For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

Matrix Algebra

Matrix Algebra


I greet you this day,
First: read the notes. Second: view the videos. Third: solve the questions/solved examples. Fourth: check your solutions with my thoroughly-explained solutions. Fifth: check your answers with the calculators as applicable.
I wrote the codes for the calculators using JavaScript, a client-side scripting language. Please use the latest Internet browsers. The calculators should work.
Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. You may contact me.
If you are my student, please do not contact me here. Contact me via the school's system.
Thank you for visiting!!!

Samuel Dominic Chukwuemeka (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S

Objectives

Students will:
(1.) Discuss matrices.
(2.) Discuss the types of matrices.
(3.) Solve problems on the equality of matrices.
(4.) Add matrices.
(5.) Subtract matrices.
(6.) Multiply matrices.
(7.) Determine the transpose of a matrix.
(8.) Determine the minor of a matrix element.
(9.) Determine the cofactor of a matrix element.
(10.) Calculate the determinant of a matrix.
(11.) Write a linear system as an augmented matrix.
(12.) Write an augmented matrix as a linear system.
(13.) Differentiate between row echelon matrix and row reduced echelon matrix.
(14.) Write a matrix in a row echelon form.
(15.) Write a matrix in a reduced row echelon form.
(16.) Row-reduce matrices to identity matrices.
(17.) Perform row operations on a matrix.
(18.) Determine the adjoint of a matrix.
(19.) Determine the inverse of a matrix.
(20.) Solve linear systems using Cramer's rule.
(21.) Solve linear systems using Gaussian Elimination / Gauss - Jordan method.
(22.) Solve linear systems using Matrix Inverse method.
(23.) Solve problems in Matrices using technology - TI-84 / TI-84 Plus calculator.
(24.) Solve problems in Matrices using technology - Google Spreadsheet.
(25.) Solve problems in Matrices using technology - Microsoft Excel Spreadsheet.
(26.) Solve problems in Matrices using technology - Matrix Calculators.
(27.) Discuss the application of Matrices in Cryptology.


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


Student Projects

General Project Requirements

(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.

Definitions / Types of Matrices / Examples

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.

principal diagonal of a square matrix

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 $

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".

A matrix is said to be in row echelon form (ref) or echelon form if each row has more leading zeros that the rows before it.
A row with a leading zero is a row that has the first element as $0$
A row with two leading zeros is a row that has the first and the second element as $0$
A row with three leading zeros is a row that has the first, the second, and the third element as $0$
A good example of a row echelon matrix is the upper triangular matrix. (Matrices $A_{ref}$ and $D_{ref}$).

Examples of row echelon matrices are:

$ A_{ref} = \begin{bmatrix} 2 & -1 \\[3ex] 0 & 5 \end{bmatrix} \:\:\:\:\:\:\:\:\:\: B_{ref} = \begin{bmatrix} 2 & 0 \\[3ex] 0 & 5 \end{bmatrix} \:\:\:\:\:\:\:\:\:\: C_{ref} = \begin{bmatrix} 2 & -1 \\[3ex] 0 & 0 \end{bmatrix} \\[10ex] D_{ref} = \begin{bmatrix} 2 & 5 & 4 \\[3ex] 0 & 3 & 2 \\[3ex] 0 & 0 & -3 \end{bmatrix} \:\:\:\:\:\:\:\:\:\: E_{ref} = \begin{bmatrix} 2 & 5 & 4 \\[3ex] 0 & 0 & 2 \\[3ex] 0 & 0 & 0 \end{bmatrix} \:\:\:\:\:\:\:\:\:\: F_{ref} = \begin{bmatrix} 2 & 5 & 4 \\[3ex] 0 & 3 & 2 \\[3ex] 0 & 0 & 0 \end{bmatrix} $

The pivot is the first non-zero element in each row of a row echelon matrix.
Based on the row echelon matrices above;

Row Echelon Matrix $A_{ref}$ $B_{ref}$ $C_{ref}$ $D_{ref}$ $E_{ref}$ $F_{ref}$
Pivots $2, 5$ $2, 5$ $2$ $2, 3, -3$ $2, 2$ $2, 3$

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;

Row Echelon Matrix $A_{ref}$ $B_{ref}$ $C_{ref}$ $D_{ref}$ $E_{ref}$ $F_{ref}$
Rank $2$ $2$ $1$ $3$ $2$ $2$

A matrix is said to be in reduced row echelon form (rref) or row reduced echelon form (rref) if:
(1.) the leading entry in any non-zero row is $1$
(2.) all entries in the column above or below a leading $1$ is zero
(3.) the leading entry of each non-zero row row after the first row occurs to the right of the leading entry of the previous row.
(4.) Any zero row (rows in which all the elements are zeros) is the last row.
A good example of a reduced row echelon matrix is the identity matrix. (Matrices $A_{rref}$ and $E_{rref}$).

Examples of reduced row echelon matrices are:

$ A_{rref} = \begin{bmatrix} 1 & 0 \\[3ex] 0 & 1 \end{bmatrix} \:\:\:\:\:\:\:\:\:\: B_{rref} = \begin{bmatrix} 1 & 0 \\[3ex] 0 & 0 \end{bmatrix} \:\:\:\:\:\:\:\:\:\: C_{rref} = \begin{bmatrix} 0 & 1 \\[3ex] 0 & 0 \end{bmatrix} \\[10ex] D_{rref} = \begin{bmatrix} 1 & 1 \\[3ex] 0 & 0 \end{bmatrix} \:\:\:\:\:\:\:\:\:\: E_{rref} = \begin{bmatrix} 1 & 0 & 0 \\[3ex] 0 & 1 & 0 \\[3ex] 0 & 0 & 1 \end{bmatrix} \:\:\:\:\:\:\:\:\:\: F_{rref} = \begin{bmatrix} 1 & 0 & 3 \\[3ex] 0 & 1 & 0 \\[3ex] 0 & 0 & 0 \end{bmatrix} \\[10ex] G_{rref} = \begin{bmatrix} 1 & -5 & 0 \\[3ex] 0 & 0 & 1 \\[3ex] 0 & 0 & 0 \end{bmatrix} \:\:\:\:\:\:\:\:\:\: H_{rref} = \begin{bmatrix} 1 & 0 & 0 \\[3ex] 0 & 1 & 2 \\[3ex] 0 & 0 & 0 \end{bmatrix} \:\:\:\:\:\:\:\:\:\: J_{rref} = \begin{bmatrix} 1 & 0 & 0 \\[3ex] 0 & 1 & 0 \\[3ex] 0 & 0 & 0 \end{bmatrix} \\[10ex] K_{rref} = \begin{bmatrix} 1 & 0 \\[3ex] 0 & 1 \\[3ex] 0 & 0 \end{bmatrix} $

The Inverse of a matrix is defined as the matrix whose product with the original matrix gives an identity matrix.

Singular Matrix: A matrix whose determinant is zero.
It has no inverse.
Because it has no inverse, it is also known as a Non-invertible Matrix

Non-singular Matrix: A matrix whose determinant is non-zero.
It has an inverse.
Because it has an inverse, it is also known as an Invertible Matrix


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.

Introduction

What is a Matrix?

Matrix is a Latin word for "womb".

What does a woman's womb contain?
How were you born? You came from a womb.

This means that a matrix contains something. Right?
The singular word is Matrix.
The plural word is Matrices.
Matrix is an ordered rectangular array of numbers or variables enclosed within brackets.
The idea of matrix was "probably" first used by Chinese Mathematicians to solve linear systems.

Some textbooks/resources may say otherwise.
History is subjective. Avoid going into that argument.

Some prominent mathematicians in the field of Matrices are:
(1.) Liu Hui
(2.) Isaac Newton
(3.) Seki Kowa
(4.) Gottfried Leibnitz
(5.) Gabriel Cramer (Cramer's Rule)
(6.) Carl Friedrich Gauss (Guass - Jordan method also known as Gaussian Elimination method)
(7.) Wilhelm Jordan (Guass - Jordan method also known as Gaussian Elimination method)
(8.) Olga Taussky-Todd (used matrices to analyze the vibration in airplanes during the second world war)

How do we represent Matrices?
A matrix is represented by an upper-case alphabet.
The elements of a matrix are represented by lower-case corresponding alphabets with subscripts.
Say we have a matrix, Matrix $A$;
The elements of Matrix $A$ are represented as $a_{ij}$
$a_{ij}$ means the element in the $ith$ row and $jth$ column

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

$a_{11}$ is the element in the $1^{st}$ row and $1^{st}$ column
$a_{12}$ is the element in the $1^{st}$ row and $2^{nd}$ column
$a_{21}$ is the element in the $2^{nd}$ row and $1^{st}$ column
$a_{22}$ is the element in the $2^{nd}$ row and $2^{nd}$ column

As you can see;
We have two rows: Row $1$, $R_1$ and Row $2$, $R_2$
The rows are the horizonal lines.
The elements, $a_{11}$ and $a_{12}$ are in Row $1$
The elements, $a_{21}$ and $a_{22}$ are in Row $2$

Similarly,
We have two columns: Column $1$, $C_1$ and Column $2$, $C_2$
The columns are the vertical lines.
The elements, $a_{11}$ and $a_{21}$ are in Column $1$
The elements, $a_{12}$ and $a_{22}$ are in Column $2$

Two rows and Two columns gives us a $2 * 2$ matrix
This leads us to....

Order of a Matrix
This is also known as the Size or Dimension of a matrix.
It is written as $m * n$
where:
$m$ is the number of rows of the matrix
$n$ is the number of columns of the matrix
It is read as $m\:\: by\:\: n$
So, the order of a matrix is read as: number of rows by number of columns

Examples
Write the size of each matrix.
List the elements.

(1.) $ P = \begin{bmatrix} -3 \\ -7 \end{bmatrix} $
$2 * 1$ matrix
$p_{11} = -3$
$p_{21} = -7$

(2.) $ R = \begin{bmatrix} -1 & -3 & -5 \\ 2 & 4 & -1 \end{bmatrix} $
$2 * 3$ matrix
$r_{11} = -1$
$r_{12} = -3$
$r_{13} = -5$
$r_{21} = 2$
$r_{22} = 4$
$r_{23} = -1$

(3.) $ G = \begin{bmatrix} 1 & -7 & 5 \\ -1 & 0 & 2 \\ -2 & 12 & -3 \end{bmatrix} $
$3 * 3$ matrix
$g_{11} = 1$
$g_{12} = -7$
$g_{13} = 5$
$g_{21} = -1$
$g_{22} = 0$
$g_{23} = 2$
$g_{31} = -2$
$g_{32} = 12$
$g_{33} = -3$

Matrix Operations

We can perform several arithmetic operations on matrices.
We can:
(1.) Perform operations on the Equality of matrices
(2.) Add matrices
(3.) Subtract matrices
(4.) Multiply matrices

Equality of Matrices

For two matrices to be equal, they must have the same size.
If they are equal, then their corresponding elements must be equal.
Say:
$ C = \begin{bmatrix} c_{11} & c_{12} \\[2ex] c_{21} & c_{22} \end{bmatrix} $ and $ D = \begin{bmatrix} d_{11} & d_{12} \\[2ex] d_{21} & d_{22} \end{bmatrix} $

If $C = D$; then

$ c_{11} = d_{11} \\[2ex] c_{12} = d_{12} \\[2ex] c_{21} = d_{21} \\[2ex] c_{22} = d_{22} $

Addition and Subtraction of Matrices

Two matrices can be added or subtracted if they have the same size.
The sum will be a matrix whose elements are the sums of the corresponding elements of the individual matrices.
The difference will be a matrix whose elements are the differences of the corresponding elements of the individual matrices.
The size of the sum will be the same size as the size of the individual matrices.
The size of the difference will be the same size as the size of the individual matrices.

$ C + D = \begin{bmatrix} c_{11} + d_{11} & c_{12} + d_{12} \\[2ex] c_{21} + d_{21} & c_{22} + d_{22} \end{bmatrix} $


Similarly,
$ C - D = \begin{bmatrix} c_{11} - d_{11} & c_{12} - d_{12} \\[2ex] c_{21} - d_{21} & c_{22} - d_{22} \end{bmatrix} $


We can write the operations of addition and subtraction in a condensed form as: $ C \pm D = \begin{bmatrix} c_{11} \pm d_{11} & c_{12} \pm d_{12} \\[2ex] c_{21} \pm d_{21} & c_{22} \pm d_{22} \end{bmatrix} $


Examples - Equality, Addition, and Subtraction of Matrices

For Questions (1.) and (2.); tell whether each statement is true or false.
Give reasons.
(1.) $ \begin{bmatrix} 2 & 5 \\[2ex] 7 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 7 \\[2ex] 4 & 5 \end{bmatrix} $

False.
They have the same size. However, the corresponding elements are not equal.
$5 \ne 7$
$7 \ne 4$
$4 \ne 5$


(2.) $ \begin{bmatrix} -3 \\[2ex] 2 \\[2ex] 4 \end{bmatrix} = \begin{bmatrix} -3 & 2 & 4 \end{bmatrix} $

False.
They do not have the same size.
The first matrix is a $3 * 1$ matrix while the second matrix is a $1 * 3$ matrix.


For Questions (3.) through (6.); determine the value of the variables.

(3.) $ \begin{bmatrix} 7 & 1 \\[2ex] 3 & 12 \end{bmatrix} = \begin{bmatrix} c & 1 \\[2ex] d & e \end{bmatrix} $

$c = 7$
$d = 3$
$e = 12$


(4.) $ \begin{bmatrix} 2 & 4 & p-2 \\[2ex] 2c - 2 & 3 & 2 \\[2ex] 2d & -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 5 \\[2ex] 3 & e & 2 \\[2ex] 4 & -3 & 2 \end{bmatrix} $

$ p - 2 = 5 \\[2ex] p = 5 + 2 \\[2ex] p = 7 \\[3ex] 2c - 2 = 3 \\[2ex] 2c = 3 + 2 \\[2ex] 2c = 5 \\[2ex] c = \dfrac{5}{2} \\[5ex] 2d = 4 \\[2ex] d = 2 \\[3ex] e = 3 $


(5.) $ \begin{bmatrix} m & -2 \\[2ex] 3 & n \end{bmatrix} - \begin{bmatrix} -2 & p \\[2ex] 1 & 3 \end{bmatrix} = \begin{bmatrix} -7 & -12 \\[2ex] 2c & 4 \end{bmatrix} $

$ m - (-2) = -7 \\[2ex] m + 2 = -7 \\[2ex] m = -7 - 2 \\[2ex] m = -9 \\[3ex] -2 - p = -12 \\[2ex] -2 + 12 = p \\[2ex] 10 = p \\[2ex] p = 10 \\[3ex] 3 - 1 = 2c \\[2ex] 2 = 2c \\[2ex] 2c = 2 \\[2ex] c = 1 \\[3ex] n - 3 = 4 \\[2ex] n = 4 + 3 \\[2ex] n = 7 $


(6.) $ \begin{bmatrix} c + 2 & 3d + 1 & 5e \\[2ex] 4p & 0 & 3 \end{bmatrix} + \begin{bmatrix} 3c & 2d & 5e \\[2ex] 2p & 5 & 6 \end{bmatrix} = \begin{bmatrix} 10 & -14 & 90 \\[2ex] 10 & 5 & 9 \end{bmatrix} $

$ c + 2 + 3c = 10 \\[2ex] 4c + 2 = 10 \\[2ex] 4c = 10 - 2 \\[2ex] 4c = 8 \\[2ex] c = 2 \\[3ex] 3d + 1 + 2d = -14 \\[2ex] 5d + 1 = -14 \\[2ex] 5d = -14 - 1 \\[2ex] 5d = -15 \\[2ex] d = -3 \\[3ex] 5e + 5e = 90 \\[2ex] 10e = 90 \\[2ex] e = 9 \\[3ex] 4p + 2p = 10 \\[2ex] 6p = 10 \\[2ex] p = \dfrac{10}{6} \\[5ex] p = \dfrac{5}{3} $


(7.) $ \begin{bmatrix} 1 & 2 & 5 & -1 \\[2ex] 3 & 0 & 2 & -4 \end{bmatrix} + \begin{bmatrix} 8 & 10 & -5 & 3 \\[2ex] -2 & -1 & 4 & 5 \end{bmatrix} $

$ \begin{bmatrix} 1 & 2 & 5 & -1 \\[2ex] 3 & 0 & 2 & -4 \end{bmatrix} + \begin{bmatrix} 8 & 10 & -5 & 3 \\[2ex] -2 & -1 & 4 & 5 \end{bmatrix} \\[2ex] $
$ = \begin{bmatrix} 1 + 8 & 2 + 10 & 5 + (-5) & -1 + 3 \\[2ex] 3 + (-2) & 0 + (-1) & 2 + 4 & -4 + 5 \end{bmatrix} \\[2ex] $
$ = \begin{bmatrix} 9 & 12 & 0 & 2 \\[2ex] 1 & -1 & 6 & 1 \end{bmatrix} $


(8.) $ \begin{bmatrix} 2 & 8 & 12 & 2 \\[2ex] 7 & 4 & -1 & 5 \\[2ex] 1 & 3 & 0 & 10 \end{bmatrix} - \begin{bmatrix} 1 & 3 & 7 & 9 \\[2ex] 2 & -3 & 2 & 4 \\[2ex] 8 & 5 & -3 & 26 \end{bmatrix} $

$ \begin{bmatrix} 2 & 8 & 12 & 2 \\[2ex] 7 & 4 & -1 & 5 \\[2ex] 1 & 3 & 0 & 10 \end{bmatrix} - \begin{bmatrix} 1 & 3 & 7 & 9 \\[2ex] 2 & -3 & 2 & 4 \\[2ex] 8 & 5 & -3 & 26 \end{bmatrix} \\[2ex] $
$ = \begin{bmatrix} 2 - 1 & 8 - 3 & 12 - 7 & 2 - 9 \\[2ex] 7 - 2 & 4 -(-3) & -1 - 2 & 5 - 4 \\[2ex] 1 - 8 & 3 - 5 & 0 -(-3) & 10 - 26 \end{bmatrix} \\[2ex] $
$ = \begin{bmatrix} 1 & 5 & 5 & -7 \\[2ex] 5 & 7 & -3 & 1 \\[2ex] -7 & -2 & 3 & -16 \end{bmatrix} $


(9.) $ \begin{bmatrix} 4m - 6n \\[2ex] 6c - 3d \\[2ex] 2e + 5f \\[2ex] -4x + 2y \end{bmatrix} - \begin{bmatrix} 5m + 6n \\[2ex] 2c - 5d \\[2ex] -4e - 7f \\[2ex] 5x - 3y \end{bmatrix} $

$ \begin{bmatrix} 4m - 6n \\[2ex] 6c - 3d \\[2ex] 2e + 5f \\[2ex] -4x + 2y \end{bmatrix} - \begin{bmatrix} 5m + 6n \\[2ex] 2c - 5d \\[2ex] -4e - 7f \\[2ex] 5x - 3y \end{bmatrix} \\[2ex] $
$ = \begin{bmatrix} 4m - 6n - (5m + 6n) \\[2ex] 6c - 3d - (2c - 5d) \\[2ex] 2e + 5f - (-4e - 7f) \\[2ex] -4x + 2y - (5x - 3y) \end{bmatrix} \\[2ex] $
$ = \begin{bmatrix} 4m - 6n - 5m - 6n \\[2ex] 6c - 3d - 2c + 5d \\[2ex] 2e + 5f + 4e + 7f \\[2ex] -4x + 2y - 5x + 3y \end{bmatrix} \\[2ex] $
$ = \begin{bmatrix} -m - 12n \\[2ex] 4c + 2d \\[2ex] 6e + 12f \\[2ex] -9x + 5y \end{bmatrix} $


(10.) ACT Which of the following matrices is equal to $ \begin{bmatrix} 9 & 8 \\[2ex] -4 & 7 \end{bmatrix} + \begin{bmatrix} -6 & 6 \\[2ex] 5 & 4 \end{bmatrix} $

$ \begin{bmatrix} 9 & 8 \\[2ex] -4 & 7 \end{bmatrix} + \begin{bmatrix} -6 & 6 \\[2ex] 5 & 4 \end{bmatrix} \\[5ex] = \begin{bmatrix} 9 + (-6) & 8 + 6 \\[2ex] -4 + 5 & 7 + 4 \end{bmatrix} \\[5ex] = \begin{bmatrix} 9 - 6 & 14 \\[2ex] 1 & 11 \end{bmatrix} \\[5ex] = \begin{bmatrix} 3 & 14 \\[2ex] 1 & 11 \end{bmatrix} $



Multiplication of Matrices

We have two forms of the Multiplication of Matrices:

(I.) Scalar Multiplication
This is the multiplication of a matrix by a scalar.
The matrix can be any size.
The scalar can be a number or a variable.
When you multiply a matrix by a matrix, multiply "all" elements of the matrix by the scalar.

Say we have a $3 * 2$ Matrix $C$ and a scalar, $k$;

$ C = \begin{bmatrix} c_{11} & c_{12} \\[2ex] c_{21} & c_{22} \\[2ex] c_{31} & c_{32} \end{bmatrix} \\[7ex] kC = \begin{bmatrix} k * c_{11} & k * c_{12} \\[2ex] k * c_{21} & k * c_{22} \\[2ex] k * c_{31} & k * c_{32} \end{bmatrix} $


Examples - Scalar Multiplication

(1.) ACT Which of the following matrices is equal to $ 4\begin{bmatrix} -1 & 2 \\[2ex] 0 & -4 \end{bmatrix} $

$ A.\:\: \begin{bmatrix} -4 & -8 \end{bmatrix} \\[3ex] B.\:\: \begin{bmatrix} 4 -16 \end{bmatrix} \\[3ex] C.\:\: \begin{bmatrix} 3 & 6 \\[2ex] 4 & 0 \end{bmatrix} \\[4ex] D.\:\: \begin{bmatrix} -\dfrac{1}{4} & \dfrac{1}{2} \\[3ex] 0 & -1 \end{bmatrix} \\[4ex] E.\:\: \begin{bmatrix} -4 & 8 \\[2ex] 0 & -16 \end{bmatrix} $


$ 4\begin{bmatrix} -1 & 2 \\[2ex] 0 & -4 \end{bmatrix} = \begin{bmatrix} 4 * -1 & 4 * 2 \\[2ex] 4 * 0 & 4 * -4 \end{bmatrix} = \begin{bmatrix} -4 & 8 \\[2ex] 0 & -16 \end{bmatrix} $


(2.) JAMB If $ P = \begin{bmatrix} 3 & -2 & 4 \\[2ex] 5 & 0 & 6 \\[2ex] 7 & 5 & -1 \end{bmatrix} $, then $-2P$ is


$ -2P = \begin{bmatrix} -2(3) & -2(-2) & -2(4) \\[2ex] -2(5) & -2(0) & -2(6) \\[2ex] -2(7) & -2(5) & -2(-1) \end{bmatrix} = \begin{bmatrix} -6 & 4 & -8 \\[2ex] -10 & 0 & -12 \\[2ex] -14 & -10 & 2 \end{bmatrix} $





(II.) Matrix Multiplication

This is the multiplication of a matrix by another matrix.
Bring it to Matrix Algebra
Say you have two matrices: $A$ and $B$
$\boldsymbol{AB}$ is possible (or conformable for multiplication) if the number of columns of $A$ is equal to the number of rows of $B$
The order of $\boldsymbol{AB}$ will then be the the number of rows of $A$ by the number of rows of $B$
The multiplication process is $row \:\:*\:\: column$, so multiply:
each element in the first row of $A$ and the corresponding element in the first column of $B$
break - breathe in...out ☺
each element in the first row of $A$ and the corresponding element in the second column of $B$
break - breathe in...out ☺
each element in the first row of $A$ and the corresponding element in the third column of $B$
break - breathe in...out ☺
till you are done with all the columns of $B$
then multiply:
each element in the second row of $A$ and the corresponding element in the first column of $B$
break - breathe in...out ☺
each element in the second row of $A$ and the corresponding element in the second column of $B$
break - breathe in...out ☺
each element in the second row of $A$ and the corresponding element in the third column of $B$
break - breathe in...out ☺
till you are done with all the columns of $B$
then multiply:
each element in the third row of $A$ and the corresponding element in the first column of $B$
break - breathe in...out ☺
each element in the third row of $A$ and the corresponding element in the second column of $B$
break - breathe in...out ☺
each element in the third row of $A$ and the corresponding element in the third column of $B$
break - breathe in...out ☺
till you are done with all the columns of $B$
till you are done with all the rows of $A$...that is it!

So, you have to multiply all the rows of $A$ and all the columns of $B$
first row of $A$ and first column of $B$
break
first row of $A$ and second column of $B$
break
first row of $A$ and third column of $B$
break
till you are done with all the columns of $B$
second row of $A$ and first column of $B$
break
second row of $A$ and second column of $B$
break
second row of $A$ and third column of $B$
break
till you are done with all the columns of $B$
third row of $A$ and first column of $B$
break
third row of $A$ and second column of $B$
break
third row of $A$ and third column of $B$
break
till you are done with all the columns of $B$

Notable Notes when Multiplying Matrices, say Matrix $A$ and Matrix $B$

(1.) Write the first matrix to be on the $LHS$, the second matrix to be on the $Middle$, and the product to be on the $RHS$
Example: $A\:\:\:\:\:\:\:\:\:\: B\:\:\:\:\:\:\:\:\:\: AB$

(2.) Write the order of matrix $A$ beneath it, and the order of matrix $B$ beneath it.

(3.) Check to see whether $A$ and $B$ can multiply
Is the number of columns of $A$ equal to the number of rows of $B$

(4.) If yes, go ahead and write the order of the product beneath it.

(5.) Do the multiplication process as described earlier.


Examples - Matrix Multiplication

(1.) ACT The table below shows the numbers of rows and columns in each of $5$ matrices.

Matrix Number of rows Number of columns
$A$
$B$
$C$
$D$
$E$
$m$
$m$
$k$
$m$
$n$
$n$
$m$
$n$
$k$
$m$

For distinct values of $k, m, \:\:and\:\: n$, which of the following matrix products is NOT possible?

$ F.\:\: ED \\[3ex] G.\:\: DC \\[3ex] H.\:\: CE \\[3ex] J.\:\: AE \\[3ex] K.\:\: AC $

Let us analyze each product

Remember that:
The number of columns of the first matrix should be equal to the number of rows of the second matrix for the product to be possible

$ ED \\[3ex] E \hskip 5em D \\[3ex] n * m \hskip 3em m * k \\[3ex] m = m \\[3ex] ED \:\:is\:\: possible \\[5ex] DC \\[3ex] D \hskip 5em C \\[3ex] m * k \hskip 3em k * n \\[3ex] k = k \\[3ex] DC \:\:is\:\: possible \\[5ex] CE \\[3ex] C \hskip 5em E \\[3ex] k * n \hskip 3em n * m \\[3ex] n = n \\[3ex] CE \:\:is\:\: possible \\[5ex] AE \\[3ex] A \hskip 5em E \\[3ex] m * n \hskip 3em n * m \\[3ex] n = n \\[3ex] AE \:\:is\:\: possible \\[5ex] AC \\[3ex] A \hskip 5em C \\[3ex] m * n \hskip 3em k * n \\[3ex] n \ne k \\[3ex] AC \:\:is\:\: not\:\: possible $


Minor, Cofactor, Adjoint, Determinant of a Matrix

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} $

minor of a matrix

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} a & b \\[3ex] c & d \end{bmatrix} \\[5ex] det A = a * d - c * b \\[3ex] det A = ad - cb $

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} $

minor of a matrix

$ 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 remaining 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} $

$ 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} $

$ 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} $

$ 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} $

$ 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} $

$ 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} $

$ 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} $

$ 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} $

$ 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} $

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} $

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

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

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 $

Inverse of a Matrix - First Method

Inverse of a Matrix

The Inverse of a matrix is defined as the matrix whose product with the original matrix gives an identity matrix.
The inverse of a matrix, say $A$:

(1.) is denoted by $A^{-1}$

(2.) can be positive or negative

(3.) cannot be zero

(4.) is one of the methods used to solve a linear system based on the Matrix Inverse method.

The inverse of a matrix can be found using any two methods:

(1.) Formula method
(2.) Row Reduction method

We shall be discussing the Formula method in this section

Formula Method for Finding the Inverse of a Matrix

$ A^{-1} = \dfrac{adj\: A}{det\: A} \\[5ex] $ Based on a $2 * 2$ matrix:
Let us use the easier forms - Please refer to the section on Minor, Cofactor, Adjoint, Determinant of a Matrix
We shall be using those formulas
Given that:

$ A = \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \\[7ex] adj\:\: A = \begin{bmatrix} d & -b \\[3ex] -c & a \end{bmatrix} \\[7ex] det\: A = ad - cb \\[5ex] \therefore A^{-1} = \dfrac{ \begin{bmatrix} d & -b \\[3ex] -c & a \end{bmatrix}}{ad - cb} \\[10ex] $

Based on a $3 * 3$ matrix:
Let us use the easier forms - Please refer to the section on Minor, Cofactor, Adjoint, Determinant of a Matrix
We shall be using those formulas
Given that:

$ A = \begin{bmatrix} a & b & c \\[3ex] d & e & f \\[3ex] g & h & i \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} \\[10ex] det\: A = aei + bgf + cdh - ahf - bdi - cge \\[5ex] \therefore A^{-1} = \dfrac{ \begin{bmatrix} ei - hf & hc - bi & bf - ec \\[3ex] gf - di & ai - gc & dc - af \\[3ex] dh - ge & gb - ah & ae - db \end{bmatrix}}{det\: A = aei + bgf + cdh - ahf - bdi - cge} \\[10ex] $

Based on the Formula Method:
If the determinant is zero, the matrix will not have an inverse because it will lead to an undefined value (the determinant is the denominator of the formula).
This leads to us to more types of matrices.

Singular Matrix: A matrix whose determinant is zero.
It has no inverse.
Because it has no inverse, it is also known as a Non-invertible Matrix

Non-singular Matrix: A matrix whose determinant is non-zero.
It has an inverse.
Because it has an inverse, it is also known as an Invertible Matrix

Augmented Matrix

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".

To write a linear system as an augmented matrix:
(1.) Move all variables to the Left Hand Side ($LHS$) and the constants to the Right Hand Side ($RHS$)

(2.) Complete all missing variables. If there is any missing variable, complete it by writing it and putting a zero as the coefficient

(3.) Complete any missing constant. Put zero as appropriate.

(4.) "Line up" the variables in order.

(5.) Write the coefficient matrix - the matrix of coefficients.

(6.) Write the matrix of constants.

(7.) Use a vertical line to separate the coefficient matrix from the matrix of constants.

Examples - Augmented Matrices

(1.) ACT Which of the following augmented matrices represents the system of linear equations below?

$ 3x + 5y = 20 \\[3ex] 2x - y = 9 \\[3ex] F.\:\: \left[ \begin{array}{cc|c} 3 & 5 & -20 \\[3ex] 2 & -1 & -9 \end{array} \right] \\[3ex] G.\:\: \left[ \begin{array}{cc|c} 3 & 5 & 20 \\[3ex] 2 & -1 & 9 \end{array} \right] \\[3ex] H.\:\: \left[ \begin{array}{cc|c} 3 & 5 & 20 \\[3ex] 2 & 0 & 9 \end{array} \right] \\[3ex] J.\:\: \left[ \begin{array}{cc|c} 3 & 5 & 20 \\[3ex] 2 & 1 & 9 \end{array} \right] \\[3ex] K.\:\: \left[ \begin{array}{cc|c} 3 & 2 & 20 \\[3ex] 5 & -1 & 9 \end{array} \right] $


The option is clearly $G$ $ \left[ \begin{array}{cc|c} 3 & 5 & 20 \\[3ex] 2 & -1 & 9 \end{array} \right] $


(2.) Write the linear system in augmented matrix form.

$ 4x + y = -2 \\[3ex] 3y = 2x + 8 $


Variables are complete
But, the variables in $eqn.(2)$ are not lined up.
Line them up
Then, write in augmented matrix form

$ 4x + y = -2 ...eqn.(1) \\[3ex] 3y = 2x + 8 ...eqn.(2) \\[3ex] Arrange\:\: eqn.(2) \\[3ex] 2x + 8 = 3y \\[3ex] 2x - 3y = -8 \\[3ex] Write\:\: augmented\:\: matrix \\[3ex] \left[ \begin{array}{cc|c} 4 & 1 & -2 \\[3ex] 2 & -3 & -8 \end{array} \right] $


(3.) Write the linear system in augmented matrix form.

$ 3x - 2y + p = 9 \\[3ex] x + y + 4p = 4 \\[3ex] 2x + 3p = -1 $


$ 3x - 2y + p = 9 ...eqn.(1)...okay \\[3ex] x + y + 4p = 4 ...eqn.(2) ...okay \\[3ex] 2x = -1 - 3p ...eqn.(3) ...fix \\[3ex] 2x + 3p = -1 \\[3ex] 2x + 0y + 3p = -1 \\[3ex] Write\:\: augmented\:\: matrix \\[3ex] \left[ \begin{array}{ccc|c} 3 & -2 & 1 & 9 \\[3ex] 1 & 1 & 4 & 4 \\[3ex] 2 & 0 & 3 & -1 \end{array} \right] $


Row Operations

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 $LHS$

Examples - Row Operations

(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] $

Check
$LHS$ $RHS$
$3x + 9y$
$3(2) + 9(0)$
$6 + 0$
$6$
$6$
$2x + y$
$2(2) + 0$
$4 + 0$
$4$
$4$


(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] $


Cramer's Rule

Cramer's Rule is also known as the Method of Determinants.
It is used to solve a linear system.

Before Using Cramer's Rule:
(1.) Complete all incompletes. Make sure all the variables are present in all the equations.
If there is any missing variable, write it, and put "zero" as the coefficient.

(2.) All the variables should be on the LHS (Left Hand Side) of the equations, and all the constants should be on the RHS (Right Hand Side) of the equations.

(3.) "Line up" all the variables.

(4.) Write the matrix of coefficients (also known as the coefficient matrix).

(5.) Write the matrix of variables (also known as variable matrix).

(6.) Write the matrix of constants (also known as constant matrix).
Write the coefficient matrix, the variable matrix, and the constant matrix this way:
[Matrix of coefficients][Matrix of variables] = [Matrix of constants]

Given: a $2 * 2$ Linear System
To: set it up as a $2 * 2$ matrix
And: solve using Cramer's Rule

Linear System
$ ax + by = e ...eqn.(1) \\[3ex] cx + dy = f ...eqn.(2) \\[3ex] $ Steps:
Step (1.) ... okay
Step (2.) ... okay
Step (3.) ... okay
Step (4.) ... let's do it
$ Matrix\:\: of\:\: Coefficients = \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \\[5ex] $ Step (5.) ... let's do it
$ Matrix\:\: of\:\: Variables = \begin{bmatrix} x \\[3ex] y \end{bmatrix} \\[5ex] $ Step (6.) ... let's do it
$ Matrix\:\: of\:\: Constants = \begin{bmatrix} e \\[3ex] f \end{bmatrix} \\[5ex] $ Get them all together
$ \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} e \\[3ex] f \end{bmatrix} $

Using Cramer's Rule
$ x = \dfrac{ \begin{vmatrix} e & b \\[3ex] f & d \end{vmatrix}} {\begin{vmatrix} a & b \\[3ex] c & d \end{vmatrix}} \:\:\:and\:\:\: y = \dfrac{ \begin{vmatrix} a & e \\[3ex] c & f \end{vmatrix}} {\begin{vmatrix} a & b \\[3ex] c & d \end{vmatrix}} $

What do you notice?
Note all the responses of the students and discuss.

Observations from the Method of Determinants
(1.) Determinants are used to solve the linear system.

(2.) The denominators are the same. That is a nice thing ☺☺☺

(3.) The denominator is the determinant of the coefficient matrix.

(4.) For the numerators:
To find $x$, replace the column for $x$ (the coefficients of $x$) with the column of the constants.
Then, write the column for $y$ (the coefficients of $y$), and find the determinant.

$(x, y)$ means that $x$ comes first before $y$
So, the coefficients of $x$ which is replaced by the constants (because we want to find $x$) is written first, before the coefficients of $y$

To find $y$, write the column for $x$ (the coefficients of $x$ first, then replace the column for $y$ (the coefficients of $y$) with the column of the constants, and find the determinant.

$(x, y)$ means that $x$ comes first before $y$
So, the coefficients of $x$ is written first, before the coefficients of $y$ which is replaced by the constants (because we want to find $y$)

As always with Mr. C, you should Check your solutions


Student: What if we have a $3 * 3$ matrix?
How do we solve it using Cramer's Rule?
Teacher: Good question. Let's discuss it.


Given: a $3 * 3$ Linear System
To: set it up as a $3 * 3$ matrix
And: solve using Cramer's Rule

Linear System
$ ax + by + cp = d ...eqn.(1) \\[3ex] ex + fy + gp = h ...eqn.(2) \\[3ex] kx + my + np = r ...eqn.(3) \\[3ex] $ Steps:
Step (1.) ... okay
Step (2.) ... okay
Step (3.) ... okay
Step (4.) ... let's do it
$ Matrix\:\: of\:\: Coefficients = \begin{bmatrix} a & b & c \\[3ex] e & f & g \\[3ex] k & m & n \end{bmatrix} \\[7ex] $ Step (5.) ... let's do it
$ Matrix\:\: of\:\: Variables = \begin{bmatrix} x \\[3ex] y \\[3ex] p \end{bmatrix} \\[7ex] $ Step (6.) ... let's do it
$ Matrix\:\: of\:\: Constants = \begin{bmatrix} d \\[3ex] h \\[3ex] r \end{bmatrix} \\[5ex] $ Get them all together
$ \begin{bmatrix} a & b & c \\[3ex] e & f & g \\[3ex] k & m & n \end{bmatrix} \begin{bmatrix} x \\[3ex] y \\[3ex] p \end{bmatrix} = \:\:\: \begin{bmatrix} d \\[3ex] h \\[3ex] r \end{bmatrix} $

Ask the students to write $x$, $y$, and $p$
Correct and re-teach as necessary.

Using Cramer's Rule
$ x = \dfrac{ \begin{vmatrix} d & b & c \\[3ex] h & f & g \\[3ex] r & m & n \end{vmatrix}} {\begin{vmatrix} a & b & c \\[3ex] e & f & g \\[3ex] k & m & n \end{vmatrix}} \:\:\:and\:\:\: y = \dfrac{ \begin{vmatrix} a & d & c \\[3ex] e & h & g \\[3ex] k & r & n \end{vmatrix}} {\begin{vmatrix} a & b & c \\[3ex] e & f & g \\[3ex] k & m & n \end{vmatrix}} \:\:\:and\:\:\: p = \dfrac{ \begin{vmatrix} a & b & d \\[3ex] e & f & h \\[3ex] k & m & r \end{vmatrix}} {\begin{vmatrix} a & b & c \\[3ex] e & f & g \\[3ex] k & m & n \end{vmatrix}} $

Teacher: Guess what, folks ...speaking like an American lol
Student: What?
Teacher: We have another application of Cramer's Rule
We can use Cramer's Rule to classify Linear Systems
Do you remember the Classification of Linear Systems?
How many methods did we cover? Name them
We shall be discussing another method of classification here

Cramer's Rule is also used to classify Linear Systems
In Finding any variable:

(1.) If the determinant of the numerator is zero or non-zero and the determinant of the denominator is non-zero, the system is consistent-independent.
It has only one solution.

(2.) If the determinant of the numerator is zero and the determinant of the denominator is zero, the system is consistent-dependent.
It has many solutions.

(3.) If the determinant of the numerator is non-zero and the determinant of the denominator is zero, the system is inconsistent.
It has no solutions.

$2 * 2$ Matrix

For Cryptology using a $2 * 2$ matrix;

(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 (positive and/or negative integers) only.
Zeros, 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.


Some examples of encoding matrices that you can use are:
$ \begin{bmatrix} 5 & 3 \\[3ex] 2 & 1 \end{bmatrix} $ $ \begin{bmatrix} 7 & 4 \\[3ex] 2 & 1 \end{bmatrix} $ $ \begin{bmatrix} 3 & 4 \\[3ex] 1 & 1 \end{bmatrix} $ $ \begin{bmatrix} 1 & 1 \\[3ex] 2 & 3 \end{bmatrix} $ $ \begin{bmatrix} 3 & 4 \\[3ex] 2 & 3 \end{bmatrix} $ $ \begin{bmatrix} 1 & 2 \\[3ex] 2 & 3 \end{bmatrix} $


Given: a scenario between Mom and Sam
Mom: Sam my son, what did you learn in church today?
Sam: Mom, the priest said something very important.
He told us to always ..............................
124 48 47 16 55 22 80 29 20 8
and his message is encoded with the encoding matrix
5 2 3 1

If you were the Mom, and your son wrote this to you, what would you do? ☺ ☺ ☺

To: decode the message

The Cipher Key is:
$A$ $B$ $C$ $D$ $E$ $F$
$1$ $2$ $3$ $4$ $5$ $6$
$G$ $H$ $I$ $J$ $K$ $L$
$7$ $8$ $9$ $10$ $11$ $12$
$M$ $N$ $O$ $P$ $Q$ $R$
$13$ $14$ $15$ $16$ $17$ $18$
$S$ $T$ $U$ $V$ $W$ $X$
$19$ $20$ $21$ $22$ $23$ $24$
$Y$ $Z$ $ $ $?$ $!$ $.$
$25$ $26$ $27, 0$ $28$ $29$ $30, 0$

$1^{st}$ Step:
Because this question deals with $2 * 2$ matrix, write the encoding matrix in a $2 * 2$ matrix form
5 2 3 1
Write the elements per column (not per row)
Encoding Matrix, A
$ A = \begin{bmatrix} 5 & 3 \\[3ex] 2 & 1 \end{bmatrix} $


$2^{nd}$ Step:
Because this question deals with $2 * 2$ matrix, write the encoded message in matrix form such that the number of rows is $2$
Ask students why they have to do it that way?
124 48 47 16 55 22 80 29 20 8
Encoded Message, B
Write the elements per column (not per row)
$ B = \begin{bmatrix} 124 & 47 & 55 & 80 & 20 \\[3ex] 48 & 16 & 22 & 29 & 8 \end{bmatrix} $


$3^{rd}$ Step:
Determine the Decoding Matrix, $A^{-1}$
We shall use both methods - The Formula Method and the Row Reduction Method.
You may use any one you wish.

$ \underline{Formula\:\: Method} \\[3ex] A^{-1} = \dfrac{adj\: A}{\det\: A} \\[5ex] For\:\: A = \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} = \begin{bmatrix} 5 & 3 \\[3ex] 2 & 1 \end{bmatrix} \\[5ex] a = 5 \\[3ex] b = 3 \\[3ex] c = 2 \\[3ex] d = 1 \\[3ex] adj\: A = \begin{bmatrix} d & -b \\[3ex] -c & a \end{bmatrix} \\[5ex] -b = -1 * 3 = -3 \\[3ex] -c = -1 * 2 = -2 \\[3ex] \therefore adj\: A = \begin{bmatrix} 1 & -3 \\[3ex] -2 & 5 \end{bmatrix} \\[5ex] det\: A = ad - cb = 5(1) - 2(3) = 5 - 6 = -1 \\[3ex] A^{-1} = \dfrac{adj\: A}{\det\: A} = \begin{bmatrix} \dfrac{1}{-1} & \dfrac{-3}{-1} \\[5ex] \dfrac{-2}{-1} & \dfrac{5}{-1} \end{bmatrix} \\[10ex] A^{-1} = \begin{bmatrix} -1 & 3 \\[3ex] 2 & -5 \end{bmatrix} \\[7ex] \underline{Row\:\: Reduction\:\: Method} \\[3ex] A \ \ | \ \ I = I \ \ | \ \ A^{-1} \\[3ex] \left[ \begin{array}{cc|cc} 5 & 3 & 1 & 0 \\[3ex] 2 & 1 & 0 & 1 \end{array} \right] \underrightarrow{-2R_1 + 5R_2} \left[ \begin{array}{cc|cc} 5 & 3 & 1 & 0 \\[3ex] 0 & -1 & -2 & 5 \end{array} \right] \\[3ex] $
$ \underrightarrow{3R_2 + R_1} \left[ \begin{array}{cc|cc} 5 & 0 & -5 & 15 \\[3ex] 0 & -1 & -2 & 5 \end{array} \right] \begin{matrix} \underrightarrow{R_1 \div 5} \\ \underrightarrow{-R_2} \end{matrix} \left[ \begin{array}{cc|cc} 1 & 0 & -1 & 3 \\[3ex] 0 & 1 & 2 & -5 \end{array} \right] $

$ A^{-1} = \begin{bmatrix} -1 & 3 \\[3ex] 2 & -5 \end{bmatrix} $


$4^{th}$ Step:
Determine the Decoded Message
Decoded Message = $A^{-1} * B$

Ask students if the matrices are conformable for multiplication

Order of Matrix $A^{-1} = 2 * 2$
Order of Matrix $B = 2 * 5$
Number of columns of $A^{-1}, (2)$ = Number of rows of $B, (2)$
The matrices can multiply.
Order of Product, $A^{-1}B$ = Number of rows of $A^{-1}$ by Number of columns of $B$
Order of $A^{-1}B = 2 * 5$

$ \begin{bmatrix} -1 & 3 \\[3ex] 2 & -5 \end{bmatrix} * \begin{bmatrix} 124 & 47 & 55 & 80 & 20 \\[3ex] 48 & 16 & 22 & 29 & 8 \end{bmatrix} \\[5ex] -1 * 124 + 3 * 48 = -124 + 144 = 20 \\[3ex] -1 * 47 + 3 * 16 = -47 + 48 = 1 \\[3ex] -1 * 55 + 3 * 22 = -55 + 66 = 11 \\[3ex] -1 * 80 + 3 * 29 = -80 + 87 = 7 \\[3ex] -1 * 20 + 3 * 8 = -20 + 24 = 4 \\[5ex] 2 * 124 + (-5) * 48 = 248 - 240 = 8 \\[3ex] 2 * 47 + (-5) * 16 = 94 - 80 = 14 \\[3ex] 2 * 55 + (-5) * 22 = 110 - 110 = 0 \\[3ex] 2 * 80 + (-5) * 29 = 160 - 145 = 15 \\[3ex] 2 * 20 + (-5) * 8 = 40 - 40 = 0 $

$ A^{-1} * B = \begin{bmatrix} 20 & 1 & 11 & 7 & 4 \\[3ex] 8 & 14 & 0 & 15 & 0 \end{bmatrix} $


$5^{th}$ Step:
Determine the Message using the Cipher Key

Decoded Message: 20 8 1 14 11 0 7 15 4 0
Message:                 T H A N K   G O D .

The Priest's basic message is THANK GOD.


Given: a secret message, THANK GOD.
To Encode: the secret message using a $2 * 2$ encoding matrix

$1^{st}$ Step:
Write the decoded message using the Cipher Key
Convert the characters to digits using the Cipher Key.
Message:                 T H A N K   G O D .
Decoded Message: 20 8 1 14 11 0 7 15 4 0
You may use $27$ for the space, and $30$ for the period if you wish.


$2^{nd}$ Step:
Because this question deals with $2 * 2$ matrix, write the decoded message in matrix form such that the number of rows is $2$
Ask students why they have to do it that way?
Make sure the number of digits is divisible by $2$ (an even number).
20 8 1 14 11 0 7 15 4 0
It has $10$ digits. Divide it by $2$.
$10$ digits divided by $2$ rows gives $5$ digits in each row.
Write the elements per column (not per row)
Decoded Message
$ \begin{bmatrix} 20 & 1 & 11 & 7 & 4 \\[3ex] 8 & 14 & 0 & 15 & 0 \end{bmatrix} $


$3^{rd}$ Step:
Find an encoding matrix:
(1.) that has only positive integers. The elements of the encoding matrix has no zeros, negative integers, fractions and/or decimals.
(2.) that is invertible. The encoding matrix has an inverse.
(3.) such that the elements of the inverse of that matrix has only integers.
The inverse of the encoding matrix is the decoding matrix.
The elements of the decoding matrix should be only integers (positive and negative integers).
The elements of the decoding matrix should not have zeros, fractions, or decimals.
You may use the calculators on the Matrix Algebra website, Microsoft Excel Spreadsheet, or Google Spreadsheet to find a suitable encoding matrix.
Encoding Matrix: 5 2 3 1


$4^{th}$ Step:
Because this question deals with $2 * 2$ matrix, write the encoding matrix as a $2 * 2$ matrix
Encoding Matrix: 5 2 3 1
Write the elements per column (not per row)
Encoding Matrix
$ \begin{bmatrix} 5 & 3 \\[3ex] 2 & 1 \end{bmatrix} $


$5^{th}$ Step:
Write the Encoded Message
Encoded Message = Encoding Matrix * Decoded Message
Let us make sure they can multiply.
Order of Encoding Matrix = $2 * 2$
Order of Decoded Message = $2 * 5$
Number of columns of Encoding Matrix = Number of rows of Decoded Message
The matrices can multiply.
Order of Product = Number of rows of Encoding Matrix by Number of columns of Decoded Message
Order of Product = $2 * 5$

$ \begin{bmatrix} 5 & 3 \\[3ex] 2 & 1 \end{bmatrix} * \begin{bmatrix} 20 & 1 & 11 & 7 & 4 \\[3ex] 8 & 14 & 0 & 15 & 0 \end{bmatrix} \\[7ex] 5(20) + 3(8) = 100 + 24 = 124 \\[3ex] 5(1) + 3(14) = 5 + 42 = 47 \\[3ex] 5(11) + 3(0) = 55 + 0 = 55 \\[3ex] 5(7) + 3(15) = 35 + 45 = 80 \\[3ex] 5(4) + 3(0) = 20 + 0 = 20 \\[5ex] 2(20) + 1(8) = 40 + 8 = 48 \\[3ex] 2(1) + 1(14) = 2 + 14 = 16 \\[3ex] 2(11) + 1(0) = 22 + 0 = 22 \\[3ex] 2(7) + 1(15) = 14 + 15 = 29 \\[3ex] 2(4) + 1(0) = 8 + 0 = 8 $

$ Encoded \: Message = \begin{bmatrix} 124 & 47 & 55 & 80 & 20 \\[3ex] 48 & 16 & 22 & 29 & 8 \end{bmatrix} \\[3ex] $
Release and write the elements per column (not per row)
Encoded Message = 124 48 47 16 55 22 80 29 20 8

$3 * 3$ Matrix

For Cryptology using a $3 * 3$ matrix;

(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 (positive and/or negative integers) only.
Zeros, 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.


Some examples of encoding matrices that you can use are:
$ \begin{bmatrix} 1 & 1 & 1 \\[3ex] 2 & 1 & 2 \\[3ex] 2 & 3 & 1 \end{bmatrix} $ $ \begin{bmatrix} 3 & 1 & 2 \\[3ex] 1 & 1 & 3 \\[3ex] 3 & 2 & 5 \end{bmatrix} $ $ \begin{bmatrix} 1 & 2 & 2 \\[3ex] 1 & 1 & 3 \\[3ex] 2 & 2 & 5 \end{bmatrix} $ $ \begin{bmatrix} 1 & 2 & 2 \\[3ex] 1 & 1 & 1 \\[3ex] 2 & 2 & 1 \end{bmatrix} $ $ \begin{bmatrix} 1 & 2 & 1 \\[3ex] 1 & 2 & 5 \\[3ex] 1 & 3 & 2 \end{bmatrix} $


Given: Please complete this sentence.
If you ....................................., the world will be a good place to live.
The "blank" is this encoded message:
49 83 91 25 50 30 33 41 91 28 51 52 17 26 40 33 48 84
encoded with the encoding matrix
1 2 2 1 1 3 1 2 1

What do we need to do to make this beautiful world created by GOD, a good place to live?
It is one of the commandments given to us by JESUS CHRIST.
Ask students for their responses.

To: decode the message

The Cipher Key is:
$A$ $B$ $C$ $D$ $E$ $F$
$1$ $2$ $3$ $4$ $5$ $6$
$G$ $H$ $I$ $J$ $K$ $L$
$7$ $8$ $9$ $10$ $11$ $12$
$M$ $N$ $O$ $P$ $Q$ $R$
$13$ $14$ $15$ $16$ $17$ $18$
$S$ $T$ $U$ $V$ $W$ $X$
$19$ $20$ $21$ $22$ $23$ $24$
$Y$ $Z$ $ $ $?$ $!$ $.$
$25$ $26$ $27, 0$ $28$ $29$ $30, 0$

$1^{st}$ Step:
Because this question deals with $3 * 3$ matrix, write the encoding matrix in a $3 * 3$ matrix form
1 2 2 1 1 3 1 2 1
Write the elements per column (not per row)
Encoding Matrix, A
$ A = \begin{bmatrix} 1 & 1 & 1 \\[3ex] 2 & 1 & 2 \\[3ex] 2 & 3 & 1 \end{bmatrix} $


$2^{nd}$ Step:
Because this question deals with $3 * 3$ matrix, write the encoded message in matrix form such that the number of rows is $3$
Ask students why they have to do it that way?
49 83 91 25 50 30 33 41 91 28 51 52 17 26 40 33 48 84
Encoded Message, B
Write the elements per column (not per row)
$ B = \begin{bmatrix} 49 & 25 & 33 & 28 & 17 & 33 \\[3ex] 83 & 50 & 41 & 51 & 26 & 48 \\[3ex] 91 & 30 & 91 & 52 & 40 & 84 \end{bmatrix} $


$3^{rd}$ Step:
Determine the Decoding Matrix, $A^{-1}$
We shall use both methods - The Formula Method and the Row Reduction Method.
You may use any one you wish.

$ \underline{Formula\:\: Method} \\[3ex] A^{-1} = \dfrac{adj\: A}{\det\: A} \\[5ex] For\:\: A = \begin{bmatrix} a & b & c \\[3ex] d & e & f \\[3ex] g & h & i \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\[3ex] 2 & 1 & 2 \\[3ex] 2 & 3 & 1 \end{bmatrix} \\[5ex] a = 1 \\[3ex] b = 1 \\[3ex] c = 1 \\[3ex] d = 2 \\[3ex] e = 1 \\[3ex] f = 2 \\[3ex] g = 2 \\[3ex] h = 3 \\[3ex] i = 1 \\[3ex] 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} \\[10ex] ei - hf = 1(1) - 3(2) = 1 - 6 = -5 \\[3ex] hc - bi = 3(1) - 1(1) = 3 - 1 = 2 \\[3ex] bf - ec = 1(2) - 1(1) = 2 - 1 = 1 \\[3ex] gf - di = 2(2) - 2(1) = 4 - 2 = 2 \\[3ex] ai - gc = 1(1) - 2(1) = 1 - 2 = -1 \\[3ex] dc - af = 2(1) - 1(2) = 2 - 2 = 0 \\[3ex] dh - ge = 2(3) - 2(1) = 6 - 2 = 4 \\[3ex] gb - ah = 2(1) - 1(3) = 2 - 3 = -1 \\[3ex] ae - db = 1(1) - 2(1) = 1 - 2 = -1 \\[5ex] \therefore adj\: A = \begin{bmatrix} -5 & 2 & 1 \\[3ex] 2 & -1 & 0 \\[3ex] 4 & -1 & -1 \end{bmatrix} \\[10ex] det\: A = aei + bgf + cdh - ahf - bdi - cge \\[3ex] aei = 1 * 1 * 1 = 1 \\[3ex] bgf = 1 * 2 * 2 = 4 \\[3ex] cdh = 1 * 2 * 3 = 6 \\[3ex] ahf = 1 * 3 * 2 = 6 \\[3ex] bdi = 1 * 2 * 1 = 2 \\[3ex] cge = 1 * 2 * 1 = 2 \\[5ex] \therefore det\: A = 1 + 4 + 6 - 6 - 2 - 2 = 1 \\[5ex] A^{-1} = \dfrac{adj\: A}{\det\: A} = \begin{bmatrix} \dfrac{-5}{1} & \dfrac{2}{1} & \dfrac{1}{1} \\[5ex] \dfrac{2}{1} & \dfrac{-1}{1} & \dfrac{0}{1} \\[5ex] \dfrac{4}{1} & \dfrac{-1}{1} & \dfrac{-1}{1} \end{bmatrix} \\[16ex] \therefore A^{-1} = \begin{bmatrix} -5 & 2 & 1 \\[3ex] 2 & -1 & 0 \\[3ex] 4 & -1 & -1 \end{bmatrix} \\[10ex] \underline{Row\:\: Reduction\:\: Method} \\[3ex] A \ \ | \ \ I = I \ \ | \ \ A^{-1} \\[3ex] \left[ \begin{array}{ccc|ccc} 1 & 1 & 1 & 1 & 0 & 0 \\[3ex] 2 & 1 & 2 & 0 & 1 & 0 \\[3ex] 2 & 3 & 1 & 0 & 0 & 1 \end{array} \right] \begin{matrix} \underrightarrow{-2R_1 + R_2} \\ \underrightarrow{-2R_1 + R_3} \end{matrix} \left[ \begin{array}{ccc|ccc} 1 & 1 & 1 & 1 & 0 & 0 \\[3ex] 0 & -1 & 0 & -2 & 1 & 0 \\[3ex] 0 & 1 & -1 & -2 & 0 & 1 \end{array} \right] \\[2ex] $
$ \begin{matrix} \underrightarrow{R_2 + R_1} \\ \underrightarrow{R_2 + R_3} \end{matrix} \left[ \begin{array}{ccc|ccc} 1 & 0 & 1 & -1 & 1 & 0 \\[3ex] 0 & -1 & 0 & -2 & 1 & 0 \\[3ex] 0 & 0 & -1 & -4 & 1 & 1 \end{array} \right] \underrightarrow{R_3 + R_1} \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & -5 & 2 & 1 \\[3ex] 0 & -1 & 0 & -2 & 1 & 0 \\[3ex] 0 & 0 & -1 & -4 & 1 & 1 \end{array} \right] \\[3ex] \begin{matrix} \underrightarrow{-R_2} \\ \underrightarrow{-R_3} \end{matrix} \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & -5 & 2 & 1 \\[3ex] 0 & 1 & 0 & 2 & -1 & 0 \\[3ex] 0 & 0 & 1 & 4 & -1 & -1 \end{array} \right] \\[5ex] A^{-1} = \begin{bmatrix} -5 & 2 & 1 \\[3ex] 2 & -1 & 0 \\[3ex] 4 & -1 & -1 \end{bmatrix} $


$4^{th}$ Step:
Determine the Decoded Message
Decoded Message = $A^{-1} * B$

Ask students if the matrices are conformable for multiplication

Order of Matrix $A^{-1} = 3 * 3$
Order of Matrix $B = 3 * 6$
Number of columns of $A^{-1}, (3)$ = Number of rows of $B, (3)$
The matrices can multiply.
Order of Product, $A^{-1}B$ = Number of rows of $A^{-1}$ by Number of columns of $B$
Order of $A^{-1}B = 3 * 6$

$ \begin{bmatrix} -5 & 2 & 1 \\[3ex] 2 & -1 & 0 \\[3ex] 4 & -1 & -1 \end{bmatrix} * \begin{bmatrix} 49 & 25 & 33 & 28 & 17 & 33 \\[3ex] 83 & 50 & 41 & 51 & 26 & 48 \\[3ex] 91 & 30 & 91 & 52 & 40 & 84 \end{bmatrix}\\[2ex] $
$ -5 * 49 + 2 * 83 + 1 * 91 = -245 + 166 + 91 = 12 \\[3ex] -5 * 25 + 2 * 50 + 1 * 30 = -125 + 100 + 30 = 5 \\[3ex] -5 * 33 + 2 * 41 + 1 * 91 = -165 + 82 + 91 = 8 \\[3ex] -5 * 28 + 2 * 51 + 1 * 52 = -140 + 102 + 52 = 14 \\[3ex] -5 * 17 + 2 * 26 + 1 * 40 = -85 + 52 + 40 = 7 \\[3ex] -5 * 33 + 2 * 48 + 1 * 84 = -165 + 96 + 84 = 15 \\[5ex] 2 * 49 + (-1) * 83 + 0 * 91 = 98 - 83 + 0 = 15 \\[3ex] 2 * 25 + (-1) * 50 + 0 * 30 = 50 - 50 + 0 = 0 \\[3ex] 2 * 33 + (-1) * 41 + 0 * 91 = 66 - 41 + 0 = 25 \\[3ex] 2 * 28 + (-1) * 51 + 0 * 52 = 56 - 51 + 0 = 5 \\[3ex] 2 * 17 + (-1) * 26 + 0 * 40 = 34 - 26 + 0 = 8 \\[3ex] 2 * 33 + (-1) * 48 + 0 * 84 = 66 - 48 + 0 = 18 \\[5ex] 4 * 49 + (-1) * 83 + (-1) * 91 = 196 - 83 - 91 = 22 \\[3ex] 4 * 25 + (-1) * 50 + (-1) * 30 = 100 - 50 - 30 = 20 \\[3ex] 4 * 33 + (-1) * 41 + (-1) * 91 = 132 - 41 - 91 = 0 \\[3ex] 4 * 28 + (-1) * 51 + (-1) * 52 = 112 - 51 - 52 = 9 \\[3ex] 4 * 17 + (-1) * 26 + (-1) * 40 = 68 - 26 - 40 = 2 \\[3ex] 4 * 33 + (-1) * 48 + (-1) * 84 = 132 - 48 - 84 = 0 $

$ A^{-1} * B = \begin{bmatrix} 12 & 5 & 8 & 14 & 7 & 15 \\[3ex] 15 & 0 & 25 & 5 & 8 & 18 \\[3ex] 22 & 20 & 0 & 9 & 2 & 0 \end{bmatrix} $


$5^{th}$ Step:
Determine the Message using the Cipher Key

Decoded Message: 12 15 22 5 0 20 8 25 0 14 5 9 7 8 2 15 18 0
Message:              L O V E   T H Y   N E I G H B O R .

The message is LOVE THY NEIGHBOR.


Given: a secret message, LOVE THY NEIGHBOR.
To Encode: the secret message using a $3 * 3$ encoding matrix

$1^{st}$ Step:
Write the decoded message using the Cipher Key
Convert the characters to digits using the Cipher Key.
Message:              L O V E   T H Y   N E I G H B O R .
Decoded Message: 12 15 22 5 0 20 8 25 0 14 5 9 7 8 2 15 18 0
You may use $27$ for the space, and $30$ for the period if you wish.


$2^{nd}$ Step:
Because this question deals with $3 * 3$ matrix, write the decoded message in matrix form such that the number of rows is $3$
Ask students why they have to do it that way?
Make sure the number of digits is divisible by $3$.
12 15 22 5 0 20 8 25 0 14 5 9 7 8 2 15 18 0
It has $18$ digits. Divide it by $3$.
$18$ digits divided by $3$ rows gives $6$ digits in each row.
Write the elements per column (not per row)
Decoded Message
$ \begin{bmatrix} 12 & 5 & 8 & 14 & 7 & 15 \\[3ex] 15 & 0 & 25 & 5 & 8 & 18 \\[3ex] 22 & 20 & 0 & 9 & 2 & 0 \end{bmatrix} $


$3^{rd}$ Step:
Find an encoding matrix:
(1.) that has only positive integers. The elements of the encoding matrix has no zeros, negative integers, fractions and/or decimals.
(2.) that is invertible. The encoding matrix has an inverse.
(3.) such that the elements of the inverse of that matrix has only integers.
The inverse of the encoding matrix is the decoding matrix.
The elements of the decoding matrix should be only integers (positive and negative integers).
The elements of the decoding matrix should not have zeros, fractions, or decimals.
You may use the calculators on the Matrix Algebra website, Microsoft Excel Spreadsheet, or Google Spreadsheet to find a suitable encoding matrix.
Encoding Matrix: 1 2 2 1 1 3 1 2 1


$4^{th}$ Step:
Because this question deals with $3 * 3$ matrix, write the encoding matrix as a $3 * 3$ matrix
Encoding Matrix: 1 2 2 1 1 3 1 2 1
Write the elements per column (not per row)
Encoding Matrix
$ \begin{bmatrix} 1 & 1 & 1 \\[3ex] 2 & 1 & 2 \\[3ex] 2 & 3 & 1 \end{bmatrix} $


$5^{th}$ Step:
Write the Encoded Message
Encoded Message = Encoding Matrix * Decoded Message
Let us make sure they can multiply.
Order of Encoding Matrix = $3 * 3$
Order of Decoded Message = $3 * 6$
Number of columns of Encoding Matrix = Number of rows of Decoded Message
The matrices can multiply.
Order of Product = Number of rows of Encoding Matrix by Number of columns of Decoded Message
Order of Product = $3 * 6$

$ \begin{bmatrix} 1 & 1 & 1 \\[3ex] 2 & 1 & 2 \\[3ex] 2 & 3 & 1 \end{bmatrix} * \begin{bmatrix} 12 & 5 & 8 & 14 & 7 & 15 \\[3ex] 15 & 0 & 25 & 5 & 8 & 18 \\[3ex] 22 & 20 & 0 & 9 & 2 & 0 \end{bmatrix} \\[2ex] $
$ 1(12) + 1(15) + 1(22) = 12 + 15 + 22 = 49 \\[3ex] 1(5) + 1(0) + 1(20) = 5 + 0 + 20 = 25 \\[3ex] 1(8) + 1(25) + 1(0) = 8 + 25 + 0 = 33 \\[3ex] 1(14) + 1(5) + 1(9) = 14 + 5 + 9 = 28 \\[3ex] 1(7) + 1(8) + 1(2) = 7 + 8 + 2 = 17 \\[3ex] 1(15) + 1(18) + 1(0) = 15 + 18 + 0 = 33 \\[5ex] 2(12) + 1(15) + 2(22) = 24 + 15 + 44 = 83 \\[3ex] 2(5) + 1(0) + 2(20) = 10 + 0 + 40 = 50 \\[3ex] 2(8) + 1(25) + 2(0) = 16 + 25 + 0 = 41 \\[3ex] 2(14) + 1(5) + 2(9) = 28 + 5 + 18 = 51 \\[3ex] 2(7) + 1(8) + 2(2) = 14 + 8 + 4 = 26 \\[3ex] 2(15) + 1(18) + 2(0) = 30 + 18 + 0 = 48 \\[5ex] 2(12) + 3(15) + 1(22) = 24 + 45 + 22 = 91 \\[3ex] 2(5) + 3(0) + 1(20) = 10 + 0 + 20 = 30 \\[3ex] 2(8) + 3(25) + 1(0) = 16 + 75 + 0 = 91 \\[3ex] 2(14) + 3(5) + 1(9) = 28 + 15 + 9 = 52 \\[3ex] 2(7) + 3(8) + 1(2) = 14 + 24 + 2 = 40 \\[3ex] 2(15) + 3(18) + 1(0) = 30 + 54 + 0 = 84 $

$ Encoded \: Message = \begin{bmatrix} 49 & 25 & 33 & 28 & 17 & 33 \\[3ex] 83 & 50 & 41 & 51 & 26 & 48 \\[3ex] 91 & 30 & 91 & 52 & 40 & 84 \end{bmatrix} \\[2ex] $
Release and write the elements per column (not per row)
Encoded Message = 49 83 91 25 50 30 33 41 91 28 51 52 17 26 40 33 48 84

Matrices using TI-84/TI-84 Plus Calculator

This calculator will:
(1.) Solve several problems involving matrices.
To use the calculator;

  • Visit the link: TI-84 Calculator
  • Click the "Matrices" folder
  • Click the link of "whatever you want to do"
  • Follow the directions as illustrated by the "blinking light"

References

Chukwuemeka, S.D (2016, April 30). Samuel Chukwuemeka Tutorials - Math, Science, and Technology. Retrieved from https://www.chukwuemekasamuel.com

Bittinger, M. L., Beecher, J. A., Ellenbogen, D. J., & Penna, J. A. (2017). Algebra and Trigonometry: Graphs and Models ($6^{th}$ ed.). Boston: Pearson.

Sullivan, M., & Sullivan, M. (2017). Algebra & Trigonometry ($7^{th}$ ed.). Boston: Pearson.

Tan, S. (2015). Finite Mathematics for the Managerial, Life, and Social Sciences (Revised/Expanded ed.). Boston: Cengage Learning.

CAREER INDIA - Free CMAT Past Questions. (n.d). Retrieved from https://www.careerindia.com/entrance-exam/cmat-question-papers-e23.html

CrackACT - Free ACT Past Questions. (n.d.). Retrieved from http://www.crackact.com/act-downloads/

CSEC Math Tutor - Free Caribbean Secondary Education Certificate Examination (CSEC) Past Questions. (n.d). Retrieved from https://www.csecmathtutor.com/past-papers.html

MySchoolGist - Free West African Senior School Certificate Examination (WASSCE) Past Questions. (n.d). Retrieved from https://www.myschoolgist.com/ng/free-waec-past-questions-and-answers/

VAST LEARNERS - Free JAMB Past Questions. (n.d). Retrieved from https://www.vastlearners.com/free-jamb-past-questions/