HOW TO SOLVE LINEAR EQUATIONS OF TWO AND THREE VARIABLES BY MATRIX METHOD ~ Simplified Mathematics, integration,differentiation,matrix,determinant,function,trigonometry,limit

Today we are going to learn, how to solve system of linear equations of 2 and 3 variables using matrix method. So 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)

And A-1 = (1/Det A ) ( Ad joint A)

Where

We need the inverse of A , which we write as A^-1

To find the inverse 1st find out Co-factor Matrix of A

Ad joint A = ( 5 5 -2 1 )

Co factor A = ( 5 −2 5 1 )

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 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 are 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^-1 B ----------------------------------(1)
Where

Now we have to find the

A^-1 . , and it will exist if its determinants value |A| must be 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 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 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 discussed the method of solving the System of linear Equations with the help of Matrix Method , 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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