOTE:* I used division in some of my examples. If you prefer, you may see the division as multiplication.
For example: $\dfrac{R_1}{3}$ means $\dfrac{1}{3} * R_1$

Row operations are:
(1.) used to solve linear systems. Please review the examples in this regard. In addition, please review the "Gauss-Jordan Method".

(2.) used to determine the inverse of matrices. Please review the "Inverse - Second Method".

The steps for row operations are:
(1.) The row that is used to operate/work on another row (doctor on patient) should be on the Left Hand Side ($LHS$). That row does not change. Write it first.

(2.) The row that is being operated on/worked on (patient being worked on) should be on the Right Hand Side ($RHS$). That row changes. Write it next.

(3.) Make sure that addition operations are always used.
In other words, use the elimination-by-addition method when writing the operations.
For example: write: $-3... + 2...$ rather than $2... - 3...$

(4.) Change "applicable" numbers to zeros first before changing "other applicable" numbers to ones.
In other words, do not worry about the ones first. Work on the zeros first, then on the ones next.

(5.) Avoid fractions.

(6.) Simplify the coefficients wherever possible to avoid dealing with "big numbers".

(7.) If the coefficients in two rows, say Row $1$ and Row $2$ are:
same signs (positive-positive OR negative-negative); then a negative sign should be in front of the row on the $LHS$
different signs (positive-negative OR negative-positive); then a negative sign should not be in front of the row on the $LHS$

(1.) Reduce the augmented matrix to an identity matrix.
Check your work using a linear system

$ \left[ \begin{array}{cc|c} 3 & 9 & 6 \\[3ex] 2 & 1 & 4 \end{array} \right] $

$ \left[ \begin{array}{cc|c} 3 & 9 & 6 \\[3ex] 2 & 1 & 4 \end{array} \right] \rightarrow \left[ \begin{array}{cc|c} 1 & 0 & x \\[3ex] 0 & 1 & y \end{array} \right] \\[3ex] $ We want to change the:
$3$ to $1$
$9$ to $0$
$2$ to $0$
Leave the $1$ as is. This will change as we progress with our operations. However, we want to eventually change that number to $1$.
We do not know the values of $x$ and $y$ until we are done with the operations.

Begin with the first column.
Change the $2$ to $0$
Use Row $1$ to change $2$ to $0$. A row cannot change itself. It has to be changed by another row.
So, use the $3$ in Row $1$ to change the $2$ in Row $2$ to $0$
Row $1$ will not change. It will be on the $LHS$.
Row $2$ will change. It will be on the $RHS$
Let Row $1$ = $R_1$
Let Row $2$ = $R_2$
Elimination-by-addition method: $-2R_1 + 3R_2$
Test it: $-2 * 3 + 3 * 2 = -6 + 6 = 0$
It worked!
Do it to the rest of the columns.
First column: $-2 * 3 + 3 * 2 = -6 + 6 = \color{red}{0}$
Second column: $-2 * 9 + 3 * 1 = -18 + 3 = \color{red}{-15}$
Third column: $-2 * 6 + 3 * 4 = -12 + 12 = \color{red}{0}$
So, our first operation results in:

$ \left[ \begin{array}{cc|c} 3 & 9 & 6 \\[3ex] 2 & 1 & 4 \end{array} \right] \underrightarrow{-2R_1 + 3R_2} \left[ \begin{array}{cc|c} 3 & 9 & 6 \\[3ex] \color{red}{0} & \color{red}{-15} & \color{red}{0} \end{array} \right] \\[3ex] $ What do we notice?
The $1$ has changed to $-15$
But, do not worry! Let us deal with the $0's$ first
So, we have:

$ \left[ \begin{array}{cc|c} 3 & 9 & 6 \\[3ex] 0 & -15 & 0 \end{array} \right] \\[3ex] $ What do you think our second operation should be?
We need to change the $9$ to $0$
Use Row $2$ ($LHS$) to operate on Row $1$ ($RHS$)
Use $-15$ to change $9$ to $0$
Write: $9R_2 + 15R_1$
Test it: $9 * -15 + 15 * 9 = -135 + 135 = 0$
It works! But, wait a minute!
Do you like to deal with "big numbers"?
Because $9$ and $15$ are divisible by $3$, we can simplify.
How about we write: $3R_2 + 5R_1$? Let us test it
Test it: $3 * -15 + 5 * 9 = -45 + 45 = 0$
It also works! Which operation is easier?
So, we need to simplify the coefficients wherever possible to avoid dealing with "big numbers"

For the operation: $3R_2 + 5R_1$
First column: $3 * 0 + 5 * 3 = 0 + 15 = \color{red}{15}$
Second column: $3 * -15 + 5 * 9 = -45 + 45 = \color{red}{0}$
Third column: $3 * 0 + 5 * 6 = 0 + 30 = \color{red}{30}$
So, our second operation results in:

$ \left[ \begin{array}{cc|c} 3 & 9 & 6 \\[3ex] 0 & -15 & 0 \end{array} \right] \underrightarrow{3R_2 + 5R_1} \left[ \begin{array}{cc|c} \color{red}{15} & \color{red}{0} & \color{red}{30} \\[3ex] 0 & -15 & 0 \end{array} \right] \\[3ex] $ We are done with the $0's$
So, we have:

$ \left[ \begin{array}{cc|c} 15 & 0 & 30 \\[3ex] 0 & -15 & 0 \end{array} \right] \\[3ex] $ We need to change the $15$ in Row $1$ and $-15$ in Row $2$ to $1$
It is much easier now.
Divide Row $1$ by $15$
Divide Row $2$ by $-15$ OR Divide $-$Row $2$ by $15$

For the operation: $R_1 \div 15$
First row:
$15 \div 15 = 1$
$0 \div 15 = 0$
$30 \div 15 = 2$
For the operation: $-R_2 \div 15$

Second row:
$-0 \div 15 = 0$
$-(-15) \div 15 = 1$
$-0 \div 15 = 0$

$ \left[ \begin{array}{cc|c} 15 & 0 & 30 \\[3ex] 0 & -15 & 0 \end{array} \right] \begin{matrix} \underrightarrow{R_1 \div 15} \\ \underrightarrow{-R_2 \div 15} \end{matrix} \left[ \begin{array}{cc|c} 1 & 0 & 2 \\[3ex] 0 & 1 & 0 \end{array} \right] \\[5ex] \implies x = 2 \:\:and\:\: y = 0 \\[3ex] $ To check our work (check for correctness), write the augmented matrix as a linear system and check the solutions of the linear system.

$ \underline{Augmented\:\: matrix \rightarrow Linear\:\: system} \\[3ex] \left[ \begin{array}{cc|c} 3 & 9 & 6 \\[3ex] 2 & 1 & 4 \end{array} \right] means \\[5ex] 3x + 9y = 6 ...eqn.(1) \\[3ex] 2x + y = 4 ...eqn.(2) \\[3ex] x = 2 \:\:and\:\: y = 0 \\[3ex] $

(2.) Reduce the augmented matrix to an identity matrix.
Check your work using a linear system

$ \left[ \begin{array}{cc|c} 1 & 2 & 1 \\[3ex] 2 & 3 & -1 \end{array} \right] $

(3.) Reduce the augmented matrix to an identity matrix.
Check your work using a linear system

$ \left[ \begin{array}{ccc|c} 1 & 3 & 1 & 3 \\[3ex] 3 & 8 & 3 & 7 \\[3ex] 2 & -3 & 1 & -10 \end{array} \right] $

(4.) Reduce the augmented matrix to an identity matrix.
Check your work using a linear system

$ \left[ \begin{array}{ccc|c} 0 & -1 & 1 & 2 \\[3ex] 4 & -3 & 2 & 16 \\[3ex] 3 & 2 & 1 & 11 \end{array} \right] $

(1.) This is an individual project. It is not a group project.
Students may work together. However, each student must submit his/her own project.
Each student will model a real-world scenario that involves communication/discussion between two people, or among several people.

(2.) Each student will encode a message.

(3.) Another student (a different student) will decode the message.

(4.) If a student does not encode a message properly, another student may point out the error(s).
Pointing out the error and correcting that error may count as a decoded message.

(5.) All work/steps should be shown in decoding the message. Points will be deducted for any missing steps.
Please review the examples I did. Did I miss any step? You should not miss any step.

(6.) All messages must be good messages, and be at least two words. No colloquial expressions; bad words; profane words; insensitive words related to race, sex, tribe, religion, national origin and gender among others would be accepted. If any such words appear, the student who encoded that message will not gain any point. The student who decoded the message will not be penalized.

(7.) All work must be turned in by the due date to receive credit.
Late work will not be accepted.

Skills Measured/Acquired
(1.) Use of prior knowledge
(2.) Critical Thinking
(3.) Interdisciplinary connections/applications
(4.) Technology
(5.) Active participation through direct questioning
(6.) Research

(1.) The encoding matrix must be invertible.
This means that it must have an inverse.
This means that it must be a square matrix.
Each student should use a different encoding matrix.

(2.) The elements of the encoding matrix should be positive integers only.
Zeros, Negative integers, Fractions, and Decimals will not be used.

(3.) The elements of the encoded message must be positive integers only.
Zeros, Negative integers, Fractions, and Decimals will not be used.

(4.) The elements of the decoding matrix must be integers only.
Fractions and Decimals will not be used.

(5.) The elements of the decoded message must be non-negative integers only.
Negative integers, Fractions, and Decimals will not be used.

This calculator will solve several problems involving matrices.
To use the calculator;

(1.) Visit the link: TI-84 Calculator

(2.) Click the Matrices folder

(3.) Click the link of whatever you want to do

(4.) Follow the directions as illustrated by the blinking light