Inverse method of solving linear equations of three variables

Inverse method of solving linear equations of 3 variables

Matrix method of solving linear equations of three variables with the help of example. 

Set of given equations are 

4x + 6z = 100     --------------------------------> (1)

3x + 6y + z = 100   -----------------------------> (2)

3x + 4y + 3z  = 100   ---------------------------> (3)

The system of these equations can be transformed into Matrix formed 

AX =  B  ,  ⇒ X =   A^-1 B    -------------------------->  (*)

Where A is matrix written from the coefficients of x, y and z when these equations are in symmetric form and B is the matrix written from constants from right hand sides in column form and X is matrix of all the variables in column form. 

In order to find the solution of set of these equations , first we have to find the inverse of matrix A if it exist then we can find the solution otherwise Matrix method fails to find the solution of the set of linear equations . 

Evaluation of Determinant 

|A|  = 4 (18 - 4) -0(9 - 3) +6(12 - 18)

       =  4(14) + 0 + 6(-6) 

       = 56 - 36

        = 20

Since the determinant value of this matrix  is not equal to zero ,Therefore its inverse can be calculated.

And  formula for finding the inverse of matrix A is 

Where Adjoint A is the transpose of co factor matrix. And in order to find the co factor matrix of any matrix, we have to find co factors of all the elements present in this matrix. 

How to calculate  co factors of all the elements of the matrix A

Let us calculate these cofactors.  

Now these co factors can be written in matrix form known as co factor Matrix. 

Co factors of 1st row are  (18 - 4) , -(9 - 3), (12 - 18) 

i. e. Co factors of 1st row are 14, -6 , -6

Co factors of 2nd row are -(0 -24), (12 - 18) , -(16 - 0) 

I. e. Co factors of 2nd row are  24 , -6 , -16

Co factors of 3rd row are  (0 - 36), -(4 - 18), (24 - 0) 

i.e. Co factors of 3rd row are  -36 , 14 , 24

Co factor Matrix

Writing co factors of 1st row in 1st row of this matrix , co factors of 2nd row in 2nd row of this matrix . Similarly co factors of 3rd row in 3rd row of this matrix . 

Adjoint  Matrix

To find the Ad joint of this matrix we have to take it's transpose, Because transpose of any matrix is called Ad joint of the matrix. So writing all the elements which are in 1st row in 1st column, and  all the elements which are in 2nd row in 2nd column and  all the elements which are in 3rd row in 3rd column. 

Inverse  Matrix

Now we can find inverse of the matrix A by putting the value of inverse of A in equation  (4), we get

Now  putting the values  Matrix B and    A^-1  in (4) 

After simplification and using the properties of equality of two matrices  ( Two matrices of same order are equal iff their respective elements are equal to each other ) 

⇒  x = 10

     y  = 10

      z = 10

So this was the Matrix method of solving linear equations of three variables using inverse of matrix. Your valuables comments will be appreciated for betterment of this blog.

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Missing number Reasoning SSC CGL questions and answers

Ten most Missing number Reasoning SSC CGL questions and answers for competitive exams, reasoning questions with answers for bank exams, Reasoning tricks for competitive exams have been discussed in this article for upcoming competitive exams. 

10 Most Important Reasoning Problems for competitive Exams

Question #1

Misiing Number 

Since the question mark is in the 3rd figure. So the value of question mark will be calculated with the help of remaining four numbers in the 3rd figure in the same way, the middle numbers have been calculated in 1st and 2nd figure with the help of remaining four numbers. This question mark value can be found by this method.   

Formula:-

Sum of four numbers = Product of both the digits in middle number.

Calculation:-

7 + 4 + 7 + 12 = 30 = 5 × 6  = Multiplication  of 5 and 6 (Middle number in 1st figure ) 

15 + 7 + 12 + 8 = 42 = 6 × 7 = Multiplication  of 6 and 7 (Middle number in 2nd figure ) 

11 + 18 + 5 + ? =  ????  = 6 × 6 = Multiplication  of 6 and 6(Middle number in 3rd figure ) 

⇒ 34 + ? = 36

⇒  ? = 36 - 34

⇒  ? = 2

 Hence option (1)2 is correct.

Question #2

Since the question mark is in the 3rd figure. So the value of question mark will be calculated with the help of remaining four numbers in the  figure in the same way, the middle numbers in 1st and 2nd figure can be calculated with the help of remaining four numbers around each figure. This question mark value can be found by this method.

1st Circle

8 + 4 + 3 + 1 = 16 (Greatest number in 1st circle) 

or 

16 - (8 + 4 + 3) = The number in the middle of circle

16 - 15 = The number in the middle of circle

1  =  The number in the middle of circle

2nd Circle

4 + 3 + 5 + 8 = 20 (Greatest number in 2nd circle)

or 

20 - (4 + 3 + 5) = The number in the middle of circle

20 - 12 = The number in the middle of circle

8  =  The number in the middle of circle

3rd Circle

6 + 4 + 5 + ? = 18 

⇒ 15 + ? = 18

⇒ ? = 18 - 15

⇒ ? = 3 (Greatest number in 3rd circle)

This will be the value of question mark.

or 

18 -  (6 + 4 + 5 ) = The number in the middle of circle

18 - 15 = The number in the middle of circle

3 =  The number in the middle of circle

Hence option (1)3 is correct.

 

Question #3

Starting from 4th quadrant in anticlockwise direction , add one to the sum of squares of both the numbers in outer part of the quadrant to get the value of the number in the inner part of same quadrant.

4th Quadrant

(2 + 8)² + 1 =  10²  + 1 =  100 + 1 = 101

1st Quadrant

(3 + 8)² + 1 =  11²  + 1 = 121 + 1 = 122

2nd Quadrant

(7 + 1)²  + 1 =   8² + 1 = 64 + 1 = 65

3rd Quadrant

(5 + 4)² + 1 =   9²  + 1 = 81 + 1 = 82 (The value of question mark)

Hence option (1)82  is correct.

Question #4

Starting from 4th quadrant in anticlockwise direction , Subtract one from the sum of squares of both the numbers in outer part of the quadrant to get the value of the number in the inner part of same quadrant.

4th Quadrant

(2 + 8 )² - 1 =  10²  + 1 =  100  - 1 = 99

1st Quadrant

(3 + 8)² - 1 =  11²  + 1 = 121 - 1 = 120

2nd Quadrant

(7 + 1 )²  - 1 =   8² + 1 = 64 - 1 = 63

3rd Quadrant

(5 + 4 )² - 1 =   9²  + 1 = 81 - 1 = 80 (The value of question mark)

Hence option (2)80  is correct.

Question #5

Since the question mark is in the 3rd figure. So the value of question mark will be calculated with the help of remaining four numbers in the  figure in the same way, the middle numbers in 1st and 2nd figure can be calculated with the help of remaining four numbers around each figure. This question mark value can be found by following method.

1st Figure

63 + 34  +  8  = 105 (Greatest number in 1st figure) 

or 

105 - (63 + 34) = The number in the middle of 1st figure

105 - 97 = The number in the middle of 1st figure

8  =  The number in the middle of 1st figure

2nd Figure

30 + 10 + 5  =  45 (Greatest number in 2nd figure) 

or 

45 - (30 + 10) = The number in the middle of 2nd figure

45 - 40 = The number in the middle of 2nd figure

5  =  The number in the middle of 2nd figure

3rd Figure

31 + 62 + ? = 121 This implies ? = 28 (Greatest number in 3rd figure) 

or 

121 - (62 + 31) = The number in the middle of 3rd figure

121 - 93 = The number in the middle of 3rd figure

28  =  The number in the question mark in 3rd figure

Hence option (4)28  is correct.

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Question #6

This circle has been divided into four quadrants and each quadrant is composed of three numbers . In each quadrant two numbers are in outer quadrant of the circle and one number is in the inner quadrant of the circle. 

Starting from fourth quadrant in clockwise direction. 

Formula : - Half of sum of the squares of both the numbers in outer sector is equal to inner sector.

 4th Quadrant

( 8² + 6²)/2 = ( 64 + 36)/2 = 100/2  =50 (Number in the inner part of the circle in 4th quadrant )

 3rd Quadrant

( 7² + 5²)/2 = ( 49 + 25)/2 = 74/2  = 37  (Number in the inner part of the circle in 3rd quadrant )

 2nd Quadrant

( 10² + 8²)/2 = ( 100 + 64)/2 = 164/2  = 82 (Number in the inner part of the circle in 2nd quadrant )

 1st Quadrant

( 9² + 11²)/2 = ( 81 + 121)/2 =  202/2  = 101 (Number in the inner part of the circle in 1st quadrant  ).

And 101 will be the value of question mark.

Hence option (4)101  is correct.

Question #7

This circle has also been divided into four quadrants and each quadrant is composed of three numbers . In each quadrant two numbers are in outer sector of the circle and one number is in the inner sector of the circle. 

Starting from 1st quadrant in clockwise direction. 

Formula : -  Sum of the squares of both the numbers in outer sector is equal to inner sector.

  1st Quadrant

(3 + 5)² = (8)² = 64  (Number in the inner part of the circle in 1st quadrant )

 2nd Quadrant

(5 + 4)² = (9)² = 81  (Number in the inner part of the circle in 2nd quadrant )

 3rd Quadrant

(4 + 2)² = (6)² = 36  (Number in the inner part of the circle in 3rd quadrant )

 4th Quadrant

(6 + 4)² = (10)² = 100  (Number in the inner part of the circle in 4th quadrant ).

And 100 will be the value of question mark.

 

Question #8

This box problem consists of three rows and three columns. To find the value of question marks in 3rd column of 3rd row, we have to add 1st ,2nd and 3th columns since since in every column is same in every row.

Formula :- The sum of 1st three columns in every row is same .

7 + 8 + 15 = 30 ( 1st  row) 

9 + 20 + 1 = 30 (2nd row) 

14 + 2 + ? = 30 ( Because sum of all the three digits in 3rd row must be equal to the sum in other two rows)

Hence  16 + ? = 30

⇒ ? = 30 - 16 

⇒  ? = 14

Therefore the value of question mark is 14

Hence option (B)14  is correct.

Question #9

This box problem consists of four rows and four columns. To find the value of question marks in 3rd column of 3rd row, we have to divide the sum of  1st ,2nd and 3th rows with 4 to get the value of number in 4th row.

Formula :- The one fourth of  the sum of 1st three rows in every column.

5 + 8 + 7  = 20 ÷ 4 = 5 ( 1st column  in 4th row) 

9 + 6 + 13 = 28 ÷ 4 = 7 (2nd column in 4th row) 

(7 + 10 + 19) ÷ 4  = 36 ÷ 4 = 9 (4th column in 4th row) 

(8 + 9 + ?) ÷  4 =  8  (3rd column in 4th row) 

⇒ 17 + ?  =  8  ×  4

⇒ ? = 32 - 17

⇒ ? = 15

Hence option (D)15  is correct.

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Question #10

This box problem consists of four rows and four columns. To find the value of question marks in 4th column of  fourth row, we have to multiply 1st ,2nd and 3th rows  at the same time and then add 10 to the result so obtained. 

Formula :- Ten more than the product of 1st three rows in every column.

(5 × 6 × 7 )  + 10 = 210 + 10 = 220 (1st column  in 4th row) 

(6 × 5 × 4 )  + 10  = 120 + 10 = 130 (2nd column in 4th row)

(3 × 4 × 5 )  + 10  =  60 + 10 = 70 (3rd column in 4th row) 

(8 × 7 × 6 )  + 10  = 336 + 10 = 346 (4th column in 4th row) 

Hence option (B)346  is correct.

In this post I discussed Ten Missing number Reasoning SSC CGL all questions and answers . Comment your valuable suggestion for further improvement.

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Matrix method of solving linear equations of three variables

Learn the process of solving linear equations of three variables by matrix method  .Let us understand this method with the help of an example

Matrix method of solving linear equations of 3 variables

Matrix method of solving linear equations of three variables with the help of an example. 

The system of these equations can be transformed into Matrix form as 

AX =  B  ,  ⇒ X =   A^-1 B   ->  (*)

Where A is matrix written from the coefficients of x, y and z when these equations are in symmetric form and B is the matrix written from constants from right hand sides in column form and X is matrix of all the variables in column form. 

In order to find the solution of set of these equations , first we have to find the inverse of matrix A if it exist then we can find the solution otherwise Matrix method fails to find the solution of the set of linear equations . 

Evaluation of Determinant 

|A|  = 1 (-9 - 27) -1(6 - 63) -1(6 + 21)

       =  -36 + 57 - 27

       = -63 + 57

        = -6

Since the determinant value is not equal to zero ,Therefore its inverse can be calculated.

And  formula for finding the inverse of matrix A is 

Where Adjoint A is the transpose of co factor matrix. And in order to find the co factor matrix of any matrix, we have to find co factors of all the elements present in that matrix. 

How to calculate  co factors of all the elements of the matrix A. 

 Let us calculate these cofactors. 

Co factors of 1st row are  -36 , 57 , 27

Co factors of 2nd row are  -6, 10 , 4

Co factors of 3rd row are  6 . -11 , -5

Now these co factors can be written in matrix form known as co factor Matrix. 

Co factor Matrix

Writing co factors of 1st row in 1st row of this matrix , co factors of 2nd row in 2nd row of this matrix . Similarly co factors of 3rd row in 3rd row of this matrix . 

Adjoint  Matrix

To find the Ad joint of this matrix we have to take it's transpose, Because transpose of any matrix is called Ad joint of the matrix. So writing all the elements which are in 1st row in 1st column, and  all the elements which are in 2nd row in 2nd column and  all the elements which are in 3rd row in 3rd column. 

Now we can find inverse of the matrix A by putting the value of inverse of A in equation  (4), we get

Now  putting the values  Matrix B and    A^-1  in (4) 

After simplification and using the properties of equality of two matrices  ( Two matrices of same order are equal if and only if their respective elements are equal to each other ) 

⇒  x = -54/-6 = 9

 y =  12/-6 = 2

 z  = -24/-6  = -4

Hence

 x = 9 

 y = 2

 z = -4

So this was the Matrix method of solving linear equations of three variables using inverse of matrix. Your valuables comments will be appreciated for betterment of this blog.

Also read this post for understanding inverse of matrix using elementary row transformation

 

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Ten Logical Reasoning questions and answers for competitive exams

Ten Logical Reasoning questions and answers for competitive exams  like Bank PO, Bank clerk, SSC CGL, ssc chsl, RRB NTPC , group D etc have been discussed in this post in very easy style and shortcut way .

Logical Reasoning questions and answers for competitive Exams 

Problem # 1

All the letters in each option are written from alphabet either in order or in reverse order.

From 1st option MNOP are written in  order starting from M in alphabet. 

From 2nd option TSRQ are written in reverse order starting from Q  in alphabet. i.e.  

QRST  ⇒ QRST   ⟺  TSRQ (After reversing the orders of letters).

From 3rd option  EDCB are written in reverse order starting from B  in alphabet. i.e.  

BCDE  ⇒ BCDE   ⟺  EDCB (After reversing the orders of letters).

 From 4th option  XWVU are written in reverse order starting from U in alphabet. i.e.  

UVWX  ⇒ UVWX  ⟺ XWVU ( After reversing the orders of letters).

Since  2nd,  3rd and 4th options are following the same pattern because all the letters in these options are written in order. But letters in 1st option are  written in reverse order. And this option is odd one.

Hence Option (1)MNOP is correct option.

Problem # 2

All the letters are written alternatively from alphabet either in order or in reverse order.

From 1st option MKIG are written in reverse order starting from G with a gap of one letter in alphabet. i.e.  

G H I J K L M ⇒ G H I J K L M ( Omitting blue colored letters )  ⇒ G I K M ⟺ MKIG (After reversing the orders of  letters).

From 2nd option VTRP are written in reverse order starting from P with a gap of one letter in alphabet. i.e.  

PQRSTUV ⇒  PQRSTUV (Omitting blue colored letters)  ⇒  PRTV ⟺ VTRP (After reversing the orders of letters).

From 3rd option EGIK are NOT written in reverse order starting from K with a gap of one letter in alphabet. 

From 4th option YWUS are written in reverse order starting from S with a gap of one letter in alphabet. i.e.  

STUVWXY ⇒  STUVWXY (Omitting blue colored letters)  ⇒  SUWY ⟺ YWUP (After reversing the orders of letters).

Since 1st , 2nd and 4th options are following the same pattern because all the letters in these options are written in reverse order with a gap of one letter. But letter in 3rd option are not written in reverse order.

Hence Option (3)EGIK is correct option.

Problem # 3

In this reasoning problem 1st number (5) is associated to 100 with the help of any rule , in the same rule we have to associate 7 to a number out of four given options.

Look carefully the given numbers consists of three digits. These three digits can be utilised with the help a formula given below.

Formula :-  4 × (1st number)²  = 2nd Number

 4 × (3rd number)²  = 4th Number

⇒ {5} = 4 × 5² = 4 × 25  = 100 (2nd number)

⇒{7} = 4 × 7² = 4 × 49 = 196 (4th number)

Option (4)196 is correct option.

Problem # 4

In this problem of reasoning we have to combine all the three given  numbers in such a way that after applying any mathematical operation/operations we could get the number in the right hand side of all the three problems given above. Because in these types of reasoning problems we can change mathematical sign according to our requirements.

Replacing - sign with + , + sign with × in above conditions . Now place 1st two numbers in bracket . Then apply the BODMAS rule.

(7 + 2) × 4 = 9 × 4 = 36 

(10 + 5) × 2 = 15 × 2 = 30

Similarly 

(20 + 5) × 6 = 25 × 6 = 150

Option (4)150 is correct option.

Problem # 5

In this reasoning problem 1st number (392) is associated to 28 with the help of any rule , in the same way we have to associate 722 to a number out of four given options. 

Formula :- 2 × √(1st Number ÷ 2) = 2nd Number

2 × √(392 ÷ 2) = 2 × √(196)  = 2  × 14 = 28 . 

Similarly 

2 × √(722 ÷ 2) = 2 × √(361)  = 2  × 19 = 38

Option (3)38 is correct option.

Problem # 6

 

These words are used when a book is prepared, published and read by Reader.   To publish a book , first of all it has to be written by the Author . After this book will be with the Editor for any change or editing. Then that book will be published by Publisher. After publication it will be sold by Bookseller. And at last it will be purchased by read by Reader.

The journey of the book is as follow

1. Author
2. Editor
3. Publisher
4. Bookseller
5. Reader

So according to given problem's solution the correct option will be  (D) 4 , 1 , 3 , 5 , 2

Problem # 7

Divide the sum of both the numbers with 10 to get the number in right hand side.

( 15 + 55 ) ÷ 10 = 70 ÷ 10 = 7

( 25 + 35 ) ÷ 10 = 60 ÷ 10 = 6

( 35 + 45 ) ÷ 10 = 80 ÷ 10 = 8

Similarly ( 23 + 17 ) ÷ 10 = 40 ÷ 10 = 4

Hence option (3)4 is correct.

Problem # 8

In this reasoning problem three out of four options have been calculated by Multiplying  the 1st number with its next number(successive number) to get 2nd number. This rule is applicable to only three out of four option. And one which do not follow this rule will be the correct option.

(1)   9 × ( 9 + 1 ) = 9 × 10 = 90

(2)   13 × ( 13 + 1 ) = 13 × 14 = 182

(3)   8 × ( 8 + 1 ) = 8 × 9 = 72

(4)   12 × ( 12 + 1 ) = 12 × 13 = 156, but this is given as 144

Hence option (4)12 - 144  is odd one.

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Problem # 9

Take the difference of two successive numbers, this difference is square of continuous natural numbers.

6  -  5 = 1 = 1²

10 -  6 = 4  = 2²

19 -  10 = 9  = 3²

35 - 19 = 16  = 4²

Similarly the difference of last number and 2nd last number can be calculated. This difference will be the square of some natural number.

?  - 35 = 25  = 5²

?  = 25 + 35  = 60

Hence option (2)60  is correct.

Problem # 10

Take the difference of two successive numbers, this difference is cube of continuous natural numbers.

3085 - 1357 = 1728 = 12³ 

5282 - 3085 = 2197 = 13³

8026 - 5282 = 2744 = 14³

Similarly the difference of last number and 2nd last number can be calculated. This difference will be the cube of some natural number.

? -  8026 = 3375 = 15³

? = 3375 + 8026 = 11401

Hence option (4)11401 is correct.

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Conclusion

So these were the ten  problems regarding the post Ten Logical Reasoning questions and answers pdf for competitive exams like SSC CGL ,SSC CHSL ,CPO ,Bank exams and RRB NTPC etc which were explained in this post. Feel free to comment your valuable suggestions.

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