Log in

Connect faster with

Register

Signup faster with


|   Education without borders.
Menu
a Guest

matrix inversion technique

x+y+z=4

2x-y+z=3

x-2y+3z=5

 

Posted in Algebra, asked by yuvaraj, 6 years ago. 1782 hits.

0

Given Linear  Equations are ,x+y+z=4

                                         2x-y+z=3

                                          x-2y+3z=5

we can rewrite the given equations in the form of  3*3 matrices,AX=B

                                                                                           x=(A)-1B

                      A-1=1/det(A)[adj(A)]T

 so we have to find the det(A),adj(A),then we have transpose it.then we will calculate the remaining matrice  with matrice B,we will get the values for x,y,z.

 

 

Venkat Nagendra Thati
Venkat Nagendra Thati - 6 years ago
Ask Venkat Nagendra Thati for further help.
Report

0

In this case we use matrix linear solving method,

[A]*X=[B]

 

Matrix A =    1     1    1                      X= x                      Matrix B=   4

                   2    -1   1                            y                                       3

                   1    -2   3                            z                                       5


Therefore,

X = [A]-1[B]


Determinant of A = -9

Adjoint of A =     -1    -5   2

                         -5     2    1

                         -3     3   -3

[A]-1=    1/9    5/9    -2/9

             5/9    -2/9    -1/9

             3/9    -3/9     3/9

Hence,

X=[A]-1[B]

X =  1

       1

       2

 

Therefore,

x = 1, y= 1, z=2

                  

Gurudas Kadam
Gurudas Kadam - 6 years ago
Ask Gurudas Kadam for further help.
Report

0

X + y + z = 4

2x – y + z = 3

X - 2y + 3z = 5

 

We can find values of X, Y & Z BY THIS FORMULA,

AX=B or X = A-¹B

A = ALL THE NO. WITH VARIABLES

B = ALL THE RIGHT HAND SIDE NO.S ( AFTER EQUAL SIDE NUMBERS)

X  = ALL VARIABLES 

Matrix A =    1     1    1                      X= x                      Matrix B=   4

                     2    -1   1                            y                                       3

                     1    -2   3                            z                                       5


Therefore,

X = INVERSE OF MATRIX A THEN MULTIPLIED BY B

X = [A]-1[B]

FOR INVERSE

1/determinant of A Multiplied by ADJOINT OF A

first we find derminant

=1[(-1 x 3)-(1 x -2)] -1[(2 x 3) – (1 x 1)]+1[(2 x -2) – (-1 x 1)]

=1(-3+2) -1(6-1) +1(-4+1)   [ solve it with the help of signs ]

=-1-5-3
Determinant of A = -9

Adjoint of A =         -1    -5   2

                            -5     2    1

                            -3     3   -3

[A]-1=    1/9    5/9    -2/9

             5/9    -2/9    -1/9

             3/9    -3/9     3/9

Hence,

X=[A]-1[B]

X =  1

       1

       2

 

we get 

x = 1, y= 1, z=2

          

 

Furqan Ali
Furqan Ali - 6 years ago
Ask Furqan Ali for further help.
Report
Please register/login to answer this question. 
- Just now

0

a Guest
Just now
× Attachments/references, if any, will be shown after refreshing the page.
/homework-help
/homework-help/get_answer/324
/homework-help/save_reply/324
/homework-help/save_answer/324
/homework-help/accept_answer/324
/homework-help/unaccept_answer/324
/homework-help/unpublish_answer/324
/homework-help/delete_answer/324
/homework-help/update_answer/324
/homework-help/like/324
/homework-help/dislike/324
/homework-help/report_answer/324
/homework-help/report_question/324
/homework-help/save_bounty/324