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Matrices: Definition, Operations, Special Matrices, Determinants, and Inverses, Slides of Linear Algebra

An introduction to matrices, including their definition, operations such as addition, scalar multiplication, and matrix multiplication, special matrices like square, transpose, symmetric, diagonal, and identity matrices, determinants, and inverses.

What you will learn

  • What is scalar multiplication of a matrix?
  • How do you find the determinant of a 2x2 matrix?
  • What is a matrix?
  • How do you find the inverse of a 2x2 matrix?
  • How do you add two matrices?

Typology: Slides

2021/2022

Uploaded on 09/27/2022

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7 Matrices
7.1 Introduction
Definition. Am×nmatrix is a rectangular array of numbers having m
rows and ncolumns.
A=
a11 a12 . . . a1n
a21 a22 . . . a2n
a31 a32 . . . a3n
. . . .
. . . .
am1am2. . . amn
The element aij represents the entry in the ith row and jth column. We
sometimes denote Aby (aij)m×n.
7.2 Matrix Operations
Addition
We can only add two matrices of the same dimension i.e. same number of
rows and columns. We then add element-wise.
Example.
2 1
1 3 +4 3
5 6 =6 4
6 9 .
Scalar Multiplication
If cis a real number and A= (aij)m×nis a matrix then cA = (caij )m×n.
Example.
52 1
1 3 =10 5
5 15 .
Matrix Multiplication
Given a matrix A= (aij)m×nand a matrix B= (bij )r×s, we can only multiply
them if n=r. In such a case the multiplication is defined to be the matrix
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7 Matrices

7.1 Introduction

Definition. A m × n matrix is a rectangular array of numbers having m rows and n columns.

A =

a 11 a 12... a 1 n a 21 a 22... a 2 n a 31 a 32... a 3 n

.... .... am 1 am 2... amn

The element aij represents the entry in the ith row and jth column. We sometimes denote A by (aij )m×n.

7.2 Matrix Operations

Addition

We can only add two matrices of the same dimension i.e. same number of rows and columns. We then add element-wise.

Example. (^) ( 2 1 1 3

Scalar Multiplication

If c is a real number and A = (aij )m×n is a matrix then cA = (caij )m×n.

Example.

5

Matrix Multiplication

Given a matrix A = (aij )m×n and a matrix B = (bij )r×s, we can only multiply them if n = r. In such a case the multiplication is defined to be the matrix

C = (cij )m×s as follows:

cij =

∑^ n

k=

aikbkj.

We may view the cij th element as the dot product of the ith row of the matrix A and jth column of the matrix B.

Example. 1. ( 2 1 1 3

7.4 Determinants

Definition. The determinant of a 2 × 2 matrix is defined as follows: ∣ ∣ ∣ ∣

a b c d

∣ =^ ad^ −^ bc.

Definition. The determinant of a 3 × 3 matrix is defined as follows:

∣ ∣ ∣ ∣ ∣ ∣ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33

= a 11

a 22 a 23 a 32 a 33

∣ −^ a^12

a 21 a 23 a 31 a 33

∣ +^ a^13

a 21 a 22 a 31 a 33

= a 11 (a 22 a 33 − a 32 a 23 ) − a 12 (a 21 a 33 − a 31 a 23 ) + a 13 (a 21 a 32 − a 31 a 22 )

Definition. The definition of determinants can be generalized for a square matrix of any size. Let A = (aij ) be a square n × n matrix. Fix a row i. Then the determinant is defined to be,

det (A) =

∑^ n

j=

(−1)i+j^ aij det (Aij ),

where Aij is the minor matrix obtained by deleting row i and column j. It is called the ij Cof actor of A. T

Example. Find the determinant of

A =

Choose row i = 1.

A 11 =

det (A 11 ) = − 2

A 12 =

det (A 11 ) = − 6

A 13 =

det (A 11 ) = − 2

A 14 =

det (A 11 ) = − 2

det (A) =

∑^ n

j=

(−1)1+j^ a 1 j det (A 1 j )

= (−1)^2 (1)det (A 11 ) + (−1)^3 (1)det (A 12 ) + (−1)^4 (0)det (A 13 ) + (−1)^5 (−1)det (A 14 ) = (1)(1)(−2) + (−1)(1)(−6) + (1)(0)(−2) + (−1)(−1)(−2) = 2

We can find the inverse in another way: Write as (A|I 2 ) ( 1 1 1 0 1 2 0 1

Then use matrix row operations to get it into the form (I 2 |B). ( 1 0 a b 0 1 c d

You will find that B = A−^1. ( 1 1 1 0 1 2 0 1

R 2 =R 2 −R 1 →

R 1 =R 1 −R 2 →

Observe that the matrix obtained ( a b c d

is the inverse of A.

Inverse of a 3 × 3 matrix

The method is the same as in the 2 × 2 case:

  • Check determinant A is non-zero.
  • Rewrite as (A|I 3 ).
  • Use row operations to put in the form (I 3 |A−^1 ).

Example. Let A =

. Find A−^1.

  1. Check that det(A) 6 = 0.

det(A) = 1(0 − (−3)(1)) − (−1)(0 − (−2)(1)) + 1(0 − (−2)(−2)) = 3 + 2 − 4 = 1 6 = 0

Therefore the inverse exists.

  1. Rewrite as (A|I 3 ). (^) 

  1. Use row operations to put in the form (I 3 |A−^1 ).

 

 R^3 =R →^3 +2R^1

R 2 = − 21 R 2 →

R 3 =R 3 +5R 2 →

R 3 =− 2 R 3 →

R 2 =R 2 + 12 R 3 →

R 1 =R 1 −R 3 →

R 1 =R 1 +R 2 →

A−^1 =

Steps to find eigenvalues and eigenvectors:

  1. Form the characteristic equation

det(λI − A) = 0.

  1. To find all the eigenvalues of A, solve the characteristic equation.
  2. For each eigenvalue λ, to find the corresponding set of eigenvectors, solve the linear system of equations

(λI − A)~x = 0

Example. 1. A =

. Find the eigenvalues and eigenvectors of A.

Solution: • We will find the characteristic equation of A. We will first find the matrix λI − A.

λI − A = λ

λ 0 0 λ

λ − 4 2 1 λ − 3

The determinant of this matrix gives us the characteristic polyno- mial.

det(λI − A) = 0 ∣ ∣ ∣ ∣

λ − 4 2 1 λ − 3

∣ =^0

(λ − 4)(λ − 3) − 2 = 0 λ^2 − 7 λ + 12 − 2 = 0 λ^2 − 7 λ + 10 = 0

  • To find all the eigenvalues of A, solve the characteristic equation.

λ^2 − 7 λ + 10 = 0 (λ − 5)(λ − 2) = 0

So we have two eigenvalues λ 1 = 2, λ 2 = 5.

  • For each eigenvalue λ, to find the corresponding set of eigenvectors, we simply solve the linear system of equations given by, (λI − A)~x = 0.

case(i) λ 1 = 2.

(2I − A)~x = 0 ( 2 − 4 − 2 − 1 2 − 3

x 1 x 2

x 1 x 2

− 2 x 1 − 2 x 2 −x 1 − x 2

We get two equations: 2 x 1 + 2x 2 = 0 x 1 + x 2 = 0 Both equations give the relation x 1 = −x 2. Therefore the set of eigenvectors corresponding to λ = 2 is given by:

{

x 1 x 2

|x 1 = −x 2 }

−x 2 x 2

{ x 2

|x 2 is a real number.}

An eigenvector corresponding to λ 1 = 2 is

case(i) λ 2 = 5.

(5I − A)~x = 0 ( 5 − 4 − 2 − 1 5 − 3

x 1 x 2

x 1 x 2

x 1 − 2 x 2 −x 1 + 2x 2