Save or print this lesson:

Lesson 3: Inverse Matrices

Previous Lesson Module 8, Lesson 3 of 4 Next Lesson

Learning Objectives

Understand the concept of an inverse matrix and its properties
Determine when a matrix is invertible using the determinant
Find the inverse of 2x2 matrices using the formula method
Find the inverse of larger matrices using row operations (Gauss-Jordan method)
Verify that two matrices are inverses of each other
Apply properties of inverse matrices
Use inverse matrices to solve systems of equations
Apply inverse matrices to real-world problems including cryptography

Introduction: What is an Inverse Matrix?

Inverse Matrix: For a square matrix A, if there exists a matrix A^(-1) (read "A inverse") such that:

A · A^(-1) = I and A^(-1) · A = I

where I is the identity matrix, then A^(-1) is called the inverse of A.

Just like multiplying a number by its reciprocal gives 1 (for example, 5 × 1/5 = 1), multiplying a matrix by its inverse gives the identity matrix.

Key Facts about Inverse Matrices:

Only square matrices can have inverses (same number of rows and columns)
Not all square matrices have inverses; those that do are called invertible or nonsingular
Matrices without inverses are called singular
If an inverse exists, it is unique (there's only one inverse for any given matrix)
The inverse of A is denoted A^(-1)

Identity Matrix: A square matrix with 1s on the main diagonal and 0s elsewhere.

For 2×2: I = [1 0; 0 1]
For 3×3: I = [1 0 0; 0 1 0; 0 0 1]

The identity matrix acts like the number 1 in matrix multiplication: for any matrix A, A · I = I · A = A.

Section 1: Determining Invertibility with Determinants

Before finding an inverse, we must determine if one exists. The determinant provides a quick test.

Invertibility Test:

A square matrix A is invertible if and only if det(A) ≠ 0

Recall: Determinant of a 2×2 Matrix

For A = [a b; c d], det(A) = ad - bc

Example 1: Testing Invertibility

Determine if A = [3 5; 2 4] is invertible.

Solution:

Calculate the determinant:

det(A) = (3)(4) - (5)(2)
det(A) = 12 - 10
det(A) = 2

Since det(A) = 2 ≠ 0, the matrix is invertible.

Answer: Yes, A is invertible because det(A) = 2 ≠ 0.

Example 2: Singular Matrix

Determine if B = [6 4; 3 2] is invertible.

Solution:

Calculate the determinant:

det(B) = (6)(2) - (4)(3)
det(B) = 12 - 12
det(B) = 0

Since det(B) = 0, the matrix does NOT have an inverse.

Answer: No, B is singular (not invertible) because det(B) = 0.

Example 3: Invertibility with Variables

For what value(s) of k is the matrix A = [k 2; 3 k] NOT invertible?

Solution:

The matrix is not invertible when det(A) = 0:

det(A) = k · k - 2 · 3
k² - 6 = 0
k² = 6
k = ±√6

Answer: The matrix is not invertible when k = √6 or k = -√6.

Section 2: Finding Inverse of 2×2 Matrices (Formula Method)

For 2×2 matrices, there's a straightforward formula for finding the inverse.

Inverse of a 2×2 Matrix:

If A = [a b; c d] and det(A) = ad - bc ≠ 0, then:

A^(-1) = (1/(ad - bc)) · [d -b; -c a]

Steps to Find Inverse of 2×2 Matrix:

Calculate the determinant det(A) = ad - bc
Verify det(A) ≠ 0 (if zero, no inverse exists)
Swap positions of a and d
Change signs of b and c (make them negative)
Multiply the entire matrix by 1/det(A)

Example 4: Basic Inverse Calculation

Find the inverse of A = [4 7; 2 6].

Solution:

Step 1: Calculate the determinant:

det(A) = (4)(6) - (7)(2) = 24 - 14 = 10

Step 2: Since det(A) = 10 ≠ 0, the inverse exists.

Step 3: Apply the formula:

A^(-1) = (1/10) · [6 -7; -2 4]

Step 4: Simplify:

A^(-1) = [6/10 -7/10; -2/10 4/10] = [3/5 -7/10; -1/5 2/5]

Answer: A^(-1) = [3/5 -7/10; -1/5 2/5] or [0.6 -0.7; -0.2 0.4]

Example 5: Inverse with Negative Entries

Find the inverse of B = [5 -3; -2 1].

Solution:

Step 1: Calculate the determinant:

det(B) = (5)(1) - (-3)(-2) = 5 - 6 = -1

Step 2: Since det(B) = -1 ≠ 0, the inverse exists.

Step 3: Apply the formula:

B^(-1) = (1/(-1)) · [1 -(-3); -(-2) 5]
B^(-1) = -1 · [1 3; 2 5]
B^(-1) = [-1 -3; -2 -5]

Answer: B^(-1) = [-1 -3; -2 -5]

Example 6: Inverse of Identity Matrix

Find the inverse of I = [1 0; 0 1].

Solution:

Calculate the determinant:

det(I) = (1)(1) - (0)(0) = 1

Apply the formula:

I^(-1) = (1/1) · [1 0; 0 1] = [1 0; 0 1] = I

Answer: I^(-1) = I. The identity matrix is its own inverse.

Example 7: Inverse with Fractions

Find the inverse of C = [1/2 1/4; 1 1/2].

Solution:

Step 1: Calculate the determinant:

det(C) = (1/2)(1/2) - (1/4)(1) = 1/4 - 1/4 = 0

Since det(C) = 0, this matrix has NO inverse.

Answer: C is singular; C^(-1) does not exist.

Example 8: Finding Inverse with Parameters

Find the inverse of A = [2 1; 3 k] in terms of k (assume the inverse exists).

Solution:

Step 1: Calculate the determinant:

det(A) = 2k - 3

Step 2: For the inverse to exist, 2k - 3 ≠ 0, so k ≠ 3/2.

Step 3: Apply the formula:

A^(-1) = 1/(2k-3) · [k -1; -3 2]

Answer: A^(-1) = [k/(2k-3) -1/(2k-3); -3/(2k-3) 2/(2k-3)], where k ≠ 3/2

Section 3: Verifying Inverse Matrices

To verify that two matrices are inverses, we multiply them in both orders and check if we get the identity matrix.

Verification Process:

Matrices A and B are inverses if and only if:

A · B = I AND B · A = I

You must check BOTH products!

Example 9: Verifying Inverses

Verify that A = [2 5; 1 3] and B = [3 -5; -1 2] are inverses.

Solution:

Check A · B:

A · B = [2 5; 1 3] · [3 -5; -1 2]
= [(2)(3)+(5)(-1) (2)(-5)+(5)(2); (1)(3)+(3)(-1) (1)(-5)+(3)(2)]
= [6-5 -10+10; 3-3 -5+6]
= [1 0; 0 1] = I

Check B · A:

B · A = [3 -5; -1 2] · [2 5; 1 3]
= [(3)(2)+(-5)(1) (3)(5)+(-5)(3); (-1)(2)+(2)(1) (-1)(5)+(2)(3)]
= [6-5 15-15; -2+2 -5+6]
= [1 0; 0 1] = I

Answer: Yes, A and B are inverses because both A·B = I and B·A = I.

Example 10: Not Inverses

Are A = [1 2; 3 4] and B = [4 -2; -3 1] inverses?

Solution:

Check A · B:

A · B = [1 2; 3 4] · [4 -2; -3 1]
= [(1)(4)+(2)(-3) (1)(-2)+(2)(1); (3)(4)+(4)(-3) (3)(-2)+(4)(1)]
= [4-6 -2+2; 12-12 -6+4]
= [-2 0; 0 -2]

This is NOT the identity matrix.

Answer: No, A and B are NOT inverses because A·B ≠ I.

Example 11: Self-Checking an Inverse

You found that the inverse of A = [3 1; 5 2] is A^(-1) = [2 -1; -5 3]. Verify your answer.

Solution:

Calculate A · A^(-1):

[3 1; 5 2] · [2 -1; -5 3]
= [(3)(2)+(1)(-5) (3)(-1)+(1)(3); (5)(2)+(2)(-5) (5)(-1)+(2)(3)]
= [6-5 -3+3; 10-10 -5+6]
= [1 0; 0 1] = I

Answer: Verified! The inverse is correct.

Section 4: Row Operations Method (Gauss-Jordan Elimination)

For matrices larger than 2×2, or when the formula is cumbersome, we use the Gauss-Jordan method.

Gauss-Jordan Method for Finding Inverses:

Write the augmented matrix [A | I], where I is the identity matrix
Use row operations to transform the left side to I: [I | B]
The right side B is the inverse A^(-1)
If you cannot transform A to I, then A has no inverse

Allowed Row Operations:

Multiply a row by a nonzero constant
Add a multiple of one row to another row
Swap two rows

Example 12: Gauss-Jordan Method for 2×2

Find the inverse of A = [1 2; 3 7] using row operations.

Solution:

Step 1: Set up the augmented matrix [A | I]:

[1 2 | 1 0]
[3 7 | 0 1]

Step 2: Eliminate below the first pivot (make column 1 entry in row 2 equal to 0):

R2 - 3R1 → R2:
[1 2 | 1 0]
[0 1 | -3 1]

Step 3: Eliminate above the second pivot (make column 2 entry in row 1 equal to 0):

R1 - 2R2 → R1:
[1 0 | 7 -2]
[0 1 | -3 1]

We now have [I | A^(-1)].

Answer: A^(-1) = [7 -2; -3 1]

Example 13: Gauss-Jordan with Scaling

Find the inverse of B = [2 1; 4 3] using row operations.

Solution:

Step 1: Set up [B | I]:

[2 1 | 1 0]
[4 3 | 0 1]

Step 2: Make the first pivot equal to 1 by scaling R1:

(1/2)R1 → R1:
[1 1/2 | 1/2 0]
[4 3 | 0 1]

Step 3: Eliminate below the pivot:

R2 - 4R1 → R2:
[1 1/2 | 1/2 0]
[0 1 | -2 1]

Step 4: Eliminate above the second pivot:

R1 - (1/2)R2 → R1:
[1 0 | 3/2 -1/2]
[0 1 | -2 1]

Answer: B^(-1) = [3/2 -1/2; -2 1]

Example 14: 3×3 Matrix Inverse

Find the inverse of A = [1 0 1; 0 2 1; 1 1 1] using row operations.

Solution:

Step 1: Set up [A | I]:

[1 0 1 | 1 0 0]
[0 2 1 | 0 1 0]
[1 1 1 | 0 0 1]

Step 2: Eliminate first column below pivot:

R3 - R1 → R3:
[1 0 1 | 1 0 0]
[0 2 1 | 0 1 0]
[0 1 0 | -1 0 1]

Step 3: Make second pivot equal to 1:

(1/2)R2 → R2:
[1 0 1 | 1 0 0]
[0 1 1/2 | 0 1/2 0]
[0 1 0 | -1 0 1]

Step 4: Eliminate second column below pivot:

R3 - R2 → R3:
[1 0 1 | 1 0 0]
[0 1 1/2 | 0 1/2 0]
[0 0 -1/2 | -1 -1/2 1]

Step 5: Make third pivot equal to 1:

-2R3 → R3:
[1 0 1 | 1 0 0]
[0 1 1/2 | 0 1/2 0]
[0 0 1 | 2 1 -2]

Step 6: Back-substitute to clear column 3:

R2 - (1/2)R3 → R2 and R1 - R3 → R1:
[1 0 0 | -1 -1 2]
[0 1 0 | -1 0 1]
[0 0 1 | 2 1 -2]

Answer: A^(-1) = [-1 -1 2; -1 0 1; 2 1 -2]

Example 15: Detecting a Singular Matrix with Row Operations

Try to find the inverse of C = [1 2; 2 4] using row operations.

Solution:

Step 1: Set up [C | I]:

[1 2 | 1 0]
[2 4 | 0 1]

Step 2: Try to eliminate below the pivot:

R2 - 2R1 → R2:
[1 2 | 1 0]
[0 0 | -2 1]

We get a row of zeros on the left side. This means we cannot transform C to I.

Answer: C is singular; C^(-1) does not exist. (Note: det(C) = 4 - 4 = 0 confirms this.)

Section 5: Properties of Inverse Matrices

Inverse matrices have several important properties that are useful for computations and proofs.

Key Properties of Inverses:

Inverse of an Inverse: (A^(-1))^(-1) = A
Inverse of a Product: (AB)^(-1) = B^(-1)A^(-1) (reverse order!)
Inverse of a Transpose: (A^T)^(-1) = (A^(-1))^T
Inverse of a Scalar Multiple: (kA)^(-1) = (1/k)A^(-1) for k ≠ 0
Determinant of an Inverse: det(A^(-1)) = 1/det(A)
Uniqueness: If B and C are both inverses of A, then B = C

Example 16: Using the Product Rule

If A = [2 1; 1 1] and B = [3 -1; -1 1], find (AB)^(-1).

Solution:

Method 1 (Using the property):

First find A^(-1) and B^(-1), then use (AB)^(-1) = B^(-1)A^(-1).

det(A) = 2(1) - 1(1) = 1
A^(-1) = [1 -1; -1 2]

det(B) = 3(1) - (-1)(-1) = 2
B^(-1) = (1/2)[1 1; 1 3] = [1/2 1/2; 1/2 3/2]

Now compute B^(-1)A^(-1):

(AB)^(-1) = [1/2 1/2; 1/2 3/2] · [1 -1; -1 2]
= [0 1/2; -1 2]

Answer: (AB)^(-1) = [0 1/2; -1 2]

Example 17: Verifying the Inverse of an Inverse

If A^(-1) = [3 1; 2 1], what is (A^(-1))^(-1)?

Solution:

By the property (A^(-1))^(-1) = A, we need to find the original matrix A.

Since A^(-1) = [3 1; 2 1], we find the inverse of this matrix:

det(A^(-1)) = 3(1) - 1(2) = 1
(A^(-1))^(-1) = [1 -1; -2 3]

Answer: (A^(-1))^(-1) = [1 -1; -2 3], which is the original matrix A.

Example 18: Scalar Multiple Property

If A = [2 1; 1 1] and A^(-1) = [1 -1; -1 2], find (3A)^(-1).

Solution:

Use the property (kA)^(-1) = (1/k)A^(-1):

(3A)^(-1) = (1/3)A^(-1)
(3A)^(-1) = (1/3)[1 -1; -1 2]
(3A)^(-1) = [1/3 -1/3; -1/3 2/3]

Answer: (3A)^(-1) = [1/3 -1/3; -1/3 2/3]

Example 19: Determinant Property

If det(A) = 5, find det(A^(-1)).

Solution:

Use the property det(A^(-1)) = 1/det(A):

det(A^(-1)) = 1/5

Answer: det(A^(-1)) = 1/5 = 0.2

Section 6: Applications of Inverse Matrices

Inverse matrices are powerful tools for solving systems of equations, transformations, and encryption.

Solving Matrix Equations:

If AX = B and A is invertible, then:

X = A^(-1)B

This allows us to solve systems of linear equations efficiently.

Example 20: Solving a System with Inverse Matrices

Solve the system: 2x + y = 5 and x + y = 3 using inverse matrices.

Solution:

Step 1: Write in matrix form AX = B:

[2 1; 1 1] · [x; y] = [5; 3]

Step 2: Find A^(-1):

det(A) = 2(1) - 1(1) = 1
A^(-1) = [1 -1; -1 2]

Step 3: Multiply both sides by A^(-1):

X = A^(-1)B = [1 -1; -1 2] · [5; 3]
X = [(1)(5)+(-1)(3); (-1)(5)+(2)(3)]
X = [2; 1]

Answer: x = 2, y = 1

Example 21: Cryptography - Encoding a Message

Use the encoding matrix E = [1 2; 1 3] to encrypt the message "GO" (G=7, O=15 in numerical code).

Solution:

Represent the message as a column vector:

M = [7; 15]

Encode by multiplying C = EM:

C = [1 2; 1 3] · [7; 15]
C = [(1)(7)+(2)(15); (1)(7)+(3)(15)]
C = [37; 52]

Answer: The encrypted message is [37; 52].

Example 22: Cryptography - Decoding a Message

Decrypt the message [37; 52] using the encoding matrix E = [1 2; 1 3].

Solution:

Step 1: Find the decoding matrix E^(-1):

det(E) = 1(3) - 2(1) = 1
E^(-1) = [3 -2; -1 1]

Step 2: Decode by multiplying M = E^(-1)C:

M = [3 -2; -1 1] · [37; 52]
M = [(3)(37)+(-2)(52); (-1)(37)+(1)(52)]
M = [111-104; -37+52]
M = [7; 15]

Converting back: 7 = G, 15 = O

Answer: The decrypted message is "GO".

Example 23: Geometric Transformations

A transformation matrix T = [0 -1; 1 0] rotates points 90° counterclockwise about the origin. What transformation reverses this rotation?

Solution:

The inverse transformation T^(-1) reverses the rotation.

Find T^(-1):

det(T) = 0(0) - (-1)(1) = 1
T^(-1) = [0 1; -1 0]

This matrix rotates points 90° clockwise (or equivalently, 270° counterclockwise).

Answer: T^(-1) = [0 1; -1 0] reverses the rotation.

Example 24: Business Application - Input-Output Models

An economy has two sectors. The technology matrix is A = [0.2 0.3; 0.4 0.1]. If final demand is D = [100; 200], find total production X using (I - A)^(-1)D.

Solution:

Step 1: Calculate I - A:

I - A = [1 0; 0 1] - [0.2 0.3; 0.4 0.1]
I - A = [0.8 -0.3; -0.4 0.9]

Step 2: Find (I - A)^(-1):

det(I-A) = (0.8)(0.9) - (-0.3)(-0.4) = 0.72 - 0.12 = 0.6
(I-A)^(-1) = (1/0.6)[0.9 0.3; 0.4 0.8]
(I-A)^(-1) = [1.5 0.5; 2/3 4/3]

Step 3: Calculate X:

X = (I-A)^(-1)D = [1.5 0.5; 2/3 4/3] · [100; 200]
X = [150+100; 66.67+266.67]
X = [250; 333.33]

Answer: Total production needed: Sector 1 = 250 units, Sector 2 ≈ 333 units.

Example 25: Network Flow Problem

In a network, traffic flows are related by the equation [2 1; 1 2][f1; f2] = [300; 400], where f1 and f2 are flows. Find the flows.

Solution:

Step 1: Find the inverse of [2 1; 1 2]:

det = 2(2) - 1(1) = 3
A^(-1) = (1/3)[2 -1; -1 2]

Step 2: Solve for [f1; f2]:

[f1; f2] = (1/3)[2 -1; -1 2] · [300; 400]
[f1; f2] = (1/3)[600-400; -300+800]
[f1; f2] = (1/3)[200; 500]
[f1; f2] = [66.67; 166.67]

Answer: f1 ≈ 66.67 units, f2 ≈ 166.67 units

Check Your Understanding

1. Is the matrix A = [4 2; 6 3] invertible? Why or why not?

Answer: No, A is not invertible.

det(A) = 4(3) - 2(6) = 12 - 12 = 0. Since the determinant is zero, A is singular.

2. Find the inverse of B = [3 5; 1 2] using the formula method.

Answer: B^(-1) = [2 -5; -1 3]

det(B) = 3(2) - 5(1) = 1, so B^(-1) = [2 -5; -1 3]

3. Verify that A = [1 3; 0 1] and A^(-1) = [1 -3; 0 1] are inverses.

Answer: Verified.

A·A^(-1) = [1 3; 0 1]·[1 -3; 0 1] = [1 0; 0 1] = I

4. If A^(-1) = [2 1; 3 2], what is det(A)?

Answer: det(A) = 1

det(A^(-1)) = 2(2) - 1(3) = 1, so det(A) = 1/det(A^(-1)) = 1/1 = 1

5. Find the inverse of C = [1 0; 2 1] using row operations.

Answer: C^(-1) = [1 0; -2 1]

[1 0|1 0; 2 1|0 1] → R2-2R1→R2 → [1 0|1 0; 0 1|-2 1]

6. If A and B are invertible matrices, what is (AB)^(-1)?

Answer: (AB)^(-1) = B^(-1)A^(-1)

The inverse of a product is the product of inverses in reverse order.

7. Find the inverse of D = [5 3; 3 2].

Answer: D^(-1) = [2 -3; -3 5]

det(D) = 5(2) - 3(3) = 1, so D^(-1) = [2 -3; -3 5]

8. If (2A)^(-1) = [1 2; 3 4], find A^(-1).

Answer: A^(-1) = [2 4; 6 8]

Since (2A)^(-1) = (1/2)A^(-1), we have A^(-1) = 2·(2A)^(-1) = 2[1 2; 3 4] = [2 4; 6 8]

9. Solve the system using inverses: 3x + 2y = 7 and 2x + y = 4.

Answer: x = 1, y = 2

A = [3 2; 2 1], det(A) = -1, A^(-1) = [-1 2; 2 -3]
X = A^(-1)B = [-1 2; 2 -3]·[7; 4] = [1; 2]

10. For what value of k does the matrix [k 4; 2 k] NOT have an inverse?

Answer: k = ±2

det = k² - 8 = 0, so k² = 8, giving k = ±2√2. Wait, let me recalculate: det = k·k - 4·2 = k² - 8. For no inverse, k² = 8, so k = ±2√2 ≈ ±2.83

Key Takeaways

An inverse matrix A^(-1) satisfies A·A^(-1) = I and A^(-1)·A = I, where I is the identity matrix
Only square matrices can have inverses, and not all square matrices are invertible
A matrix is invertible if and only if det(A) ≠ 0; if det(A) = 0, the matrix is singular
For 2×2 matrices: if A = [a b; c d], then A^(-1) = (1/(ad-bc))·[d -b; -c a]
To verify inverses, check that both A·B = I and B·A = I (must check both orders)
The Gauss-Jordan method finds inverses by row-reducing [A | I] to [I | A^(-1)]
Important properties: (AB)^(-1) = B^(-1)A^(-1) (reverse order!) and (A^(-1))^(-1) = A
The identity matrix I is its own inverse: I^(-1) = I
Inverse matrices solve matrix equations: if AX = B, then X = A^(-1)B
Applications include solving systems, cryptography, geometric transformations, and economic models
det(A^(-1)) = 1/det(A) for any invertible matrix A
If a row of all zeros appears during Gauss-Jordan elimination, the matrix is not invertible

Previous: Matrix Operations Back to Module 8 Next: Matrix Equations