Echelon method

Question

Solve the following system of equations by using the inverse of the coefficient matrix if it exists and by the echelon method if the inverse doesn't exist.

 

6x+y=48

41x+7y+329

Answer 1

You first want to set up the system of equations in the form AX = B

Matrix A: [ 6     1 ]

                41    7

 

Matrix B: [  48  ]

                 329

 

Set up the equation AX = B

 

[  6      1  ] [ x ]  = [  48  ]

  41     7       y          329

 

Now, find the inverse of the coefficient matrix.  To find the inverse of a 2 x 2 matrix:

 

C = [ a   b ]

         c   d

 

C^-1 = (1/detC) * [d    -b ]

                          -c     a

 

The inverse of matrix A:

detA = 7*6 - 41*1 = 42 - 41 = 1

 

A^-1 = (1/1) [ 7      -1 ]

                   -41     6

 

A^-1 = [   7      -1 ]

            -41     6

 

Now, back to your equation:

AX = B

 

Multiply the left sides of both equations by A^-1:

A^-1AX = A^-1B

 

Multiplication of A^-1 and A gives the identity matrix:

 

IX = A^-1B

 

We can drop the identity matrix out of the equation:

X = A^-1B

 

Our answer is then the multiplication of A^-1 and B:

 

[ x ] = [ 7     -1 ] [ 48 ]

  y        -41    6     329

 

[ x ] = [7*48 + -1*329  ]

  y        -41*48 + 6*329

 

[ x ]  = [  336 + -329   ]

  y         -1968 + 1974

 

[ x ] = [ 7 ]

  y         6

 

x = 7 and y = 6