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

## Solving the system of Linear Equations of two variables

To Solve the system of Linear Equations using 2×2 Matrix Method

x - 5y = 4

2x + 5y = −2

x - 5y = 4

2x + 5y = −2

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.

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 = 4and 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

2x + y - 3z = 0

x + y + z = 2

Converting this system of Linear equations into Matrix Form

AX = B

X = A

Where

Now we have to find the 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

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 A12 = -5

co-factor of A13 = 1

co-factor of A22 = 0

co-factor of A23 = -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 Inverse of a Matrix is the scalar multiplication of Ad Joint Matrix of Matrix A and reciprocal of Determinant value of the Matrix A.

^{-1}B ------------------------(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 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 Inverse of a Matrix 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

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