HOW TO SOLVE LINEAR EQUATIONS OF TWO AND THREE VARIABLES BY MATRIX METHOD

Using Matrix method to solve system of linear equation , we must know some topics such as co-factor of element,   Transpose of matrix , Ad joint of a Matrix,       Multiplication of two Matrices ,Determinant  value  of a Matrix , Inverse of a matrix etc.  Let us understand the  process of finding the solution of system of linear equations with the help of some examples.
To Solve the system  of Linear Equations using 2×2 Matrix Method
x - 5y = 4
2x + 5y = −2

Writing this system of equation in Matrix form
AX = B

where X = A-1 B------------------------(1)

Where

We need the inverse of A ( i. e. A-1 ),  To find the inverse of Matrix A ,1st find out Co-factor Matrix of A

As we know the Ad joint Matrix of any matrix can be found by taking the transpose of the Co Factor matrix.

Now let us find the determinant of  A

A∣ = 5 − (-10) = 15 , which is non Zero,

Therefore A-1
Exists     ,So
Now putting the value of  ( A-1 ) inverse of Matrix A in equation (1)

Now putting the elements of Matrix X , and
By the equality of two Matrices ,their elements in respective positions must be equal to each others,

Hence x= 2/3  and y = -2/3

How to check the correctness of a solution
put x=  2/3  and y = -2/3  in the given  equations
x - 5y = 4 implies 2/3 - 5(-2/3) = 2/3+10/3 = 4

and 2x + 5y  = -2 implies 2×(2/3) + 5(-2/3) = 4/3-10/3 = -6/3= -2

therefore both equations are satisfied by these values of 'x' and 'y'

To Solve the system of Linear Equations using 3×3 Matrix Method

x  -  y  + z  = 4
2x + y - 3z  =  0
x + y  + z   =  2

Converting this system of Linear equations into Matrix Form

AX    =  B
X  =  A-1B ------------------------(1)
Where

Now we have to find the  A-1. , and it will exist if its determinants value |A| is non Zero. Let us find |A|

|A|= 1{1×1-(-3)×1}-(-1){2×1-(3)×1}+1(2×1-1×1)

|A|= 1 {1+3}+1 {2+3}+1 {2-1}

|A|= 4+5+1 = 10

Since |A| is non Zero , Therefore
A-1 (inverse of A ) exists .

We have to find the Co factors of all the elements of matrix A . Let us find out the co factors of all the elements of matrix row wise.

Co-Factors of 1st Row

co-factor of A11  =   4
co-factor of A12  =  -5
co-factor of A13  =   1

Co-Factors of 2nd Row

co-factor of A21  =   2
co-factor of A22  =   0
co-factor of A23  =  -2

Co-Factors of 3rd Row

co-factor of A31  =  2

co-factor of A32  =  5
co-factor of A33  =  3

Now co Factor matrix of A can be written as follows

Now to find the Ad joint of this Matrix, Take the transpose of this Matrix,

As we know the is the scalar  multiplication of Ad Joint Matrix of Matrix A and reciprocal of Determinant value of the Matrix A.

Putting the values of Inverse of A and Matrix B in equation (1), So after Multiplication of these two matrices ,we get

$\displaystyle={\left(\begin{matrix}{22}\\-{16}\\-{16}\end{matrix}\right)}$
Now putting the values of elements  of   Matrix X, and equating the elements in their respective positions .

Now using the property of equality of two Matrices , All the elements in their respective position are equal to each other , we get

x = 2  , y = -1, z = 1  is the solution of the system of linear Equations.

Conclusion

In this post I have discussed the method of solving the System of linear Equations with the help of Matrix Method ,method to find inverse of a matrix .,Matrices and Determinants in Most imp,How to find inverse of 3x3 Matrix,how to solve determinant,how to solve system of equation by matrix,matrix method of solving system of equations of three variables,determinant of 3x3 matrix calculator, matrices and determinants in most imp,If you liked the post please share it with your friends , And in case of any improvement please make use of Comment Box . To keep on supporting me follow my Blog . Thanks for your valuable time to read this post .We shall meet you in next post ,till then BYE.

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