Matrix And Determinant - Problem On Determinant

Slide 1

The topic for today’s lecture is “Matrix and Determinant - Problem on determinant”
Matrices and determinants are important topics in algebra and have various applications in mathematics and other fields
In this lecture, we will focus on solving problems related to determinants

Slide 2

Determinants are mathematical objects that are associated with square matrices
A determinant of a 2x2 matrix is calculated by multiplying the elements in the main diagonal and subtracting the product of the elements in the off-diagonal
For a matrix A = [[a, b], [c, d]], the determinant is given by |A| = ad - bc

Slide 3

The determinant of a 3x3 matrix can be calculated using a formula known as the “Sarrus’ rule”
Let A be a 3x3 matrix: A = [[a, b, c], [d, e, f], [g, h, i]]
The determinant of A is given by |A| = aei + bfg + cdh - ceg - afh - bdi

Slide 4

Determinants play a crucial role in solving systems of linear equations
Given a system of equations represented in matrix form as AX = B, we can determine whether the system has a unique solution, no solution, or infinitely many solutions by examining the determinant of matrix A

Slide 5

If the determinant of matrix A is non-zero (|A| != 0), then the system of equations has a unique solution
If the determinant of matrix A is zero (|A| = 0), then either the system has no solution or infinitely many solutions
We can further determine the nature of the solutions by analyzing the consistency of the system

Slide 6

Let’s solve a problem to illustrate the application of determinants in solving systems of equations:
Consider the system of equations:
- 3x + 2y - z = 1
- 2x - 2y + 4z = -2
- x + 3y - 3z = 4

Slide 7

Representing the system in matrix form, we have:
A = [[3, 2, -1], [2, -2, 4], [1, 3, -3]]
X = [[x], [y], [z]]
B = [[1], [-2], [4]]

Slide 8

To determine whether the system has a unique solution, we can calculate the determinant of matrix A
Applying the Sarrus’ rule, we have:
|A| = 3*(-2)(-3) + 241 + (-1)23 - (-1)(-2)(-1) - 343 - 22*(-3)

Slide 9

Simplifying the determinant expression, we find:
|A| = -18 + 8 - 6 + 2 + 36 - 12 = 10

Slide 10

Since the determinant of matrix A is non-zero (|A| = 10), the system of equations has a unique solution
We can proceed to solve the system using various methods such as Gaussian elimination or matrix inversion

Slide 11

Let’s continue solving the system of equations:
Now that we know the system has a unique solution, we can use the method of matrix inversion to find the solution
The solution will be given by X = A^-1 * B, where A^-1 is the inverse of matrix A

Slide 12

To find the inverse of a 3x3 matrix, we can use the formula:
A^-1 = (1/|A|) * adj(A), where adj(A) is the adjugate of matrix A
The adjugate of a matrix A is obtained by finding the transpose of the cofactor matrix of A

Slide 13

Let’s calculate the adjugate of matrix A:
For each element of A, calculate its cofactor by taking the determinant of the minor matrix obtained by removing the row and column containing that element
Then, take the transpose of the resulting cofactor matrix

Slide 14

Using the formula for adjugate, we find the adj(A) as:
adj(A) = [[-3, 7, -1], [13, -4, 2], [-5, -3, -1]]

Slide 15

Now, we can find the inverse of A by multiplying adj(A) by (1/|A|), where |A| = 10
(1/|A|) * adj(A) = (1/10) * [[-3, 7, -1], [13, -4, 2], [-5, -3, -1]]

Slide 16

Simplifying the expression, we get the inverse of A as:
A^-1 = [[-0.3, 0.7, -0.1], [1.3, -0.4, 0.2], [-0.5, -0.3, -0.1]]

Slide 17

Now that we have the inverse of matrix A, we can find the solution by multiplying A^-1 with B
X = A^-1 * B = [[-0.3, 0.7, -0.1], [1.3, -0.4, 0.2], [-0.5, -0.3, -0.1]] * [[1], [-2], [4]]

Slide 18

By performing the matrix multiplication, we get the solution as:
X = [[1], [-2], [3]]

Slide 19

Therefore, the system of equations:
- 3x + 2y - z = 1
- 2x - 2y + 4z = -2
- x + 3y - 3z = 4
Has a unique solution: x = 1, y = -2, z = 3

Slide 20

Determinants are important tools in linear algebra and have various applications in mathematics, physics, and engineering
They can also be used to solve problems involving areas and volumes, transformations, and eigenvalues
It is crucial to understand the concepts and techniques related to determinants and matrices to excel in higher-level mathematics courses and applications.

Slide 21

Let’s solve another problem involving determinants:
Consider the system of equations:
- x + y + z = 6
- 2x + 3y + 2z = 10
- 3x + 4y + 3z = 14

Slide 22

Representing the system in matrix form, we have:
A = [[1, 1, 1], [2, 3, 2], [3, 4, 3]]
X = [[x], [y], [z]]
B = [[6], [10], [14]]

Slide 23

Let’s calculate the determinant of matrix A to determine the nature of the solution
|A| = 1 * (33 - 24) - 1 * (23 - 23) + 1 * (24 - 23)

Slide 24

Simplifying the determinant expression, we find:
|A| = 1 * (9 - 8) - 1 * (6 - 6) + 1 * (8 - 6) = 1

Slide 25

Since the determinant of matrix A is non-zero (|A| = 1), the system of equations has a unique solution
We can proceed to solve the system using matrix inversion or any other appropriate method

Slide 26

To find the inverse of matrix A, we can use the formula:
A^-1 = (1/|A|) * adj(A)

Slide 27

Using the formula for adjugate, we find the adj(A) as:
adj(A) = [[3, -1, 1], [-2, 2, -1], [-1, 1, -1]]

Slide 28

Now, we can find the inverse of A by multiplying adj(A) by (1/|A|), where |A| = 1
(1/|A|) * adj(A) = 1 * [[3, -1, 1], [-2, 2, -1], [-1, 1, -1]]

Slide 29

Simplifying the expression, we get the inverse of A as:
A^-1 = [[3, -1, 1], [-2, 2, -1], [-1, 1, -1]]

Slide 30

Now that we have the inverse of matrix A, we can find the solution by multiplying A^-1 with B
X = A^-1 * B = [[3, -1, 1], [-2, 2, -1], [-1, 1, -1]] * [[6], [10], [14]]
By performing the matrix multiplication, we get the solution as:
X = [[1], [2], [3]]
Therefore, the system of equations:
- x + y + z = 6
- 2x + 3y + 2z = 10
- 3x + 4y + 3z = 14
Has a unique solution: x = 1, y = 2, z = 3