MATRICES

A matrix, is a set of real or complex numbers (or elements) arranged in rows and columns to form a rectangular array OR

Let P be an arbitrary field.  A rectangular array of the form

where a_ij  are scalar in P  is called a MATRIX

The above matrix can be written a_ij,

    i  = 1,2,3,–,–,–,m     j = 1,2, 3 –, –, –,n

The horizontal elements are called are called rows, while the vertical elements are called columns . An m  n  matrix  has m rows  (horizontal lines) and columns (vertical lines). The entry in the  row and jth column is denoted by the double subscript notation a_ij .i is called the row subscript  because it gives the position in the horizontal lines and j the column subscript because it gives the position in the vertical lines

            A matrix having m rows and n column is of order m  n. If m = n, the entries a₁₁, a₂₂, a₃₃ –  –  –  a_nn  are the main (principal) diagonal entries

Example

           

           

 

 

 

 

 

OPERATION WITH MATRIX

1.         EQUALITY OF MATRIX :–  Two matrices are said to be equal if corresponding element are equal Therefore, the two matrices must also be of the same order or

Two matrices A = [aij] and B = [bij] are equal if they have the same order (m × n) and

 aij = bij

Example

            Solve for a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃, a₃₁, a₃₂, a₃₃, in the following matrix equation 

                     

    Solution

  Because two matrices are equal if corresponding elements are equal, we can conclude that

            a₁₁   =  1,         a₁₂   =  2,         a₁₃    =   4

            a₂₃   =   3         a₂₂   =  7,         a₂₃    =  9

            a₃₁   =   5         a₃₂   =  6,         a₃₃    =  0

Example

Find the values of f, g, x, y in the problems

 

f +3 =  2;   f = –1

2g+1 = 1;  g = 0

x–3 = –3;   x = 0

2y–2 = 2;    y = 2

 

             2f + 3 = f – 5;     f = – 8

             2g – 2 = g + 1;  g = 3

                x+1 = 2x+3;   x = – 2

                                        y = 5

Example

Find x and y

                     

Solution

            2x + 1   = 5 ;       x = 2

            3y – 5   = 4 ;        y = 3

ILLUSTRATIVE EXAMPLE

4.      The matrices         

5.      The  matrices    

2.         ADDITION AND SUBTRACTION OF MATRICES

If A = [aij] and B =[bij] are matrices of m×n then their sum is m×n  given by

A+B = [aij + bij]

Note:   Before addition or subtraction can be performed with matrices, the order must be the same. The sum and difference of two matrices of different order is undefined.

            Example

           

           

3.                      SCALAR MULTIPLICATION

If A = [aij] is an m×n matrix and k is a scalar, then the scalar multiplication of A by k is the m×n matrix given by kA = [kaij]

Example

           

Example

            Find (i) 3A – 2B   (ii)   A + 2 (B – A)

Given that                   

If A, B and C are m×n matrices, d and k are scalar, then the following properties are true

1.         A+B = B+A                    Commutative Property of Addition

2.         A+(B+C) = (A+B)+C     Associative Property of Addition

3.         (dk)A = d(KA)                Associative Property of Scalar Multiplication

4.         IA = A                             Scalar Identity

5.         d(A+B) = dA+dB            Distributive Property

6.         (d+k) A = dA+kA            Distributive Property

 

4.                                     MATRIX MULTIPLICATION

            If A and B are matrices of m×p and p×n respectively, then their product is a matrix c of order m×n in c, whose entry in the ith row and jth column  is the sum of the product formed by multiplying each entry in the ith row of A by the corresponding entry in jth column of B

             C = AB = [cij]

Where cij = ai₁bij  +  ai₂ b₂j  +  ai₃ b₃j + – – – – + ain  bnj

      This  definition indicates a row – by – column multiplication.

 

Note:   Two matrices can be multiplied together only when the number of columns in the first is equal to the number of the row in the second

 

               A                               B          =          AB

            m×p                            p×n                  m×n

                            equal

                            order of AB

 

 

 

Example

           

 

It should be noted that A.B  B.A

So multiplication of matrices is not commutative

Properties of Matrix Multiplication

If A, B and C are matrices and k is a scalar then the following properties are true

1.         A(BC) = AB(C)                        Associative property

2.         A(B+C) = AB+AC                   Distributive property

3.         (A+B) C = AC+BC                  Distributive property

4.         K(AB) = (KA)B = A(KB)         

Example                 Application of Matrix Multiplication

            The Nigeria football team submit kit list to their sponsors

                                                Female’s team Male’s team

                        Jerseys             24                                30

                        Boots               90                                76

                        Balls                 30                                34

       Each Jersey cost N42, and each boots cost N8 and each ball cost N60.  Use matrices to find the total cost of football for each of them

 

Solution

            The kit list can be written in matrix form as

                       

and the cost per item can be written in matrix form as

                       

The total cost of equipment for each item is given by the  product

           

Thus, the total cost of kits for the female’s team is N3528., and the cost of kits for the men team is N3908

 

Example A firm sell five different models of ford cars through three retails outlets.  The inventory of each model in the three outlets is given by the matrix B

                       

and the wholesale and retail price for each model is given by the matrix H

 

(a)                What is the wholesale price of the inventory at outlet 1

(b)               What is the retail price of the inventory at outlet 3?

(c)                Compute BH and interpret the result

Solution

           

whole price at outlet 1

            = N[4×480+2×1200+1×1450+3×2650+2×3050]      

= N3360+N2400+N1450+N7950+N6100

            = N21260.00

The whole price at outlet 1 is N21260.00

Interpretation:–  The entries are the wholesale and retail price of inventory at each outlet.

 

TRANSPOSE OF A MATRIX

If the rows and column of a matrix are interchanged i.e. the first row be comes the first column, the second row becomes the second column, the third row becomes the third column, e.t.c.  The new matrix so formed is called the TRANSPOSE OF THE ORIGINAL MATRIX.  If A is the original matrix, its transpose is denoted by A¹ (In some texts  or A^T are used instead of A¹)

 

Example

SPECIAL MATRICES

1.         Square Matrix:– Is a matrix of order m×m i.e. the number of rows is the same as the  number of columns

                         

2.         symmetric matrix

(a)        A square matrix A, such that A = A¹ is said to be symmetric

            e.g        A =

            The matrix A is symmetric matrix

(b)        A square matrix A such that A = – A¹ is said to be skew–symmetric

                        e.g     

In skew – symmetric matrix, the elements along the principal diagonal are zeros

Note: If A¹, B¹ are the transposes of A and B and if k is any scalar, then the following hold true

(a)                (A¹)¹  = A

(b)               (A+B)¹ = A¹+B¹

(c)                (KA)¹ = KA¹

3.  DIAGONAL   MATRIX :–   A square matrix if all the element other than those in the principal or main  diagonal are zero ,  then the matrix is called DIAGONAL MATRIX .  The matrices below  are examples of diagonal  matrices

      

4.   UNIT OR IDENTITY MATRIX:–   Is a diagonal matrix in which the element on main diagonal are unity or one. A unit matrix is denoted by I

              

5.         NULL MATRIX:–  If a matrix has all its element equal to zero.  It is denoted by 0

  

6.         ORTHOGONAL MATRIX:–  A matrix A is said to orthogonal matrix when AA¹ = I and its transpose is the same as its inverse i.e.  A¹  = A^–1

           

7.         IDEMPOTENT MATRIX:–  A symmetric matrix for A×A = A  is an idempotent matrix      

            Identity matrix is both symmetric and idempotent

8          TRIANGULAR MATRICES

            (a)  UPPER TRIANGULAR MATRIX:–  A square matrix  is upper triangular if it

                  has all zero entries  below its main diagonal

                         

(b)   LOWER TRIANGULAR MATRIX: A square matrix is lower triangular if it has

                   all zero entries above its main diagonal

                       

9.         COLUMN VECTOR OR COLUMN MATRIX:–  A matrix  with a single column

             and m – rows

                       

10.       ROW VECTOR OR ROW VECTOR:–  A matrix with a  single row and n – columns

                         are example of a row vector.

 

DETERMINANT

Determinant of a 2×2 matrix

Consider the expression  is know as a determinant of the second – order and quantities a₁, a₂, b₁, b₂ are called elements

 

Example :   Evaluate the determinant

Solution

(3×8) – 4x = 0

    24 – 4x = 0

        4x = 24

                        x = 6

Example

Solve for x

Solution

(x–1)(x–2) – 6 = 0

x² + 3x + 2 – 6 = 0

x² + 3x – 4 = 0

(x – 4) (x + 1) = 0

x = 4 or x = –1

Note:

(a)                The determinant of a given matrix A is written as det (A) or /A/ or  

(b)               When it is necessary to write out the determinant in full, the array of number is    enclosed in a pair of vertical lines (instead of the bracket used in matrices)

PROPERTIES OF DETERMINANT

1.         The value of a determinant remains unchanged if rows are changed to column or columns to rows (transpose)

                       

           

2.         If two rows ( or columns ) are identical, the value of the determinant is zero

           

3. If the element of any one row (or column ) are all multiplied by a common factor, the determinant is multiplied by that factor

4 .        Interchanging  any two rows or columns of a matrix will change the sign , but not the  absolute value of the determinant

          

5.         If the element of any row (or column ) are increased or decreased by equal multiples of the corresponding element of any other row (or column), the value of the determinant is unchanged.

                         

6.         If all the element of any row or column are, the determinant  is zero

                       

7.         The determinant of a triangular matrix i.e. a matrix with zero elements  everywhere above or below the principal diagonal, is equal to the product of the elements on the principal diagonal.

             

Note:–  The above properties of determinant are true for matrices of any other

Example

DETERMINANT OF 3×3 MATRIX

Given the matrix

           

is known as the third order determinant (3×3 determinant) denoted by   or

             = a, (b₂c₃ – b₃c₂) – b₁(a₂ c₃ – a₃ c₂) + c₁ (a₂ b₃ – b₂ a₃)

The quantities a_1,a₂, a₃, b₁, b₂,  b₃, c₁, c₂, c₃ are known as the elements  of the determinant and determinant obtained by omitting the row and column containing particular element is known as the MINOR of that element

            Thus,  

Example

           

            = 5(18 – 25) + 2 (12 + 20) +3 (–10 – 12)

            = –35 +64 – 66

            = – 37

 

Example

      Solve the equation   

Solution

By testing (x–1) is a factor

Using long division

             

Solving this gives

            x = 1 or x = –5 +

SARRUS RULE:–   This method only works for 3×3 matrices. Given a matrix A of order 3×3.

To apply sarrus rule, copy the first and second column of A to form fourth and fifth columns.  The determinant of A is then obtained by adding the products of the three “DOWNWARD DIAGONALS” and subtracting the products of the three “upwards diagonals” as shown

                   

            a₁         b₁         c₁         a₁         b₁

            a₂         b₂         c₂         a₂         b₂

            a₃         b₃         c₃         a₃         b₃

 

Thus, the determinant of the 3×3 matrix A is given by the following

a₁b₂c₃ + b₁c₂a₃ + c₁a₂b₃ – a₃b₂c₁ – b₃c₂ a₁ – c₃ a₂ b₁

Example

            Using the Sarrus evaluate

           

Solution

            Since A is a 3×3 matrix, we can Sarrus rule.

            Start by recopying the first and second columns and then computing the six diagonal product

                       

1          3          5          1          3

7          9          11        7          9

13        15        18        13        15

 

 

/A/  = (1×9×18) + (3×11×13) + (5×7×15) – (13×9×5) – 15×11×1) – (18×7×3)

       = 162 + 429 + 525 – 588 – 165 – 378

       = – 12

DETERMINANT OF 4×4 MATRIX

Given a matrix A of order 4×4

           

is the expansion of the fourth order determinant by the first row

Example

     Evaluate the determinant of the matrix A

Expanding each of the third order determinants, we get

 

Using these values, we have

             = 4(194) –2 (182)+4(118)–8(171)= 776 – 364+472–1368 =  –484

 

 

 

 

SOLUTION OF A SYSTEM OF LINEAR EQUATION BY DETERMINANT

CRAMER’S RULE

This rules uses determinant to write the solution of a system of linear equation. The rule is named after a man called Gabriel Cramer (1704 – 1752)

            Consider the system of linear equation

a₁x + b₁y = c₁

a₂x + b₂y = c₂

has a solution given by

provided a₁b₂ – a₂b₂  0

Each numerator and denominator in the solution can be expressed as a determinant

Example

Using cramer’s rule to solve the system of linear equatin

            x – 3y = 1

            2x + y = 23

Solution

CRAMER’S RULE FOR 3X3 SYSTEM

Consider the equation

 

Example

Use Cramer’s rule to solve the following system of linear equation

            4x – 2y + 3z = 2

            2x + 2y + 5z = 16

            8x – 5y – 2z = 4

Solution

 

Note:

            The following points is to be taken into consideration when solving system of linear equation

1.         If  /A/ = 0, the matrix is singular and there linear dependence among the equation

            No unique solution or infinitely many solutions

2.         If /A/  0 the matrix is non – singular and there is no linear dependence among the equation. A unique solution can be found.

MINORS AND COFACTORS OF A SQUARE MATRIX

           Given that a square matrix A, then MINOR Nij of the entry aij is the determinant of the matrix obtained by deleting ith row and jth column of A.  The cofactor cij of the entry aij is given by  Cij = (–1)^i+j Mij

 

 

 

 

Illustration of Minors

           

 

 

 

 = a₁₁ /M₁₁/ + a₁₂(–1) /M₁₂/ + a₁₃ /M₁₃/ defines the determinant o the matrix.

COFACTORS

    The cofactor of an element of n ordered determinant is the (n–1) ordered determinant obtained by deleting the row and element containing the element and multiplying by (–1)^i+j or by + 1 or – 1 according to the pattern

n×n matrix

4×4 matrix

3×3 matrix

2×2 matrix

 

A cofactor is always designated by capital letter corresponding to the element to which it belong

 

 

 

 

 

 

 

Example

            Given the matrix A =

       Determine the cofactor

            (a)    C₂₃      (b)    C₃₁     (c)    C₃₃

Solution

 

COFACTOR AND ADJOINT MATRIX

       A cofactor matrix is a matrix in which element  aij is replaced with its cofactor /Cij/

           

An Adjoint Matrix  is the transpose  of a cofactor matrix cofactor matrix

           

Note:  The product A. Adj (A) is always a diagonal matrix; that is a matrix in which all the element are zero except those on the “principal diagonal” (the diagonal which goes from the top left corner to the bottom right –hand corner).

 

 

 

 

Example

Find the cofactor matrix and adjoint

Matrix Adj(A), given the matrix

           

 

 

Solution

Solution

 

 

INVERSE MATRICES

Given that  A and B are two non – singular square of the same order such that AB = BA = I B is called the INVERSE OF A (B = A^–1)

            Therefore

                        AA^–1 = I (Unit Matrix)

Since A (Adj(A) = /A/I

Multiply both sides by A

     A^–1 A (Adj(A) = A^–1 /A/

            I(Adj(A)) = A^–1/A/

     If /A/  0

Then A^–1 =

 

INVERSE OF 2×2 MATRIX

Consider a matrix A =

it can be proved that its inverse A^–1is given by

INVERSE OF 3×3 MATRIX

Step to obtain the inverse of   3×3 matrix

1.                  Determine the determinant of the matrix

2.                  Obtain the cofactor of each element

3.                  Form the cofactor matrix

4.                  Transpose the cofactor matrix to obtain the Adjoint of the given natrix

5.                  Divide the Adjoint by the determinant to obtain the inverse of the given matrix.

Example

Find the inverse of the matrix

/A/ = 3 (2+66) – 5 (–8–54) + 7 (–44 +9)

      = 204 – 310 – 245

      = – 269

 

 

INVERTIBLE AND NOW – INVERTIBLE MATRIX

            If the determinate of a square matrix (2×2, 3×3, . . . ) is not equal to zero, that is, the matrix is non – singular, then we can obtain the inverse of A (i.e it is an invertible matrix) if the determine is zero, then the inverse cannot be obtain (it is non – invertible)

 

SOLVING  A SYSTEM OF EQUATION USING AN INVERSE

SYSTEM OF LINEAR EQUATION FOR 2 VARIABLES 

ax + by = e

cx +dy = f

in matrix form

SYSTEM OF LINEAR EQUATION FOR 3 VARIABLE

ax + by + cz = b₁

dx + ey + fz = b₂

gz + hy + iz = b₃

    In matrix form

Therefore Ax = B

            A^–1 Ax = A^–1 = B          {multiply both side by A^–1}

                   X = A^–1 B {Solution}

Where A^–1 is the inverse matrix

Example

            Solve for x and y in the following system

                        3x – 4y = –4

                        9x + 2y = 9

Solution

             

           

Example

    Using inverse method, solve the following system of linear equation

             –x – 2y + z = – 2

            3x + 6y – z = 12

            –2x – 6y + 2z = – 8

Solution 

           

 

Cofactor Matrix

           

Example

     Given the following system of linear equation

            4x₁ + x₂ – 5x₃ = 8

            –2x₁ + 3x₂ + x₃ = 12

            3x₁ – x₂ + 4x₃ = 5

i.                     Express the equations in matrix form

ii.                   Use the method of matrix inversion to solve for

x₁, x₂, x₃ solution

Solution

            In matrix form

 

            /A/ = 4(12+1) –1(–8–3) –5 (2–9)

            /A/ = 4(13) –1(–11) –5(–7)

/A/  = 98

form the cofactor matrix