HOW TO FIND THE INVERSE OF 2×2 AND 3×3 MATRIX USING SHORTCUT METHOD

Hello and Welcome to this post ,Today we are going to discuss the shortest and easiest methods of finding the Inverse of 2×2 matrix and 3×3 Matrix. Usually when we have to find the Inverse  of any  Matrix  then we follow the following steps .

1 Check whether the determinant value of the given Matrix is Non Zero.

 2  Find out   the co-factors of all the elements of the Matrix.

 3 Put these co-factors in co-factor Matrix.

 4 Find the Ad joint of this matrix by taking the  Transpose of a Matrix  of the co-factor matrix.

  5 Now  Multiply  Ad Joint of Matrix   with the reciprocal of               Determinant value of  the given Matrix.

This Method is very confusing, Long  and time Consuming.  So Let us have a New,  Easy and Shortcut Method .

Method For 2×2 Matrix

If we have to find the Inverse of  2×2 Matrix then Follows these steps.

 1 Interchange the position of the elements which are  a11  and a22 .

2  Change the Magnitude of the elements  which are in position a12 and  a21   .

3  Divide  every elements of the given Matrix with its Determinant value.

Example

 To find the Inverse of this matrix just interchange the position of elements a₁₁ and a₂₂  i.e   Interchange the positions of elements  5  and -3 and in second step change the magnitude of the elements which are  in positions a12 and  a21   i.e. change the sign of 9 and 4.

Now divide each elements with determinants value of the matrix which is  (5)(-3) - (9)(4) = -15 -36 = -51

So The Inverse of the given Matrix A  will  be 

Then after interchanging the positions of 8 and 2 change the magnitude of  7 and -6 and divide every elements with its determinant value (8)( 2) - (7)*(-6) = 16+42 = 58

The Inverse of  B is

 After interchanging the position of -3 and -6 and changing the magnitude of  -4 and -5 and at last dividing every elements with its determinant value (-3)×(-6) - (-4)×(-5) = 18 - 20 =  -2

The Inverse of  C is   

This video Explains all about Inverse of 2×2 and 3×3 Matrix

READ  MATRIX , DIFFERENT TYPES OF MATRICES AND DETERMINANTS

Method for 3×3  Matrix        

Ist of all  Write the given Matrix in five columns by adding the 4^th column as repetition of 1^st column and 5^th column as repetition of 2^nd column, then

C₁    C2     C3      C4     C5
5      -1       4       5      -1
2       3       5       2        3
5      -2      6        5      -2

Now Expanding this Matrix to 5×5 Matrix by adding 4th Row as repetition of 1st Rows and adding 5 Row as repetition of 2nd column as what we received in last step.

R₁        5             -1             4            5          -1

R₂        2              3              5           2            3

R₃        5             -2             6            5          -2

R₄        5             -1             4            5          -1

R5        2              3              5           2            3

 Now to find the Inverse  of the given Matrix ,we have to find the cofactor of every elements

1 Find the co-factor of 1st element of Row 1 i. e. 5, determinant value of the Matrix (RED below ) obtained by eliminating the 1st Row and 1st Column which will be (3×6)-{(5)×(-2)} = 28,write these co-factor value in 1st column of 1st Row. (we are evaluating co-factors row wise and writing Column wise)

R₁      5      -1      4        5          -1
R₂    2       3      5        2           3
R₃    5      -2      6       5          -2
R₄    5      -1      4       5          -1
R5     2       3      5       2           3

2 Now Find the co-factor of 2nd element of 1st Row i. e. -1,which is equal to determinant value of the Matrix  (RED below) obtained by eliminating the 1st Row and 2nd Column which will be 5*5-(2)*(6) =13,write this co-factor value in  2nd Row of 1st column .(we are evaluating co-factors row wise and writing Column wise) 

R₁        5             -1             4            5          -1

R₂        2              3              5           2            3

R₃        5             -2             6            5          -2

R₄        5             -1             4            5          -1

R5        2              3              5           2            3 

3 Now Find the co-factor of 3rd element of 1st Row  i.e. 4, which is equal to determinant value of the Matrix  (RED below ) obtained by eliminating the 1st Row and 3rd Column which will be 2*(-2)-(3)*(5) = -19,write this co-factor value in  3rd Row of 1st column.(we are evaluating co factors row wise and writing Column wise)

R₁           5            -1            4             5         -1

R₂           2             3             5            2           3

R₃           5            -2             6            5          -2

R₄           5            -1             4            5          -1

R5           2             3             5            2           3

4  Now Find the co-factor of 1st element of 2nd Row  i. e. 2, which is equal to determinant value of the Matrix  (RED below ) obtained by eliminating the 2nd  Row and 1st Column which will be -2*(4)-(6)*(-1) = -2,write this co-factor value in  2nd Column of 1st Row .(we are evaluating co factors row wise and writing Column wise) .

R₁         5         -1           4             5           -1

R₂         2          3            5             2            3

R₃         5         -2            6            5           -2

R₄         5         -1            4            5           -1

R5         2          3            5            2             3

5 Now Find the co-factor of 2nd element of 2nd Row i.e 3, which is equal to determinant value of the matrix (RED below ) obtained by eliminating the 2nd Row and 2nd column which will be 6*(5)-(5)*(4) = 10,write this co-factor value in 2nd Row of 2nd column

  R₁        5           -1          4          5          -1

  R₂        2            3           5         2           3

  R₃        5           -2          6          5          -2

  R₄        5           -1          4          5          -1

  R5        2            3          5          2           3

6 Find the co-factor of 3rd element of 2nd Row i. e.5, which is equal to determinant value of the Matrix (RED ) obtained by eliminating the 2nd Row and 2nd Column which will be 5× (-1)-(-2) × (5) = 5,write this co-factor value this 2nd Column of 3rd Row, write this co-factor value in 2nd Column of 1st Row . (we are evaluating co factors row wise and writing Column wise ) .

 R₁         5          -1           4           5          -1

 R₂         2           3           5           2            3

 R₃         5          -2           6           5          -2

 R₄         5          -1           4           5          -1

 R5         2           3           5            2           3

Similarly for 1st , 2nd, 3rd element the co-factor values will be as follows

      For  A₃1 i.e  5

R₁       5            -1             4            5          -1

R₂       2             3              5           2           3

R₃       5            -2             6            5          -2

R₄       5            -1             4            5          -1

R5       2             3              5           2           3

For A₃₂ i.e   -2
R₁          5         -1            4            5          -1 
R2          2          3            5            2           3

R₃          5         -2            6            5          -2

R₄          5          -1           4            5          -1

R5          2           3           5            2           3

  for  A₃₃  i.e. 6

R₁          5           -1           4           5          -1

R₂          2            3           5           2           3

R₃          5           -2           6           5         -2

R₄          5           -1           4           5          -1

R5          2            3           5           2           3 

so we have  -17 ,-17 and 17 as co-factors of 3rd Row, write these co factors in 3rd column .

(we are evaluating co factors row wise and writing Column wise)

Ad joint   A   =   

                                        
⎾ 28          -2          -17 ⏋
⎹⎸ 13          10         -17 ⎹
⎿ -19         5            17  ⏌

Now divide with the determinant value of given 3×3 Matrix , which will be 5(28)-1(-13) + 4(-19) = 140 + 13 -7 6 = 77.

Now divide each element of Ad joint Matrix obtained in previous step with determinant value 77,

Then   A⁻¹  =   

Conclusion

This  post was regarding short cut methods of finding Inverse of  2×2 and 3×3 Matrices , If you liked this post ,Please  share your precious views on this topic and share this post with your friends to benefit them. we shall Meet in the next post ,till then BYE . 

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HOW TO KNOW THE DIVISIBILITY TEST OF A NUMBER || DIVISIBILITY RULES FOR NUMBERS

How to know whether a given number, however large is divisible by 2, divisible by 3,4,5,6 and 10 .There are fixed divisibility test and divisibility rules for checking the divisibility of any given numbers. For different numbers there are  different rules to divide with . So in this post  we are going to discuss these divisibility rules with examples  one by one. Although there are divisibility test  for fraction also ,but this post will be restricted to natural numbers.

Divisibility rules for 2

To check whether the given number, however large  is divisible by 2 ,we have to check its right most digit/Unit place digit, if it is even number or zero.Then the given number is definitely divisible by 2

For Example

12 is divisible by 2 as its right most digit is 2 which is Even .

3548 is divisible by 2 as its right most digit is 8 which is Even 

999998 is divisible by 2 as its right most digit is 8 which is Even .

65989564 is divisible by 2 as its right most digit is 4 which is Even. 

22222229 is not divisible by 2 as its right most digit is 9 which is not A Even number.

589423100780      is divisible by 2 as its right most digit is 0 .

357913579571536   is divisible by 2 as its right most digit is 6 which is Even although all the remaining digits are odd.

Divisibility rules for 3

To check whether the given number, however large  is divisible by 3,we have to check the SUM of all its digits, if its sum is divisible by 3 then the  given number is divisible by 3, If sum of  all the digits of given number is again a large number then add the result so obtained and apply the rule again which is said earlier.

For Example

15 is divisible by 3 as sum of all its digits 1 + 5 = 6 is divisible by 3 .

833 is not divisible by 3 as sum of all its digits 8 + 3 + 3 = 14 = 1 + 4 = 5 is not divisible by 3 .

2678 is not divisible by 3 as sum of all its digits 2 + 6 + 7 + 8 = 23 = 3 + 3 = 5 is not divisible by3 .

98784552 is not divisible by 3 as sum of all its digits 9 + 8+ 7+ 8+ 4 + 5 + 5 + 2 = 48 = 4 + 8 = 12 = 1 + 2 = 3 is not divisible by 3.

359875269 is divisible by 3 as sum of all its digits 3 + 5 + 9 +8 +7 + 5 + 2 + 6 + 9 = 54 = 5 + 4 = 9 is divisible by 3.

3597841137 is divisible by 3 as sum of all its digits 3 + 5 + 9 + 7 + 8 + 4 + 1 + 1 + 3 + 7 = 48 = 12   is divisible by 3 .

Divisibility rules for 4

If the last two digits of a number is divisible by 4 ,Then the number is divisible by 4 . The number having two or more zeros at the end is also divisible by 4.

For Example

568928 : Here in this number last two digits are 28 ,which are divisible by 4,Hence the given number is divisible by 4.

134826900 : As ther are two zeros at the end,so the given number is divisible by 4.

13444255452 : As the last two digits number (52) is divisible by 4 ,the given number is divisible by 4.

35888875698549 : As the last two digits number (49) is not divisible by 4 ,the given number is not divisible by 4

97971349999567776 : As the last two digits number (76) is divisible by 4 ,the given number is divisible by 4.

44444444444444444449 : As the last two digits number (49) is not divisible by 4 ,the given number is not divisible by 4.

Divisibility rules for 5

To check whether the given number, however large is divisible by 5 ,we have to check its right most digit/Unit place digit,if it is 5 or zero. Then the given number is definitely divisible by 5.I.e if the given number ends with 0 or 5 then it is divisible by by 5.

For Example

35 is divisible by 5 as its right most digit is 5.

97835 is divisible by 5 as its right most digit is 5.

6854940 is divisible by 5 as its right most digit is 0.

35000000355 is divisible by 5 as its right most digit is 5.

3579515465855 is divisible by 5 as its right most digit is 5.

12345678888880 is divisible by 5 as its right most digit is 0.

568954975311525 is divisible by 5 as its right most digit is 5.


55555555555556 is not divisible by 5 as its right most digit is 6.

66666666666666665 is divisible by 5 as its right most digit is 5.

Divisibility rule for 6

For any number to be divisible by 6 ,it must be divisible by both 2 and 3 ,then the given number is divisible by 6, Therefore
1) The number should ends up with an even digits or zero and
2) The sum of its digit should be divisible by 3.


For Example :

56898 is divisible by 6 as sum of its digits is 5 + 6 + 8 + 9 + 8 = 36 = 3 + 6 = 9 so it is divisible by 3 and last digit is even ,as the given number is divisible by both 2 and 3 ,so it is divisible by 6.

3578952  As the last digit is even so the given number is divisible by 2 and sum of all its digits is 3 + 5 +7 + 8 + 9 + 5 + 2 = 39 = 4 + 2 = 6  which is also divisible by 3 ,which implies the given number is divisible by both 2 and 3. Therefore the given number is divisible by 6 as well.

25689879798 is divisible by 6 as sum of its digits is 2 + 5+ 6 + 8 + 9 + 8 + 7 + 9 + 7 + 9 + 8 =78 = 15 = 1 + 5 = 6 so it is divisible by 3 and last digit is even ,as the given number is divisible by both 2 and 3 ,so it is divisible by 6.

35658999962 is not divisible by 6 as sum of its digits is 3 + 5 + 6 +  5 + 8 + 9 + 9 + 9 + 9 + 6 + 2 = 71 = 7 + 1 = 8 so it is not divisible by 3 and last digit is not even ,so it is not divisible by 6.

35789248956 As the last digit is even so the given number is divisible by 2 and sum of all its digits is 3 + 5 + 7+ 8+ 9+ 2+ 4 + 8+ 9 + 5 + 6 = 66 =  6 + 6 = 12 = 1 + 2 = 3 which is also divisible by 3 ,which implies the given number is divisible by both 2 and 3. Therefore the given number is divisible by 6 as well.

Divisibility rule for 10

It is the most easiest number to identify whether it is divisible by 10 , if the given number however large ends up with 0 then it is divisible by 0.

For Example

4546546546540 ,44545454560, 445474456110 5555598959550 are divisible by 10 as all the given numbers ends with 0.

and 445645489,454545,456445555 ,454545577 and 545454 are not divisible by 10 as all these numbers do not ends up with 0.

Conclusion

Thanks for giving your valuable time to read this post of the divisibility rules and divisibility test rules for  divisibility rule for 2 , divisibility rule for 3,    divisibility rule of 4 etc . If  you liked this post  , Then share it with your friends and family members . You can also read my others articles on Mathematics Learning and understanding Maths in easy ways.

Let us learn Multiplication , division , arithmatic and simplification short cut ,tips and  tricks after buying this Book of Magical Mathematics .

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HOW TO MULTIPLY TWO DIFFERENT NUMBERS IN FASTEST AND QUICKEST WAYS

        

                     Whenever we are to multiply two numbers then most of the time we undergo long calculations , and if we have to do easy calculation then we were lucky.

         But what to do when we have to multiply two numbers in very short time. Suppose we have to multiply 32 with 11 then it is easy, because just put right most digit as it is and then increase every digit by one to left and at last put the left most digit as it is and put 352 as answer.

Example

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HOW TO UNDERSTAND RELATIONS AND FUNCTIONS ,INVERSE OF A FUNCTION

Hello Friends Welcome 
                              Today we are going to  discuss Relations and Functions , "How to understand  Relations and  Functions, Inverse of a Function" under the topic  Relations and Functions.

Ordered-Pair Numbers :-

Ordered-pair number is written within a set of parentheses and separated by a comma.

For example, (5, 6) is an ordered-pair number; the order is designated by the first element 5 and the second element 6. The pair (3, 6) is not the same as (6,3) because they have different order. Sets of ordered-pair numbers can represent relations or functions.

Example of ordered pair :

(3,8),(2,1),(7,6)

Relation

A relation is a  set of ordered-pair numbers.

consider the following table

Number of Students	1	2	3	4	5	6
Marks Obtained	96	98	97	78	77	86

In the above table the numbers of students and marks obtained by them  is a relation and can be written as a set of ordered-pair numbers.

A= {(1, 96), (2, 98), (3, 97), (4, 88),(5,77),(6,86)}

When we collect all the elements written in 1st column of the ordered pairs and placed in a set then the set so formed is called  Domain of the relation.
The domain of A= {1, 2, 3, 4,5,6}

As all the elements written in 2nd column of the ordered pairs and placed in a set then the set so formed is called  Range of the relation.

The range of A = { 96, 98, 97,88 , 77, 86}

Function

A function is a relation in which every first element in ordered pairs have unique second element associated with them. Second  elements may or may not be same.

we can better understand this concept with the help of this video

Example

 {(1, 2), (2, 3), (3, 4), (4, 5),(5,6)}  is an example of function 

 { (1, 2), (2, 3), (3, 4), (4, 5),(5,6) } is a function because all the  first elements are different.

Example

{(1, 3), (3, 3), (2, 1), (4, 2)}  is an example of function 

 {(1, 3), (2, 3), (2, 1), (4, 2)}  is a function because all the first elements are different.

Example

{ (1, 6), (2, 5), (1, 9), (4, 3) }  is not an  example of function 

As in  {(1, 6), (2, 5), (1, 9), (4, 3)}  the element "1 "   appeared twice .

Example

{(2, 15), (3, 15), (4, 15), (5, 13),(6,18)}  is  an  example of function 

As in  {(2, 15), (3, 15), (4, 15), (5, 15)}   all the first elements are different.

Example

{(1, 1), (-1, 1),(2,4),(-2,4), (3, 9), (-3, 9),(4,16),(-4,16)}  is an  example of function although   the element "1" and "-1" ,"2" and "-2" , "3" and "-3"  ,"4", "-4" have same images. This is an example of many one function.

Question:-   Find x and y if: 

(i) (5x + 3, y) = (4x + 5,  2)

(ii) (x – y, x + y) = (8, 12)
(iii) ( 2x-y , y+5 ) = ( -2,3 )

Solution

(1)  Given  (5x + 3 , y) = (4x + 5, 2)
So By the equality of ordered pair elements,
1st element of the ordered number written on the left hand side will be equal to the 1st element of the ordered pair number written on the  right hand side . Therefore 

5x + 3 = 4x + 5   and y =  2 
5x-4x = 5 - 3   and y = 2 
x = 2 and y = 2

(ii) So By the equality of ordered pair elements
x – y = 8 and  x + y = 12

Solving these two equations for x and y 

 2x =20  and    10+ y =12 

x=10   y = 2

(iii) So By the equality of ordered pair elements
2x-y  =-2  , y+5 = 3 
2x = -2+y  , y = 3-5
2x = -2+y  , y = -2
Putting the value of y in 1st Equation ,we get
2x = -2 - 2
2x = -4
x = -2
so x= -2 and y =-2

Types of Relations

A relation R in a set A is called

(i) reflexive, if (a, a) ∈ R, for every a ∈ A,

(ii) symmetric, if (a, b) ∈ R implies that (a, b) ∈ R, for all a,b ∈ A.

(iii) transitive, if (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R, for all a, b,c ∈ A.

If you want to Read  HOW  TO  UNDERSTAND BINARY  OPERATIONS , RELATIONS  AND  FUNCTIONS || COMMUTATIVE || ASSOCIATIVE  click here 

Equivalence Relation

A relation R in a set A is said to be an equivalence  relation if R is reflexive, symmetric and transitive.

1 ) Let B be the set of all triangles in a plane with R a relation in B given by

R = {(T1, T2) : T1 is congruent to T2}. Then R is an equivalence relation.

2 ) Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7}  by

R = {(a, b) : both a and b are either odd or even}. Then R is an equivalence

one-one Function

A function f : X → Y is defined to be one-one (or injection ), if the images of distinct elements of X under f are distinct, i.e., for every x, y ∈ X, f (x) = f (y) implies x = y. Otherwise, f is called many-one.

Onto Function

A function f : X → Y is said to be onto (or surjective), if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an

element x in X such that f (x) = y.

Example

1   Function f : R → R, given by f (x) = 2x, is one-one and Onto As all the elements  have only one and uniqe image under f.

2  Function f : N → N, given by f (x) = 2x, is one-one but not onto.Because  the elements  have only one and unique image under f Therefore it is one one function .But not all elements of N have image under f 

e. g .  1,3,5,7... are not the image of any elements of N under f so it is not onto function

Example

The function f : N → N, given by f (1) = f (2) = 1 and f (x) = x – 1,

for every x > 2, is onto but not one-one.

Solution

Since f is Not one-one, as f (1) = f (2) = 1. 

But f is Onto, as given any y ∈ N, y ≠ 1,

Choose x = y + 1 s.t.

 f (y + 1) = y + 1 – 1

f (y + 1)  = y. 

Also for 1 ∈ N, 

we are given  f (1) = 1

Inverse of a Function

A function f : X → Y is defined to be invertible, if there exists a function g : Y → X such that gof = IX and fog = IY. The function g is called the inverse of f and is denoted by f –1

Example

Let S = {1, 2, 3}. Determine whether the functions f : S → S defined as below have inverses. Find f , if it exists.

(a) f = {(1, 1), (2, 2), (3, 3)}

(b) f = {(2, 2), (3, 1), (4, 1)}

(c) f = {(1, 5), (3, 4), (2, 1)}

Solution

(a) It is to  proved that  f is one-one and onto Hence f is invertible with the inverse f –1 of  f given by f –1 = {(1, 1), (2, 2), (3, 3)} = f.

(b) Since f (3) = f (4) = 1, f is not one-one, so that f is not invertible.

(c) Here  f   is one-one and onto, so that f is invertible with

 f –1 = {(5, 1), (4, 3), (1, 2)}.

Composition of Functions

Let f : A → B and g : B → C be two functions. Then the composition of f and g, denoted by gof, is defined as the function gof : A → C given by

gof (x) = g(f (x)), ∀ x ∈ A

Example

fof(x) = (16x + 12 + 18x -12 ) / ( 24x + 18 - 24x +16)

fof(x) = (34 x ) / ( 34)

fof(x) =  x  =  I(x)

Example

Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as f (2) = 3, f (3) = 4, f (4) = f (5) = 5 and g(3) = g(4) = 7 and g(5) = g(9) = 11. Find gof = ?

Solution

We are given

 gof (2) = g (f (2)) 

               = g(3) 

               = 7

 gof (3) = g(f (3)

             = g(4)

              = 7,

gof (4) = g(f (4)) 

           = g(5) 

             = 11 

and  gof (5) = g(f (5))

                   = g (5)                     

                    = 11

So gof ={(2,7),(3,7),(4,11),(5,11)} 

Example

conclusion

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