Determinant Filter Slides UNIT 1.1
The Determinant Filter
Unlocking the gatekeeper of matrix solutions.
The Ultimate Question
What single number can tell us if a system of equations has a unique solution before we even try to solve it?
Unique Solution
det ≠ 0
No/Inf Solutions
det = 0
01. The 2x2 Formula
Given a matrix A:
\[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \]
The determinant, \(\text{det}(A)\) or \(|A|\), is:
\( ad - bc \)
The Cross-Multiply Method
Top-Left × Bottom-Right - Top-Right × Bottom-Left
Quick Filter Practice
Matrix P
\[ P = \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} \]
Calculation Step:
\( (4 \times 5) - (3 \times 2) \)
\( 20 - 6 = 14 \)
Matrix Q
\[ Q = \begin{bmatrix} 6 & 2 \\ 9 & 3 \end{bmatrix} \]
Calculation Step:
\( (6 \times 3) - (2 \times 9) \)
\( 18 - 18 = 0 \)
Singular Matrix Alert
If \( \text{det}(A) = 0 \), the matrix is singular.
This means the matrix is non-invertible. In a system of equations, it implies either parallel lines (no solution) or overlapping lines (infinite solutions).
02. Scaling Up: The 3x3
Expansion by Minors (along Row 1):
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]
\( a \begin{vmatrix} e & f \\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix} \)
The Pattern
● Pick a number from the top row.
● Ignore its row and column.
● Multiply by the 2x2 determinant left over.
● Watch your signs: + , - , +
3x3 Mission
\[ M = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{vmatrix} \]
\( 1(24 - 0) \)
\( - 2(0 - 5) \)
\( + 3(0 - 4) \)
\( 24 + 10 - 12 = 22 \)
Determinant Drill Worksheet The Determinant Drill
Mission: Filter the singular from the invertible.
NAME:
DATE:
PART 01
2x2 Matrix Decryption
Calculate the determinant for each 2x2 matrix. Show your cross-multiplication steps. Identify if the matrix is Singular (det = 0) or Invertible (det ≠ 0).
01
\[ A = \begin{bmatrix} 5 & 2 \\ 3 & 4 \end{bmatrix} \]
det(A) =
Singular Invertible
02
\[ B = \begin{bmatrix} -2 & 8 \\ 1 & -4 \end{bmatrix} \]
det(B) =
Singular Invertible
03
\[ C = \begin{bmatrix} 7 & 10 \\ 2 & 3 \end{bmatrix} \]
det(C) =
Singular Invertible
04
\[ D = \begin{bmatrix} 0 & 5 \\ 0 & 2 \end{bmatrix} \]
det(D) =
Singular Invertible
PART 02
3x3 Expansion Probe
Use Expansion by Minors along the top row to find the determinants below. Show each 2x2 sub-matrix step.
05
\[ M = \begin{bmatrix} 1 & 2 & 0 \\ 3 & -1 & 2 \\ 2 & 0 & 4 \end{bmatrix} \]
det(M) =
06
\[ N = \begin{bmatrix} 2 & 1 & 1 \\ 0 & 3 & 5 \\ 4 & 2 & 2 \end{bmatrix} \]
det(N) =
CRITICAL ANALYSIS
If you discover that det(N) = 0 in Problem 6, what does that tell you about the rows of matrix N?
Determinant Drill Key Answer Key
Teacher Resource: Determinant Drill
PART 01
2x2 Matrix Solutions
01
\[ A = \begin{bmatrix} 5 & 2 \\ 3 & 4 \end{bmatrix} \]
Calculation: (5 × 4) - (2 × 3) = 20 - 6
det(A) = 14
Result: Invertible
02
\[ B = \begin{bmatrix} -2 & 8 \\ 1 & -4 \end{bmatrix} \]
Calculation: (-2 × -4) - (8 × 1) = 8 - 8
det(B) = 0
Result: Singular
03
\[ C = \begin{bmatrix} 7 & 10 \\ 2 & 3 \end{bmatrix} \]
Calculation: (7 × 3) - (10 × 2) = 21 - 20
det(C) = 1
Result: Invertible
04
\[ D = \begin{bmatrix} 0 & 5 \\ 0 & 2 \end{bmatrix} \]
Calculation: (0 × 2) - (5 × 0) = 0 - 0
det(D) = 0
Result: Singular
PART 02
3x3 Expansion Solutions
05
\[ M = \begin{bmatrix} 1 & 2 & 0 \\ 3 & -1 & 2 \\ 2 & 0 & 4 \end{bmatrix} \]
Expansion: 1(-4-0) - 2(12-4) + 0(0 - -2)
= 1(-4) - 2(8) + 0
= -4 - 16
det(M) = -20
06
\[ N = \begin{bmatrix} 2 & 1 & 1 \\ 0 & 3 & 5 \\ 4 & 2 & 2 \end{bmatrix} \]
Expansion: 2(6-10) - 1(0-20) + 1(0-12)
= 2(-4) - 1(-20) + 1(-12)
= -8 + 20 - 12
det(N) = 0
CRITICAL ANALYSIS KEY
Answer: The fact that det(N) = 0 indicates that the rows are linearly dependent. Specifically, Row 3 is exactly twice Row 1 \((4, 2, 2 = 2 \times (2, 1, 1))\). When rows are multiples of each other, the matrix will always be singular.
Matrix Mirror Slides Lesson 1.2
The Matrix Mirror
Discovering the inverse: the matrix world's version of the reciprocal.
Reciprocal Thinking
Normal Math
To "undo" multiplying by 5, we multiply by 1/5.
\( 5 \times \frac{1}{5} = 1 \)
Matrix Math
To "undo" matrix A, we multiply by its inverse A⁻¹.
\( A \times A^{-1} = I \)
Wait, what is "I"?
01. The Identity Matrix (I)
The Identity Matrix is the "1" of the matrix world.
Identity Rule:
\( A \times I = A \)
2x2 Identity
\[ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
3x3 Identity
\[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
02. How to find A⁻¹
If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), then:
\( A^{-1} = \frac{1}{\text{det}(A)} \) \( \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)
Step 1
Find Determinant (\(ad-bc\))
Step 2
Swap a and d
Step 3
Negate b and c
When is Inversion Impossible?
Just like you can't divide by zero in normal math, you can't invert a matrix if the determinant is zero.
\( \frac{1}{0} = \text{Undefined} \implies \text{No Inverse} \)
Verification: The Proof
To check your work:
\( A \times A^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
If you don't get the identity matrix, something went wrong in your calculation!
Pro-Tip
Keep the \( \frac{1}{\text{det}} \) outside the matrix until the very end. Multiplying by a fraction inside the matrix early on makes the matrix multiplication much harder!
Inversion Lab Activity Inversion Lab
Objective: Derive and verify matrix inverses.
AGENT:
STATUS: Authorized
The Inversion Protocol
1. Calculate det(A) = ad - bc
\( A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)
TASK 01: DERIVATION ID: MTRX-82
ORIGINAL MATRIX
\[ A = \begin{bmatrix} 4 & 7 \\ 1 & 2 \end{bmatrix} \]
STEP A: CALCULATE DETERMINANT
STEP B: APPLY FORMULA
FINAL INVERSE (A⁻¹)
Write final matrix here
TASK 02: VERIFICATION ID: MTRX-33
PROVE: \( A \times A^{-1} = I \)
\[ \begin{bmatrix} 4 & 7 \\ 1 & 2 \end{bmatrix} \]
×
Insert A⁻¹
=
\[ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
Show multiplication steps (Row x Column):
TASK 03: THE SINGULAR TEST ID: MTRX-00
\[ B = \begin{bmatrix} 6 & 4 \\ 9 & 6 \end{bmatrix} \]
Why does this matrix not have an inverse? Support your answer with a calculation.
Inversion Guide Teacher Resource Instructional Guide
Teacher Notes: Matrix Inversion
Prerequisite Check
Students must be comfortable with 2x2 determinants and scalar multiplication before this lesson. Remind them that matrix multiplication is not commutative (\(AB \neq BA\)), but for inverses, \(A \times A^{-1} = A^{-1} \times A = I\).
Common Pitfalls
Forgetting to divide by the determinant at the end.
Swapping the wrong diagonal (only a and d swap).
Changing the signs of the wrong diagonal (only b and c negate).
Arithmetic errors with negative signs in the determinant.
Lab Activity Solutions
1 Derivation Task (Matrix A)
Determinant:
\( (4 \times 2) - (7 \times 1) = 8 - 7 = 1 \)
Final Inverse:
\[ A^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -7 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 2 & -7 \\ -1 & 4 \end{bmatrix} \]
2 Verification Task
\[ \begin{bmatrix} 4 & 7 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -7 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} (4 \times 2) + (7 \times -1) & (4 \times -7) + (7 \times 4) \\ (1 \times 2) + (2 \times -1) & (1 \times -7) + (2 \times 4) \end{bmatrix} \]
\[ = \begin{bmatrix} 8 - 7 & -28 + 28 \\ 2 - 2 & -7 + 8 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
3 Singular Test (Matrix B)
Calculation: \(\text{det}(B) = (6 \times 6) - (4 \times 9) = 36 - 36 = 0\).
Explanation: Since the determinant is zero, we would have to divide by zero in the inversion formula (\(\frac{1}{0}\)), which is undefined. Therefore, the matrix has no inverse and is singular. Visually, note that Row 2 is \(1.5 \times\) Row 1.
Pacing
10 mins Hook/Slides
Extension Idea
"Ask students to create their own 2x2 matrix and trade with a partner to find the inverse. They must verify the partner's work via multiplication."
Packing Equations Slides Lesson 1.3
Packing Equations
Translating messy systems into clean matrix structures: The AX = B Protocol.
The Power of Organization
Messy System
\( 2x + 3y = 8 \)
Clean Matrix
\[ AX = B \]
"We take a messy system of equations and 'pack' it into three neat boxes (matrices) to prepare for solving."
01. Components of AX = B
A
Coefficient Matrix
Contains just the numbers (coefficients) attached to the variables.
X
Variable Matrix
A column vector containing the variables (usually x, y, z).
B
Constant Matrix
A column vector containing the answers on the right side of the equations.
The Matrix Blueprint
\( 3x - 5y = 12 \)
\( x + 2y = -4 \)
\[ \begin{bmatrix} 3 & -5 \\ 1 & 2 \end{bmatrix} \]
A
\[ \begin{bmatrix} x \\ y \end{bmatrix} \]
X
=
\[ \begin{bmatrix} 12 \\ -4 \end{bmatrix} \]
B
Standard Form Only
Before you pack your matrices, every equation must be in Standard Form:
\( ax + by = c \)
Golden Rules
Variables must be in the same order (\(x\) then \(y\)).
Constants must be on the right side.
Use 0 for missing variables.
System Conversion Prep
Convert this 3-variable system into AX = B form:
\( x + y + z = 6 \)
\( 2y + 5z = -4 \)
\( 2x + 5y - z = 27 \)
A = \[ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 5 \\ 2 & 5 & -1 \end{bmatrix} \]
*Note the zero in Row 2 for the missing x-term!
System Translation Worksheet System Translation
Structural Mapping: Systems to AX = B
STUDENT:
DATE:
MISSION: Rewrite each system of linear equations in the matrix form AX = B . Clearly identify the Coefficient Matrix (A) , the Variable Matrix (X) , and the Constant Matrix (B) . Ensure all equations are in standard form before packing!
LEVEL 01: STANDARD 2X2
\( 4x - 3y = 11 \)
COEFFICIENT (A)
VARIABLE (X)
CONSTANT (B)
LEVEL 02: REORDERING REQUIRED
\( 6y + x = 14 \)
Fix Standard Form first!
Standard Form Eq 1:
Standard Form Eq 2:
=
LEVEL 03: 3X3 COMPLEXITY
\( x - 2y + 4z = 10 \)
Use zeroes for missing variables!
A
X
=
B
CHALLENGE: SCENARIO MAPPING
A bakery sells Cookies (c) for $2 and Brownies (b) for $3. Yesterday they sold 50 items total and made $120. Write the AX = B matrix equation for this situation.
Step 1: Write the Equations
Step 2: Matrix Form
Translation Key Teacher Resource Answer Key
Teacher Resource: System Translation
LEVEL 01: STANDARD 2X2
A
\[ \begin{bmatrix} 4 & -3 \\ 2 & 5 \end{bmatrix} \]
X
\[ \begin{bmatrix} x \\ y \end{bmatrix} \]
=
B
\[ \begin{bmatrix} 11 \\ -7 \end{bmatrix} \]
LEVEL 02: REORDERING
Standard Eq 1
\( x + 6y = 14 \)
Standard Eq 2
\( 2x - y = 8 \)
\[ \begin{bmatrix} 1 & 6 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 14 \\ 8 \end{bmatrix} \]
LEVEL 03: 3X3 COMPLEXITY
Note: Zeroes used for missing x in Eq 3 and missing y in Eq 2.
\[ \begin{bmatrix} 1 & -2 & 4 \\ 3 & 0 & 1 \\ 0 & 5 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ -2 \\ 0 \end{bmatrix} \]
BAKERY SCENARIO SOLUTION
Equations
\( c + b = 50 \)
\( 2c + 3b = 120 \)
Matrix Form
\[ \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} c \\ b \end{bmatrix} = \begin{bmatrix} 50 \\ 120 \end{bmatrix} \]
Inverse Power Slides Lesson 1.4
Inverse Power
The final transformation: Solving for the variable matrix in one decisive move.
The Inverse Solution
How do we get X by itself? We multiply both sides by the Inverse of A.
\( AX = B \)
\( A^{-1}(AX) = A^{-1}B \)
\( X = A^{-1}B \)
The Order Matters
DO NOT DO \( B \times A^{-1} \)
Matrix multiplication is picky. Since \(A^{-1}\) is on the left of \(A\) on the left side, it must be on the left of \(B\) on the right side.
01. The Solving Algorithm
1
PACK
Write the system as \(AX=B\)
2
INVERT
Calculate \(A^{-1}\) using the determinant formula
3
MULTIPLY
Calculate \(X = A^{-1}B\)
4
EXTRACT
The values in \(X\) are your solutions!
"No more substitution. No more elimination. Just pure matrix power."
Walkthrough: 2x2 Speed Solve
The System
\( x + 2y = 4 \)
Det(A)
\( 5 - 6 = -1 \)
The Inverse (A⁻¹)
\[ \frac{1}{-1} \begin{bmatrix} 5 & -2 \\ -3 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} -5 & 2 \\ 3 & -1 \end{bmatrix} \]
The Multiply (A⁻¹B)
\[ \begin{bmatrix} -5 & 2 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 \\ 11 \end{bmatrix} \] \[ = \begin{bmatrix} -20 + 22 \\ 12 - 11 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \]
\( x=2, y=1 \)
Why use Matrices?
Repeat Efficiency
If you have to solve the same system with 100 different constant vectors (B), you only find the inverse once.
Scale Up
Calculators and computers use matrix inversion to solve systems with thousands of variables instantly.
Flipped Mission
Search for a "Matrix Inverse Calculator" online. Use it to solve a 5x5 system of equations. How long would that take you by hand?
Matrix Solver Challenge Solver Challenge
Mission: Execute X = A⁻¹B Protocol
STUDENT:
SCORE:
Step 1
Pack AX = B
Step 2
Find A⁻¹
Step 3
X = A⁻¹B
Step 4
Verify!
\( 3x + y = 7 \)
\( 4x + 2y = 10 \)
Beginner Mode
1. Find Inverse (A⁻¹)
2. Calculate X = A⁻¹B
FINAL SOLUTION:
x =
y =
\( x - 2y = -5 \)
\( 3x + 2y = 13 \)
Intermediate
1. Find Inverse (A⁻¹)
2. Calculate X = A⁻¹B
FINAL SOLUTION:
x =
y =
The Data Center Mystery
A server tech knows three variables affect ping: bandwidth (x), load (y), and distance (z). They have the following system, but their solver is broken. Help them solve for x, y, and z .
Expert Mode
\( x + y + z = 10 \)
SHOW WORK: CALCULATE A⁻¹ AND MULTIPLY BY B
Provide full derivation steps here
x =
y =
z =
Solving Guide Teacher Resource Key Answer Key
Teacher Resource: Matrix Solver
PROBLEM 01: BEGINNER MODE
Inverse A⁻¹
det(A) = 3(2) - 1(4) = 2
A⁻¹ = \(\frac{1}{2} \begin{bmatrix} 2 & -1 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -0.5 \\ -2 & 1.5 \end{bmatrix}\)
Solution X = A⁻¹B
\( \begin{bmatrix} 1 & -0.5 \\ -2 & 1.5 \end{bmatrix} \begin{bmatrix} 7 \\ 10 \end{bmatrix} = \begin{bmatrix} 7 - 5 \\ -14 + 15 \end{bmatrix} \)
x = 2, y = 1
PROBLEM 02: INTERMEDIATE
Inverse A⁻¹
det(A) = 1(2) - (-2)(3) = 2 + 6 = 8
A⁻¹ = \(\frac{1}{8} \begin{bmatrix} 2 & 2 \\ -3 & 1 \end{bmatrix}\)
Solution X = A⁻¹B
\( \frac{1}{8} \begin{bmatrix} 2 & 2 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} -5 \\ 13 \end{bmatrix} = \frac{1}{8} \begin{bmatrix} -10 + 26 \\ 15 + 13 \end{bmatrix} \)
\( \frac{1}{8} \begin{bmatrix} 16 \\ 28 \end{bmatrix} \implies \) x = 2, y = 3.5
PROBLEM 03: EXPERT MODE (3X3)
Inverse Process
A = \(\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 2 & 0 & 1 \end{bmatrix}\)
det(A) = 1(1-0) - 1(0 - -2) + 1(0-2)
= 1 - 2 - 2 = -3
A⁻¹ = \(\frac{1}{-3} \begin{bmatrix} 1 & -1 & -2 \\ -2 & -1 & 1 \\ -2 & 2 & 1 \end{bmatrix}\)
Final Solve
X = \( \frac{1}{-3} \begin{bmatrix} 1 & -1 & -2 \\ -2 & -1 & 1 \\ -2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 10 \\ 2 \\ 13 \end{bmatrix} \)
= \( \frac{1}{-3} \begin{bmatrix} 10 - 2 - 26 \\ -20 - 2 + 13 \\ -20 + 4 + 13 \end{bmatrix} \)
= \( \frac{1}{-3} \begin{bmatrix} -18 \\ -9 \\ -3 \end{bmatrix} \implies \) x = 6, y = 3, z = 1
Pedagogical Note:
Encourage students to keep the determinant as a fraction multiplier (scalar) until the final multiplication with vector B is complete. This avoids messy decimals inside the matrix operations and makes the math much cleaner for students.
Codebreaker Slides Hill Cipher Lesson 1.5
Codebreaker
Applying matrix inverses to secure communication: The Hill Cipher.
The Secret Signal
"You have intercepted a secret alien transmission. Use the Key Matrix to decode the message and save the planet."
Plaintext
Key Matrix
Ciphertext
01. The Numeric Map
A
1
B
2
C
3
D
4
E
5
F
6
G
7
H
8
I
9
J
10
K
11
L
12
M
13
N
14
O
15
P
16
Q
17
R
18
S
19
T
20
U
21
V
22
W
23
X
24
Y
25
Z
0
"Every message is split into vectors of size 2. 'MATH' becomes vectors [M, A] and [T, H]."
02. Encryption: Matrix Multiplication
The Rule
Cipher = Key × Plain
To hide the message, multiply your word-vectors by the Encoding Key Matrix . This mixes the letters together!
Example: 'HI' [8, 9]
\[ \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} \] \[ \begin{bmatrix} 8 \\ 9 \end{bmatrix} \] = \[ \begin{bmatrix} 33 \\ 58 \end{bmatrix} \]
Wait, 33 and 58 aren't letters...
The Modulo 26 Loop
Since there are only 26 letters, we use Remainder Division (Modulo 26) to bring numbers back into range.
\( 33 \div 26 \implies \text{Rem } \mathbf{7} \text{ (G)} \)
03. Decryption: The Inverse
The Recovery Move
To get the original message back, multiply the Ciphertext by the Inverse Key Matrix.
Plain = Key⁻¹ × Cipher
!
Hill Cipher Mission Worksheet Hill Cipher Mission
Mission Status: Active Intercept
AGENT:
[ENCRYPT_LVL: 2]
A B C D E F G H I J K L M 1 2 3 4 5 6 7 8 9 10 11 12 13 N O P Q R S T U V W X Y Z 14 15 16 17 18 19 20 21 22 23 24 25 0
Phase 1: Encryption
Encode the word "MATH" using the Key Matrix K . Split "MATH" into two vectors of size 2. Multiply each by K, then apply Modulo 26.
KEY MATRIX (K)
\[ \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} \]
"MA" VECTOR
\[ \begin{bmatrix} 13 \\ 1 \end{bmatrix} \]
ENCRYPTED "MA"
KEY MATRIX (K)
\[ \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} \]
"TH" VECTOR
\[ \begin{bmatrix} 20 \\ 8 \end{bmatrix} \]
ENCRYPTED "TH"
Phase 2: Decryption
We intercepted a ciphertext: [14, 25] . Use the Inverse Key Matrix (K⁻¹) to decode it and find the 2-letter word.
1. Derive Inverse Matrix (K⁻¹)
det(K) = ...
2. Multiply K⁻¹ × [14, 25]
Result Mod 26 = ...
THE SECRET WORD IS:
SYSTEM SECURITY LOG
Why is it critical that det(K) ≠ 0 in cryptography? If we used a singular matrix (det = 0) as a key, would the encryption still work? Would the decryption still work? Explain using the concept of matrix inversion.
(Write your explanation here)
Decryption Key Teacher Resource Key Mission Answer Key
Teacher Resource: Hill Cipher Decryption
Phase 1: Encryption of "MATH"
Part A: "MA" [13, 1]
\[ \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} \begin{bmatrix} 13 \\ 1 \end{bmatrix} = \begin{bmatrix} 39+1 \\ 65+2 \end{bmatrix} = \begin{bmatrix} 40 \\ 67 \end{bmatrix} \]
40 mod 26 = 14 (N)
67 mod 26 = 15 (O)
Result: "NO"
Part B: "TH" [20, 8]
\[ \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} \begin{bmatrix} 20 \\ 8 \end{bmatrix} = \begin{bmatrix} 60+8 \\ 100+16 \end{bmatrix} = \begin{bmatrix} 68 \\ 116 \end{bmatrix} \]
68 mod 26 = 16 (P)
116 mod 26 = 12 (L)
Result: "PL"
ENCRYPTED WORD: NOPL
Phase 2: Decryption of [14, 25]
1. Derive K⁻¹
det(K) = (3x2) - (1x5) = 1
Note: We convert -1 and -5 to positive equivalents in Mod 26: -1 ≡ 25, -5 ≡ 21.
2. Multiply K⁻¹ × Cipher
\[ \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix} \begin{bmatrix} 14 \\ 25 \end{bmatrix} = \begin{bmatrix} 28 - 25 \\ -70 + 75 \end{bmatrix} = \begin{bmatrix} 3 \\ 5 \end{bmatrix} \]
3 = C
5 = E
THE SECRET WORD IS: CE (meaning "Communication Established")
CRITICAL ANALYSIS KEY
Teacher Answer: If the Key Matrix is singular (det = 0), the encryption still "works" in that it produces a ciphertext. However, the decryption is impossible using the inverse method because the inverse matrix does not exist (division by zero error). From a structural standpoint, a singular matrix collapses information into a lower dimension (e.g., multiple different plaintexts might result in the same ciphertext), meaning the original message cannot be uniquely recovered.
