BAM FORM 6 – MATRICES

 is an arrangement of number in rows and columns. Its usually denoted using capital letters
E.g A =           B =          C =

Order: A matrix is said to be of order m x n if it has m rows and n columns.

E.g. Order of A is 2 x 2

       Order of B is 3 x 3

       Order of C is 2 x 3

Types

–      Rows matrix is a matrix with only one row.

e.g. D = (2 1 6)

–      Column matrix is a matrix with only one column.

E.g.   Д =

–      Square matrix  is a matrix with equal number of rows and column

E.g.  A =                 B =

–      Identity matrix (T) is a square matrix with all elements in the  leading diagonal equals to 1 and the rest are 0.

E.g. I =  

                   I =

Property

AI = A

Null (zero) matrix (Z) is a matrix with all elements equation to 0 e.g.  Z = 

            Properties

            1) A + Z = A

            2) AZ = Z

            Operations

i) Addition/ subtraction

   This is only possible if the matrices have the same order

E.g.    A =         B =

ii) Scalar multiplication

            Given a scalar multiplication then

            A = t

            =   

E.g.: Evaluate tP – sQ

Given t = 3 s = ½, P =

                               Q =

Solution

            tP – SQ

   iii) Product
   AB is only possible if the order of A is m x n and that of B is n x p and  the resulting matrix will be of the order m x p.

e.g.      i)

            2 x 2        2 x 1

           

            ii)

                       

E.g.     

                3 x 3         3 x 1

=  

=    =

                iv)    

                           3 x 3                        3 x 3

   

E.g.    

    3 x 3                3 x 3

                       

=

Determinant of a 2 x 2 matrices

   Given A =

Determinant of A =

                          =

= ad – cb

If the determinant A = 0 then A is singular matrix

            E.g. which of the following is singular or non singular matrix

A =

8 – 9 = -1     Non – singular matrix

    B =

   = 1(6) – 2(3)

        = 6 – 6

        = 0       Singular matrix

C =

   = 8   Non – singular matrix

Inverse of a 2 x 2 matrix

            Given A =

The inverse of A = A^-1

        E.g.      A =

 = 2 (4) – 3(3) = -1

       

           A^-1 =

A^-1 =

Solving system of simultaneous equations in 2 unknowns

Given the following equations

In a matrix form

              =

Determinant method

1.  (Cramer’s rule)

            Solving for x

                 where

           and
Solving for y
where
and

            E.g. solving the following using Cramer’s rule

                        x + y = 3

                        x – y = 1

    = 

A=     Δ  ₌ 
 = 1 (-1) – (1)= -2

                 where

                  

            
Δ  ₌ 
 = 1 (-1) – (1)= -2

            

 =

2.   Inverse method

              =

             = A^-1 

         
E.g.  x + y = 3

                    x – y = 1

                =

            A =   = -1(1) – 1 (1) = -2

A^-1 =

                  A^-1 =

                  =     

                     =

                        =

                        =

                        =

                         =

Determinant of a 3 x 3 matrix

Given A =

            = a  – b  + c

             = a (ek – fh) – b (dk – fg) + c (dh – eg)

E.g  Find given B =

                   = 3   – (-1)  + 2

                   = 3 (3 + 8) + 1(0 – 20) + 2 (0 – 5)

                  = 33 – 20 – 10

                  = 3                                                                                                     

                   = 3

Transpose
                    

                 

E.g,    B =

                        B^T = 

Co factors

                        A =

 A co factor of an element in a 3 x 3 matrix is given by determine determinant of a 2 x 2 matrix which is formed by removing elements in the same row and column with given element and multiply by 1 or -1 according to the following procedure.

            Produce  

            E.g. C =
  

                        = (3 + 8) = 11

          Cof  0         -1

                           = -1 (-3 + 4) = -1

                         = + (-4 – 2) = -6

            Cof (-1)      -1

                           = -1 (0 – 20) = 20

               Cof 1  + 1

                        = 1 (9 – 10) = -1

            C0f -2

                        -1

                        = -1 (12 – 0) = -12

            Cof 2

                        1

                        = 1 (0 – 5) = -5

            Cof 4

                        -1

                        = -1 (-6 + 5) = 1

            Cof 3

                        1

                        = 1 (3 + 0) = 3

Matrix of Cofactors
   A =

        Adjoin of A = adj,    A^T = the transpose of matrix of Cofactors

Inverse of A

       

Summary

   Procedure for finding inverse of a 3 x 3 matrices

    i) Find the determinant

   ii) Find the cofactors

   iii) Form the matrix of cofactors

   iv) Form the adjoint

    v) Find the value

Solving simultaneous equation in 3 unknown

1. Determinant method (Cramer’s rule)

            Given that       a₁x + by + c_1z = p

                                    a₂x + b₂y c₂z= q

                                    a₃x + b₃y c₃z = r

            In a matrix form we have

               = 

E.g. using Cramer’s rule   Solve  x + 2y + z = 6
2x + y – z = 3
3x – y + 2z = 5

            Solution

           

                        = 1 (2 – 1) – 2 (4 + 3) + 1 (-2 – 3)

                        = 1 – 14 + -5 = -18

            

                 = 6 (2 – 1) – 2 (6 + 5) + 1 (-3

                 = 6 – 22 + – 8

                 = -24

           

            = 1  – 6  + 1

                        = 1 (6 + 5) – 6 (4 + 3) + 1 (10 – 9)

                        = 11 – 42 + 1

                        = -31 + 1 = -30

II. Inverse method

            a₁x + b₁y + c₁z = p

            a₂x + b₂y + c₂z = q

            a₃x + b₃y + c₃z = r

                   A  =

            AA^-1
 = A^-1

                 I   = A^-1

                    = A^-1

E.g: solving using inverse method

                        x + 2y + z
= 6

                        2x + y – z = 3

                        3x – y + 2z = 5

                          =

                 A =

             = 1  – 2  + 1

                        = 1 (2 – 1) – 2 (4 + 3) + 1 (-2 – 3)

                        = 1 – 14 + (-5)

                        = -13 – 5

                  = -18

                        A = 

            Cof 1 = + 1          cof 2 = -1

                        = +1 (2 – 1)                  = 1 (4 + 3)

                        = 1                              = -7

            Cof 1 = + 1                 cof 2 = -1

                        = + (-2 – 3)                = -1 (4 + 1)

                        = -5                             = -5

            Cof 1: + 1                   cof -1 = -1

                        = +1 (2 – 3)               = -1 (-1 – 6)

                        = -1                             = 7

            Cof 3: + 1                  cof -1= -1

                        = + 1 (-2 – 1)             = -1 (-1 – 2)

                        = -3                                         = 3

            Cof 2: + 1

                        = + 1 (1 – 4)

                        = -3

            Matrix of cofactors, C

        C  =   

Note: Adj A = C^T
Adj A = 

Inverse of A

    A^-1  =                                                                                                     

              A^-1   = 

= 

            =

            =  =

            =  =

Exercise

1.  a) Given A = , and B = ,  Find A + B

      b) If  A =   is singular find the value of k

      c) Find the inverse of B =

    d) Solve

2.  Using i) Determinant
ii) Inverse, solve for x, y and z

               3x – y + z = 2

               x + 5y + 2z = 6

               2x + 3y + z
= 0

3.  a) solve

i) Using determinant
ii) inverse

      b) If A =, B =  and

                        C =

                        Show that (A + B – 2C) is singular

      c) Solve x + y + z
= 6

                        3x – 2y – z
= -1

                        2x + 4y + 3z = 19  Using i) determinant

                                                                 ii) Inverse

4.  Solve,  2x – 3y + z = 3

                -x + 4y + 3z = 16

                3x + 2y – 2z = 1

5.  A transformation is given by the matrix M where M =  Find the (a) image of (-2, 5) under M (b) Inverse of M.

6.  If T is linear transformation such that T =  and T (x, y) = (3y, 5x),  Find T hence evaluate T (1, 2)

            b) Use inverse method to solve

           

7.  a) Given A = . B = ,  Find AB and BA

     b) Find the value of x,y,w and z

                        3  =  +